NI 9870 and binary coded decimal (BCD)
I have a scale that outputs the weight in 4 digit ASCII form which I would like to input into a CompactRio 9072 with a NI 9870 card and convert to a binary integer. The weight appears as the 7th to 10th character on lines that are terminated by a CR. I am thinking I will need a loop that once the 10th character is read, the 4 least significant bits of the 4 characters are converted from BCD to binary and the character counter resets to 0 when a CR is read. I need a good method for converting BCD to an integer value in either labview or vhdl. The rest of the logic for this project is in vhdl, so I would not mind adding the conversion logic there, except that all the algorithms I have found require decimal division or comparison in several clock cycles.
Solved!
Go to Solution.
Hello imr
Thank you very much for getting in touch with us!
I believe the following community example will resolve part of your issue.
Binary String to Number
https://decibel.ni.com/content/docs/DOC-11704
I have also provided some further information for those unaware of what Binary Coded Decimal represents.
What is Binary Coded Decimal (BCD)?
http://digital.ni.com/public.nsf/allkb/59C2A05C123008F286256CA90069A6F8
Thank you for choosing National Instruments!
Sincerely,
Greg S.
Similar Messages
-
I have a plc that stores data in memory as a BCD. When I set up Lookout to read the memory location it converts it to a decimal number. how can I prevent this?
Randall1,
I have some example code that shows you how to convert the number back from decimal to BCD. It requires Lookout 4.0.1 build 51 or higher.
Regards,
Michael Shasteen
Applications Engineering
National Instruments
www.ni.com/ask
Attachments:
hex_bc.lks 18 KB -
Hi
Does any body has any examples or idea HOW to convert BCD to ASCII
Please share the same
with regards
KarthikExample 1 : packed BCD 12 34 56 (3 bytes, 6 digits)
unpack into six bytes: 1 2 3 4 5 6
add 30 hex to each byte : 0x31 0x32 0x33 0x34 0x35 0x36
view as ascii : 1 2 3 4 5 6
you can skip the first step if your bcd is unpacked -
Help with binary to decimal, binary to hex, and hex to ascii or ascii to hex program
I decided to do a program that will do binary to decimal, binary to hex, and hex to ascii for a project related to a java programming course, which only needs to perform features from chapters 1-6 and 8 of Tony Gaddis's book. The functions work fine as their own main programs out side of this combined effort, so can anyone help me determine why I get the following 41 errrors saying: class, interface, or enum expected as well as any other errors that may show up afterwards because I'm stumped. My flowcharts, which have to be revised after discovering that my previous function were logically incorrect after running them in their own main are attached below as the spec sheet.
My code is as follows and I hope you don't mind the commented lines of unused code because I'm not sure where I want things and what I want at the moment:
import java.util.Scanner;
import java.io.*;
import java.lang.*;
public class BintoDectoHextoAscii
public static void main(String[] args)throws IOException
Scanner input = new Scanner(System.in);
System.out.println("Enter a binary number: ");
String binary = input.nextLine(); // store input from user
if (binary == input.nextLine())
//int i= Integer.parseInt(hex,2);
//String hexString = Integer.toHexString(i);
//System.out.println("Hexa decimal: " + hexString);
//int finaldecimalvalue = binaryToDecimal(hexString);
int finaldecimalvalue = binaryToDecimal(hexString);
if (binary != input.nextLine())
String hexInput; // The variable Bin Input declared as the datatype int to store the Binary value
// Create a Scanner object for keyboard input.
//Scanner keyboard = new Scanner(System.in);
// Get the number of binary files.
