Binary or number string

Hi...
How can I know if a string is a binary or number?
thank you.

You can only test for a String not containing a binary number like this..
boolean isNotBinary = ( someString.indexOf( "0" ) == -1 && someString.indexOf( "1 ) == -1 );As a number can be anything with "-012345678eE." and has to adhere to a specific format to qualify as a number AND the fact that the radix makes everything ambiguous determining if a String contains a number and is binary becomes horribly complicated.
Your definition of number needs to be more tightly defined, integer/long, float/double, fixed decimal and integer/long numbers with different radix need to be considered before you can test for numberness.

Similar Messages

Number to a binary base number

How can I convert a numeric data to a binary base number?
Ex.: A into 0001010
Thank you in advance

OK, only conversion to a string is needed to solve the problem. (Everything else is just a cosmetic feature of the indicator where you set the formatting. This does not change the underlying data.)
Just use "format into string" with a format code of "%07b" (=show seven binary digits and pad with zeroes to the left).
LabVIEW Champion . Do more with less code and in less time .

Find logo from binary or hex string

HI All ,
I have file 2 strings one with HEX and one binary this to string is contain the value
for attachment (logo) ,which FM or method i should use to see the logo, the picture itself ?
Regards
James
Edited by: James Herb on Feb 28, 2010 5:32 PM

Hi James,
This methods and function modules should work for you.
      CALL METHOD cl_binary_relation=>read_links_of_binrel
        EXPORTING
          is_object   = is_object
          ip_relation = 'ATTA'
        IMPORTING
          et_links    = et_links.
    catch
    cx_obl_parameter_error into icx_obl_parameter_error.
      exception_string = icx_obl_parameter_error->get_longtext( ).
    catch cx_obl_internal_error into icx_obl_internal_error .
      exception_string = icx_obl_internal_error->get_longtext( ).
    catch
    cx_obl_model_error into icx_obl_model_error.
      exception_string = icx_obl_model_error->get_longtext( ).
ENDTRY.
SORT et_links BY utctime.
LOOP AT et_links INTO et_links_s.
    v_tbx       = sy-tabix.
    document_id = et_links_s-instid_b.
    CALL FUNCTION 'SO_DOCUMENT_READ_API1'
      EXPORTING
        DOCUMENT_ID                      = document_id
      FILTER                           = 'X '
     IMPORTING
        DOCUMENT_DATA                    = document_data
     TABLES
        OBJECT_HEADER                    = object_header
        OBJECT_CONTENT                   = object_content
      OBJECT_PARA                      =
      OBJECT_PARB                      =
      ATTACHMENT_LIST                  =
      RECEIVER_LIST                    =
      CONTENTS_HEX                     =
     EXCEPTIONS
        DOCUMENT_ID_NOT_EXIST            = 1
        OPERATION_NO_AUTHORIZATION       = 2
        X_ERROR                          = 3
        OTHERS                           = 4.
    IF SY-SUBRC <> 0.
      MESSAGE ID SY-MSGID TYPE SY-MSGTY NUMBER SY-MSGNO
              WITH SY-MSGV1 SY-MSGV2 SY-MSGV3 SY-MSGV4.
    ENDIF.
ENDLOOP.

Best way to put binary-data into string?

Hi there!
What I want to do is to transfer binary data via HTTP/GET so what I have to do is to transfer binary data into a string.
Currently I do this the follwing way:
      byte[] rawSecData = new byte[4]; //any binary datas
      ByteArrayOutputStream secBOS = new ByteArrayOutputStream(4);
      DataOutputStream secDOS = new DataOutputStream(secBOS);
      for(int i=0; i < rawSecData.length; i++)
        secDOS.writeByte(rawSecData);
secDOS.flush();
String secData = secBOS.toString();
System.err.println("Lenght of resulting String: "+secData.length());
I know that this way already worked fine, however I now set up my system up again with another linux-distro and now strange things happen.
e.g. the secData string differs in lenght from run-to-run between 2 and 4 and I don know at all why?
Transferring the binary-stuff into string-stuff (e.g. short-binary 255 255, String: 65536) is not possible for me because of various reasons.
The funny thing is that I remeber that this already worked some time ago and I can figure out why it now doesnt...
Please help!

First of all thanks a lot for your help!
Yes, I already think its an encoding problem, but how can I specify the encoding in my application in a portable way. I dont have an idea what to do.
My applikation should run as applet on many different 1.1+ VMS (msjvm, netscape-1.1.5, ...).
Thanks again, lg Clemens

Binary value to string

Hi,
I want to convert binary value to string format. How can i do that if possible please give me sample code.?
One more question is i want to do reverse thing also that is convert string value to binary format.?
Waiting for reply

There are methods in Integer, Long etc for transforming to binary.
I think there are probably also methods to go from binary to an int. Check out the JavaDoc

Splitting a Number String without using split()

What is the better way of converting a number string where the numbers are seperated by a white space into an array without using the split() method or tocharArray e.g numberString = "12 13 14" numberArray[0] = 12, numberArray[1] = 13, etc

If you know the number of elements in the string, you can create a predefined fixed array. You can also search for empty spaces.
String text = "12 13 14";
int COUNT = 3;
public void doit() {
int[] arr = new int[COUNT];
int sub1 = 0;
int sub2 = 0;
for(int i=0; i<COUNT - 1; i++) {
    sub2 = text.indexOf(" ", sub2 + 1);
    arr[i] = Integer.parseInt(text.substring(sub1, sub2));
    sub1 = sub2 + 1;
arr[COUNT - 1] = text.substring(sub1, text.length());