System.out.print("Enter the Hex value: ");
hexInput = keyboard.nextLine();
System.out.println("Original String: "+ hexInput);
//String hexEquivalent = asciiToHex(demoString);
String hexEquivalent = asciiToHex(hexInput);
//Hex value of original String
System.out.println("Hex String: "+ hexEquivalent);
String asciiEquivalent = hexToASCII(hexEquivalent);
//ASCII value obtained from Hex value
System.out.println("Ascii String: "+ asciiEquivalent);String finalhexOutput = HextoAsciiConverter(hexEquivalent);
if (binary != input.nextLine() && hexInput != keyboard.nextLine())
BufferedReader binInput = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the Binary number:");
String hex = binInput.readLine();
//String finaldecimalvalue = binaryToDecimal(decimal);
//long finalhexvalue = BinaryToHexadecimal(num);
long finalhexvalue = BinaryToHexadecimal();
public static String BinaryToHexadecimal(String hex)
//public static void main(String[] args)throws IOException
//BufferedReader bf= new BufferedReader(new InputStreamReader(System.in));
//System.out.println("Enter the Binary number:");
//String hex = binInput.readLine();
long num = Long.parseLong(hex);
long rem;
while(num > 0)
rem = num % 10;
num = num / 10;
if(rem != 0 && rem != 1)
System.out.println("This is not a binary number.");
System.out.println("Please try once again.");
System.exit(0);
int i= Integer.parseInt(hex,2);
String hexString = Integer.toHexString(i);
System.out.println("Hexa decimal: " + hexString);
return num.tolong();
//int i= Integer.parseInt(hex,2);
//String hexString = Integer.toHexString(i);
//System.out.println("Hexa decimal: " + hexString);
//} // end BintoDectoHextoAsciil
//public static String HexAsciiConverter(String hextInput)
// Get the number of binary files.
//System.out.print("Enter the Hex value: ");
//hexInput = keyboard.nextLine();
//System.out.println("Original String: "+ hexInput);
//String hexEquivalent = asciiToHex(demoString);
//String hexEquivalent = asciiToHex(hexInput);
//Hex value of original String
//System.out.println("Hex String: "+ hexEquivalent);
//String asciiEquivalent = hexToASCII(hexEquivalent);
//ASCII value obtained from Hex value
//System.out.println("Ascii String: "+ asciiEquivalent);
//} // End function
private static String asciiToHex(String asciiValue)
char[] chars = asciiValue.toCharArray();
StringBuffer hex = new StringBuffer();
for (int i = 0; i < chars.length; i++)
hex.append(Integer.toHexString((int) chars[i]));
return hex.toString();
private static String hexToASCII(String hexValue)
StringBuilder output = new StringBuilder("");
for (int i = 0; i < hexValue.length(); i += 2)
String str = hexValue.substring(i, i + 2);
output.append((char) Integer.parseInt(str, 16));
return output.toString();
public static String binaryToDecimal(String binary)
//Scanner input = new Scanner(System.in);
//System.out.println("Enter a binary number: ");
//String binary = input.nextLine(); // store input from user
int[] powers = new int[16]; // contains powers of 2
int powersIndex = 0; // keep track of the index
int decimal = 0; // will contain decimals
boolean isCorrect = true; // flag if incorrect input
// populate the powers array with powers of 2
for(int i = 0; i < powers.length; i++)
powers[i] = (int) Math.pow(2, i);
for(int i = binary.length() - 1; i >= 0; i--)
// if 1 add to decimal to calculate
if(binary.charAt(i) == '1')
decimal = decimal + powers[powersIndex]; // calc the decimal
else if(binary.charAt(i) != '0' & binary.charAt(i) != '1')
isCorrect = false; // flag the wrong input
break; // break from loop due to wrong input
} // end else if
// keeps track of which power we are on
powersIndex++; // counts from zero up to combat the loop counting down to zero
} // end for
if(isCorrect) // print decimal output
System.out.println(binary + " converted to base 10 is: " + decimal);
else // print incorrect input message
System.out.println("Wrong input! It is binary... 0 and 1's like.....!");
return decimal.toint();
} // end function
The errors are as follows:
----jGRASP exec: javac BintoDectoHextoAscii.java
BintoDectoHextoAscii.java:65: error: class, interface, or enum expected
public static String BinaryToHexadecimal(String hex)
^
BintoDectoHextoAscii.java:73: error: class, interface, or enum expected
long rem;
^
BintoDectoHextoAscii.java:74: error: class, interface, or enum expected
while(num > 0)
^
BintoDectoHextoAscii.java:77: error: class, interface, or enum expected
num = num / 10;
^
BintoDectoHextoAscii.java:78: error: class, interface, or enum expected
if(rem != 0 && rem != 1)
^
BintoDectoHextoAscii.java:81: error: class, interface, or enum expected
System.out.println("Please try once again.");
^
BintoDectoHextoAscii.java:83: error: class, interface, or enum expected
System.exit(0);
^
BintoDectoHextoAscii.java:84: error: class, interface, or enum expected
^
BintoDectoHextoAscii.java:87: error: class, interface, or enum expected
String hexString = Integer.toHexString(i);
^
BintoDectoHextoAscii.java:88: error: class, interface, or enum expected
System.out.println("Hexa decimal: " + hexString);
^
BintoDectoHextoAscii.java:90: error: class, interface, or enum expected
return num.tolong();
^
BintoDectoHextoAscii.java:91: error: class, interface, or enum expected
^
BintoDectoHextoAscii.java:124: error: class, interface, or enum expected
StringBuffer hex = new StringBuffer();
^
BintoDectoHextoAscii.java:125: error: class, interface, or enum expected
for (int i = 0; i < chars.length; i++)
^
BintoDectoHextoAscii.java:125: error: class, interface, or enum expected
for (int i = 0; i < chars.length; i++)
^
BintoDectoHextoAscii.java:125: error: class, interface, or enum expected
for (int i = 0; i < chars.length; i++)
^
BintoDectoHextoAscii.java:128: error: class, interface, or enum expected
^
BintoDectoHextoAscii.java:130: error: class, interface, or enum expected
^
BintoDectoHextoAscii.java:135: error: class, interface, or enum expected
for (int i = 0; i < hexValue.length(); i += 2)
^
BintoDectoHextoAscii.java:135: error: class, interface, or enum expected
for (int i = 0; i < hexValue.length(); i += 2)
^
BintoDectoHextoAscii.java:135: error: class, interface, or enum expected
for (int i = 0; i < hexValue.length(); i += 2)
^
BintoDectoHextoAscii.java:138: error: class, interface, or enum expected
output.append((char) Integer.parseInt(str, 16));
^
BintoDectoHextoAscii.java:139: error: class, interface, or enum expected
^
BintoDectoHextoAscii.java:141: error: class, interface, or enum expected
^
BintoDectoHextoAscii.java:144: error: class, interface, or enum expected
public static String binaryToDecimal(String binary)
^
BintoDectoHextoAscii.java:150: error: class, interface, or enum expected
int powersIndex = 0; // keep track of the index
^
BintoDectoHextoAscii.java:151: error: class, interface, or enum expected
int decimal = 0; // will contain decimals
^
BintoDectoHextoAscii.java:152: error: class, interface, or enum expected
boolean isCorrect = true; // flag if incorrect input
^
BintoDectoHextoAscii.java:155: error: class, interface, or enum expected
for(int i = 0; i < powers.length; i++)
^
BintoDectoHextoAscii.java:155: error: class, interface, or enum expected
for(int i = 0; i < powers.length; i++)
^
BintoDectoHextoAscii.java:155: error: class, interface, or enum expected
for(int i = 0; i < powers.length; i++)
^
BintoDectoHextoAscii.java:159: error: class, interface, or enum expected
for(int i = binary.length() - 1; i >= 0; i--)
^
BintoDectoHextoAscii.java:159: error: class, interface, or enum expected
for(int i = binary.length() - 1; i >= 0; i--)
^
BintoDectoHextoAscii.java:159: error: class, interface, or enum expected
for(int i = binary.length() - 1; i >= 0; i--)
^
BintoDectoHextoAscii.java:166: error: class, interface, or enum expected
else if(binary.charAt(i) != '0' & binary.charAt(i) != '1')
^
BintoDectoHextoAscii.java:169: error: class, interface, or enum expected
break; // break from loop due to wrong input
^
BintoDectoHextoAscii.java:170: error: class, interface, or enum expected
} // end else if
^
BintoDectoHextoAscii.java:174: error: class, interface, or enum expected
} // end for
^
BintoDectoHextoAscii.java:180: error: class, interface, or enum expected
else // print incorrect input message
^
BintoDectoHextoAscii.java:185: error: class, interface, or enum expected
return decimal.toint();
^
BintoDectoHextoAscii.java:186: error: class, interface, or enum expected
} // end function
^
41 errors
----jGRASP wedge2: exit code for process is 1.
----jGRASP: operation complete.so can anyone help me determine why I get the following 41 errrors saying: class, interface, or enum expected as well as any other errors that may show up afterwards because I'm stumped.
Yes - YOU CAN!
My code is as follows and I hope you don't mind the commented lines of unused code because I'm not sure where I want things and what I want at the moment:
Excellent! Commenting out code is EXACTLY how you troubleshoot problems like yours.
Comment out sections of code until the problem goes away. Then start adding back ONE SECTION of code at a time until the problem occurs. When it does you have just FOUND the problem.
If you do that you wind up with code that looks like this:
import java.util.Scanner;
import java.io.*;
import java.lang.*;
public class BintoDectoHextoAscii {
public static void main(String[] args)throws IOException {
Scanner input = new Scanner(System.in);
System.out.println("Enter a binary number: ");
String binary = input.nextLine(); // store input from user
public static String BinaryToHexadecimal(String hex) { } // end function
Notice ANYTHING UNUSUAL?
You have a complete CLASS definition followed by a method definition.
Methods have to be INSIDE the class - you can NOT define methods on their own.
Write modular code.
Write EMPTY methods - just the method signature and maybe a RETURN NULL if you need to return something.
Then add calls to those empty methods.
When everything compiles and runs find you can start adding code to the methods ONE METHOD AT A TIME.
Test compile and run after you add the code for each method. -
Convert between binary and signed/unsigned decimal numbers
Hi all,
Has anyone written a code to convert:
binary number --> signed decimal
binary number --> unsiged decimal
signed decimal --> binary
unsigned decimal --> binary
Please help! I'm confusing.http://java.sun.com/j2se/1.4/docs/api/java/lang/Integer.html
See the methods parseInt and toBinaryString.
Please help! I'm confusing. You are, aren't you?
(Sorry, couldn't resist...) -
Cobol Packed Decimal and Binary Compressed Numbers
Hello!
I am having some difficulties finding the correct datatype in Oracle for numbers in flat files coming from Cobol which are packed-decimal type and binary compressed. These are not objects, hence the use of BLOB is not appropriate?!
Thaks for any help!!!
nullThe sqlloader manual (i.e. utilities reference, loader chapters) tells all about how to load packaed decimal.
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Adrian Zeller ([email protected]):
Hello!
I am having some difficulties finding the correct datatype in Oracle for numbers in flat files coming from Cobol which are packed-decimal type and binary compressed. These are not objects, hence the use of BLOB is not appropriate?!
Thaks for any help!!!<HR></BLOCKQUOTE>
null -
Checking and Converting binary, octal, decimal, and hex
Hi,
I have a classroom project which will need to validate that data entered in a text is of a certain type with a keyListener, and then convert that value to the other 3 types.
I know character.isDigit will handle the decimal case, and that I can then use Integer.otString methods to convert to binary, octal, and hex, but what about the other cases?
ThanksOK, this isn't working. If I've already established
that the string entered into, for example, the
integer-only textfield is a valid integer, I should be
able to simply use integer.parseint(s, 2) to convert
it into binary, right?Not exactly. You should be able to use Integer.parseInt(s, 2) to return the result of converting it from binary representation. You appear to think that this affects whatever "s" refers to, which is not the case. Note also, that method returns an int which is the decimal value in question.
When you are thinking of int variables, be careful not to think of them as "decimal" or "binary". They are neither. They are just numbers. "Decimal" and "binary" are text representations of numbers, so those concepts can only be applied to strings.
Integer.parseInt(s, 2);
txtBin.setText(s);So here you want to assume that the input is a string that represents a number in decimal, and you want to find the string that represents the number in binary. Like this:// convert string in decimal representation to number
int dec = Integer.parseInt(s);
// convert int to binary representation as string:
String binary = Integer.toBinaryString(dec);
// write that to the text field
txtBin.setText(binary);You could use a one-liner to do that, but you'll need the "dec" variable to set the other text boxes.
Rembering why I hate OO...All of what I said there is true in just about all computer languages, OO or otherwise.
PC² -
Ok to give a background to why i am writing this and so everyone has a better understanding of exactly what it is that i want it to do it goes like this.... My girlfriend got a message from a friend that was all in ones and zeros, and no its not because something on one of the mail servers is messed up he meant for it to be that way. But through the process of converting the 8 digit binary numbers into decimal values so that the corresponding ANSI character could be found I decided that it was taking to damn long so i would just write a prog to do it for me and then i could send back an annoying long message to that guy with ease. Only there is a catch over the summer i went from hotshot for a newb to a newb without a clue, i more or less didnt write a single line of code over 4 months and now i am having problems.
I wanted to write a program that could-
-convert long strings of chars into all 8 digit binary code
-as well as convert large blocks of 8 digit binary back into ANSI letter values and then ANSI characters
-I want it to use a frame and have a relativly simple usage procedure
Now that you know what i set out to do I can show you how far i got -
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
import java.awt.GradientPaint.*;
public class BinaryToDec extends JFrame implements ActionListener
// member variables
JButton quit;
JButton color;
JButton convert;
JLabel label11_5, labelPassingWord;
JTextField inputBinary, outputDecimal;
Color colorStore;
String drawString;
String bin;
public BinaryToDec()
drawString = "";
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
setLocation(100,250);
setTitle("Binary to Decimal");
setSize(420,200);
getContentPane().setBackground(Color.BLACK);
getContentPane().setLayout(null);
quit= new JButton("DONE");
quit.setForeground(Color.GREEN);
quit.setBackground(Color.BLACK);
quit.setBounds(100,10,75,20);
getContentPane().add(quit);
quit.addActionListener(this);
color= new JButton("Bkgnd Clr");
color.setForeground(Color.GREEN);
color.setBackground(Color.BLACK);
color.setBounds(15,10,75,20);
getContentPane().add(color);
color.addActionListener(this);
convert= new JButton("Convert to Decimal");
convert.setForeground(Color.GREEN);
convert.setBackground(Color.BLACK);
convert.setBounds(185,10,75,20);
getContentPane().add(convert);
convert.addActionListener(this);
label11_5 = new JLabel("Enter a Binary Number");
label11_5.setBounds(75,60,250,20);
label11_5.setForeground(Color.GREEN);
getContentPane().add(label11_5);
inputBinary = new JTextField(25);
inputBinary.setBounds(5,40,225,20);
inputBinary.setForeground(Color.BLACK);
getContentPane().add(inputBinary);
inputBinary.addActionListener(this);
labelPassingWord = new JLabel("Your Deciaml Equivilant");
labelPassingWord.setBounds(75,120,250,20);
labelPassingWord.setForeground(Color.GREEN);
getContentPane().add(labelPassingWord);
outputDecimal = new JTextField(25);
outputDecimal.setBounds(5,100,225,20);
outputDecimal.setForeground(Color.BLACK);
getContentPane().add(outputDecimal);
outputDecimal.addActionListener(this);
setVisible(true);
}// end of constructor
public void actionPerformed (ActionEvent evt)
if(evt.getActionCommand().equals("DONE"))
System.out.println("Quit Caught");
int choice = JOptionPane.showConfirmDialog(null , "You Pressed " + evt.getActionCommand() + " are you sure?"
, "Close Frame",JOptionPane.YES_NO_OPTION);
if(choice == 0)
this.dispose();
if (evt.getSource() == color)
Color color = Color.lightGray;
color = JColorChooser.showDialog(BinaryToDec.this,"Choose a color",color);
getContentPane().setBackground(color);
if (evt.getSource() == convert)
bin = inputBinary.getText();
int strLength = bin.length();
int pos1=0, pos2=7;
String temp = "";
String Decimal1 = "";
String Decimal2 = "";
int binLength = 8;
int count = 0;
do
temp = bin.substring(pos1,pos2);
Decimal1 = Convert(temp);
Decimal2 = Decimal2 + "" + Decimal1;
pos1 = pos1 + binLength;
pos2 = pos2 + binLength;
count++;
outputDecimal.setText(Decimal2);
repaint();
}while((strLength/8) >count);
}//end of actionPerform
public void paint( Graphics g)
super.paint(g);
public String Convert(String a)
int ctr=1, decimal=0;
for (int i = a.length()-1; i>=0; i--)
if(a.charAt(i)=='1') decimal+=ctr;
ctr*=2;
String converted = "" + decimal;
return converted;
public static void main(String args[])
System.out.println("Binary to Dec");
BinaryToDec in = new BinaryToDec();
}// end of main()
}// end of class FrameExample1
OK so other than that i decided to play with some of the paint functions i more or less stayed on task, at first i had it just convert 8 digit binary nums into decimal form using this -
int ctr=1, decimal=0;
for (int i = a.length()-1; i>=0; i--)
if(a.charAt(i)=='1') decimal+=ctr;
ctr*=2;
which i found in someone elses post and made some changes to, thank you someone whoever you are i hope you see this, anyways,.. then i decided i wanted it to check the length of a string and divide it by 8 assuming that it it a perfect block of 8 digit nums and then cut it into substring and go through the proccess of converting the binary to deciaml one piece at a time. I did it more or less, well it compiles right but whenever you put in more than one 8 digit binary number it ends up giving you some really strange decimal numbers and i cant seem to figure out why the hell it is doing this.??? I was hoping that some one could help me out.
If you read this far than thanks for taking the time on my post :)Did you try to decode that message by hand and verified that it really is encoded the way you think it is?
Here some code to try
public static void decode(String fileIn, String fileOut) {
try {
FileReader in=new FileReader(fileIn);
PrintWriter out=new PrintWriter(fileOut);
char[] buffer=new char[8];
int i;
do {
i=in.read(buffer, 0, 8);
if(i==8) {
out.write(Integer.parseInt(new String(buffer),2));
} while(i==8);
in.close();
out.close();
} catch (Exception e) {}
public static void encode(String fileIn, String fileOut) {
try {
FileReader in=new FileReader(fileIn);
PrintWriter out=new PrintWriter(fileOut);
int i;
while( (i=in.read()) >= 0 ) {
String str=Integer.toBinaryString(i);
for(int j=str.length(); j<8; j++)
out.write('0');
out.write(str);
in.close();
out.close();
} catch (Exception e) {}
} -
Help converting binary to decimal.
Hi all.
I'm taking a course in Java programming and so far its been a great experience. I'm currently stumped on one of my assignments and I am kindly requesting some assistance.
Basically I have to convert a binary entry into its decimal comparison.
so for instance an entry of 1101 would output 13.
I have the calculation formula, but my problem is finding out whether the point i'm looking at is either a 1 or a 0. I can't seem to 'strip it down' to the value.
My current thought is something like this:
binary = 1101;
// problem is here
thou = (binary % 1000) / 1000;
hund = (thou % 100) / 100;
tens = (hund % 10) / 10;
ones = (tens % 1) / 1;
// then do formular
decimal = ones * 1 + tens * 2 + hund * 4 + thou * 8;
I have searched but all the suggestions say to use a custom function or other obscure method. I don't know that stuff real well and we aren't even that far in the book either. All we've done so far is while, if .. else, condition statements.
I will include the actual text for the question, in case i'm not clear. This text also includes a 'tip' which isn't helping me. (http://www.geocities.com/kaveman2000/q-4-25.pdf)
Thanks again for any pointers.angeles1016 (and all),
thanks for your great help and code snippets.
angeles1016,
using the code you provided i noticed that the last (single digit) was giving the wrong response in certain cases. (1000, 0111, 0001, ..) i could be off one.
either way, i decided to fix it up and give you all the update.
// do class, main function stuff
if(binary/1000 == 1){
thousands += (binary/1000);
temp = binary % 1000;
if(temp/100 == 1){
hundreds += (temp/100);
temp = temp % 100;
if(temp/10 == 1){
tens += (temp/10);
temp = temp % 10;
if(temp/1 == 1){
ones += (temp/1);
// calculate binary to decimal
decimal = (ones * 1) + (tens * 2) + (hundreds * 4) + (thousands * 8);
// take result and convert to string for output
result += "Decimal: " + decimal + "\n";
// display in gui ?
// end function and class stuff(note: i suppose the statements similar to "thousands += (binary/1000);" can be simplified as "thousands = 1;") -
How to print space as thousand seperator and comma as decimal seperator
Hi All,
I have requirement where I need to print the amounts with space as thousand seperator and comma as decimal seperator.
I have a field wrshb which is of type mhnd-wrshb. currently I am printing this. In the adobe layout I have declared this coloumn as Decimal field.
Now in the output it is printing as comma as thousand seperator and dot as decimal seperator.
For example ,currently the value is printing as 32,811.41
but I want the amount as 32 811,41
I have declared the variable as char16, using write statement in the interface I moved the value from currency field to char field.
Then in debugging i checked the value comes as 32,811.41 and it goes to dump with teh reason-cannot interpret as a number.
Can anyone help me in fixing this?
Thanks and Regards,
Karthik Ganti.Hi Adam,
As per initial requirement, I have set the format such that the amount is printing in below format as required.
Locale---Italian.
Space as thousand seperator and comma as decimal seperator.
for example 1 234,45
As some of the Currencies will not have decimals, now users would like to print amount without decimals. For example in my case amount printing in KRW ( Korean currency ) is also similar to the above format which is wrong.
for example Now amount is printing as 55 000,00. But actually it should be 550 000. Similarly for JPY currency also, as it doesnot haves decimals ( checked in TCURX table ).
I have written some logic in the interface. below is the logic.
WRITE:
wa_mhnd1-wrshb to wa_item-wrshb CURRENCY WA_ITEM-WAERS.
*READ TABLE lt_tcurx INTO lwa_tcurx WITH KEY currkey = wa_item-waers BINARY SEARCH.
IF sy-subrc = 0.
IF lwa_tcurx-currdec = '0'.
REPLACE ',' WITH SPACE INTO WA_ITEM-WRSHB.
REPLACE ',' WITH SPACE INTO WA_ITEM-WRSHB.
else.
REPLACE ',' WITH SPACE INTO WA_ITEM-WRSHB.
REPLACE ALL OCCURRENCES OF '.' in wa_item-wrshb WITH ','.
endif.
ENDIF.
a. when the write statement gets executed amount will be in ,. ( 1,234.45 )format. Then my logic gets executed correctly. In this company code is CH10 ( EUR ) and KR10.
b. But sometimes after the write statement gets executed amount will be in ., format ( 1.234.45 ). In this case my logic works, but gives the wrong value. In this case company code is VN10 ( EUR )
In both the cases currency is EUR.
Will the decimal format change accordingly based on the company code code currency.Can you please tell me why write statement behaved differently.
Do I need to change any locale in the adobe form, or any other logic to be written in interface. ? I am trying it out from long time, but not able to fix it.
Can you please help me how to achieve this ?
Thanks and Regards,
Karthik Ganti. -
Linear search and binary search
Hi
can any one tell me what is linear and binary search in detail.
and what is the difference between them .
which one is useful in coding.
Thanks&Regards,
S.GangiReddy.hi,
If you read entries from standard tables using a key other than the default key, you can use a binary search instead of the normal linear search. To do this, include the addition BINARY SEARCH in the corresponding READ statements.
READ TABLE <itab> WITH KEY <k1> = <f1>... <kn> = <fn> <result> BINARY SEARCH.
The standard table must be sorted in ascending order by the specified search key. The BINARY SEARCH addition means that you can access an entry in a standard table by its key as quickly as you would be able to in a sorted table.
REPORT demo_int_tables_read_index_bin.
DATA: BEGIN OF line,
col1 TYPE i,
col2 TYPE i,
END OF line.
DATA itab LIKE STANDARD TABLE OF line.
DO 4 TIMES.
line-col1 = sy-index.
line-col2 = sy-index ** 2.
APPEND line TO itab.
ENDDO.
SORT itab BY col2.
READ TABLE itab WITH KEY col2 = 16 INTO line BINARY SEARCH.
WRITE: 'SY-SUBRC =', sy-subrc.
The output is:
SY-SUBRC = 0
The program fills a standard table with a list of square numbers and sorts them into ascending order by field COL2. The READ statement uses a binary search to look for and find the line in the table where COL2 has the value 16.
Linear search use sequential search means each and every reord will be searched to find. so it is slow.
Binary search uses logrim for searching. Itab MUST be sorted on KEY fields fro binary search. so it is very fast.
The search takes place as follows for the individual table types :
standard tables are subject to a linear search. If the addition BINARY SEARCH is specified, the search is binary instead of linear. This considerably reduces the runtime of the search for larger tables (from approximately 100 entries upwards). For the binary search, the table must be sorted by the specified search key in ascending order. Otherwise the search will not find the correct row.
sorted tables are subject to a binary search if the specified search key is or includes a starting field of the table key. Otherwise it is linear. The addition BINARY SEARCH can be specified for sorted tables, but has no effect.
For hashed tables, the hash algorithm is used if the specified search key includes the table key. Otherwise the search is linear. The addition BINARY SEARCH is not permitted for hashed tables.
Binary search must be preffered over linear sarch.
Hope this is helpful, Do reward. -
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how to transform jpeg image to binary coded image? using which properties? thanks .
well, I dont think so. but what exactly do you need, transforming it into bits or bytes?
well, I have no time to do the program for you, but this is the idea:
FileInputStream inputF=null;
File imageF;
byte content [];
try {
imageF=new File ("The image path");
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Doubt on calculating Hexadecimal and binary values
Please give me some tips to find the hexadecimal and binary values of negative numbers for eg:-
The hexadecimal value of -1 is 0xffffffff. Here I don't know how the corresponding hex value is calculated.
I calculate the hex value of numbers in this way. For eg: to get the hex value 11
1*16^1 + 1*16^0 = 17.
Similar logic to convert binaries to decimal by using the base 10.It's about 2s complement arithmatic.
Hexidecimal representation is used by people interested in the bit patterns, rather than numerical values as such so integers are treated as unsigned. But when working in decimal we're generally treating the variable as signed. Negative integers are represented by "adding" 2^32 to them so -1 becomes 2^32-1. This is stored as a word where all bits are 1s, which means in hex all nybles are 'F'. -
What is difference between TEXT mode and BINARY mode?
Hi,
What is difference between TEXT mode and BINARY mode? In TEXT mode what is "ENCODING DEFAULT".
OPEN DATASET wl_filename FOR INPUT
IN TEXT MODE ENCODING DEFAULT.
OPEN DATASET wl_filename FOR INPUT
IN BINARY MODE .Hi vishnu,
1. files like .TXT files are considered to be text,
and other files like .EXE . DOC etc are binary files.
2. In text files, after each line, there are two
internal binary characters
one for line feed and another one for newline.
3. But, we don't have any meaning for it,
when we view such files in notepad, for eg.
Hence, to only get the understandable content,
we should open such TXT files, in text mode.
(we can also open them in binary mode,
but we will get two extra characters after each line,
and such two extra characters are meaningless
for interpreting the text)
4. In binary mode, each byte by byte is considered,
without any meaning.
regards,
amit m. -
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http://docs.info.apple.com/article.html?path=iWeb/2.0/en/9903.html
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