Binary Stored as String

Hi,
I am storing Binary Data into Varchar(Max) field, while migrating to another binary field, since it is stored earlier in the table with the varchar and then insert into Binary will again convert the string into Binary data. which is corrupting data.
How to Insert Varchar(max) field to Binary Field in another table
Thanks, Parth

Hi Parth,
Regarding your description, the binary data is stored in a varchar(max) column, do you mean the values of the varchar(max) column are like ‘0x5445535420535452494E47’? So when you tried to migrate it to a binary field, the string ‘0x5445535420535452494E47’
was converted into binary other than the value the string stands for?
If my understanding is correct, you may reference the code as below, just use the built-in
COVERT with one more optional parameter, you can read more details under
Binary Styles in the covert page.
DECLARE @srcTbl TABLE(str1 VARCHAR(MAX))
DECLARE @tarTbl TABLE(str1 VARBINARY(MAX))
INSERT INTO @srcTbl SELECT '0x5445535420535452494E47'
INSERT INTO @tarTbl SELECT CONVERT(VARBINARY(MAX),str1,1) FROM @srcTbl -- 1 here is necessary
SELECT str1 FROM @srcTbl
SELECT str1 FROM @tarTbl
If you have any question, feel free to let me know.
Best regards,
Eric Zhang
TechNet Community Support

Conversion of binary to number

Hello,
I have a signal conditioner controlled through serial communication with ASCII commands. Once the start command is received, there is a continuous stream of offset binary data sent and read through the serial Read Indicator. The manual informs me that I must first condition the offset binary values by subtracting 2048 from the raw values to produce a 2's complement value, it also mentions that the data is encoded in 12 bits so negative numbers must be properly sign extended to move the data into 16 bit words used by the PC. I am not familiar with binary numbers and am unsure on where to start. If someone could provide some insight and suggestions on what my first step would be to converting the binary values into number format, that would be greatly appreciated.
Thank you in advance,
Jena

That won't work because he's not receiving the string the way you think it is. The hex display is an interpretation of the underlying byte values. It is not the same as number written out in hex as a string. Your conversion assumes he's getting the character 1, followed by the character 8, followed by the character 0, followed by the character 0, followed by a space character.
The simplest way is to use the Type Cast function, 'cause that's what it's there for. Wire in an array of the datatype you want. The thing to keep in mind is that even though the data may be fundamentally 12 bits you need to deal with multiples of 8 because a serial read will read bytes, not bits. Thus, you can convert your string to an array of I16 numbers like this:
Message Edited by smercurio_fc on 04-08-2008 01:07 PM
Attachments:
Example_VI_BD2.png ‏2 KB

Binary Search Using String

I'm trying to perform a binary search on CDs stored in an arraylist but it will only work with the titles with no spaces (Such as Summertime & Heartless but not Dance Wiv Me). Is this a bug or will it simply not work with strings with a space in them? Also when it does work it will return the correct title but the artist and price aren't in the same arraylist index as the search value that was returned.
* Program to allow customers to purchase CDs from an online store
* @author (Martin Hutton)
* @version (24/05/2009)
import java.util.*;
public class CDs
    private Scanner input;
    private Scanner in;
    private Scanner sc;
    private CdList aCd;
    CDs()
        this.aCd = new CdList();
        this.menu();
    public void menu()
        int select = 5;
        do
            //Menu Display
            System.out.println("\n\t\t--== Main Menu ==---");
            System.out.println("\n\t\t1. View CDs");
            System.out.println("\n\t\t2. Purchase CDs");
            System.out.println("\n\t\t3. Search CDs");
            System.out.println("\n\t\t4. Sort CDs Titles");
            System.out.println("\n\t\t5. Exit");
            input = new Scanner(System.in);
            select = input.nextInt();
            switch (select)
                case 1 : this.view();
                break;
                case 2 : this.purchase();
                break;
                case 3 : this.search();
                break;
                case 4 : this.sort();
                break;
                case 5 : exit();
                break;
                default : System.out.println("Error! Incorrect menu selection!");
        while ( select != 5 );
    public void view()
        System.out.printf("\f");//Clear screen
        System.out.println("\t\t--== Avaiable CDs ==--");
        System.out.println("");
        int size = aCd.getTitle().size();
        //loop to display array date
        for ( int i = 0; i < size; i++ )
            System.out.println( "\t" + i + "\t" + aCd.getTitle().get(i)+ "\t\t\t" + aCd.getArtist().get(i) + "\t\t\t" + aCd.getPrice().get(i) + "\n");
    public void purchase()
       System.out.printf("\f");//Clear screen
       double arrayPurchase[] = new double [15];
       in = new Scanner(System.in);
       sc = new Scanner(System.in);
       double total = 0.0;
       int itemindex = 0;
       System.out.println("How many CDs would you like to purchase? ");
       int amountNumbers = in.nextInt();
       for(int i = 0; i< amountNumbers; i++)
           int q = itemindex;
           System.out.println("Please enter the CD number: ");
           q = in.nextInt();
           //aCd.getPrice().get(q);
           arrayPurchase[i] = aCd.getPrice().get(q);
           total += arrayPurchase;
System.out.println("\nThe total is: £" + total );
public void search()
System.out.println("\t\t--== Search CDs ==--\n");
System.out.println("Search for a CD: ");
String lc = input.next();
Collections.sort(aCd.getTitle());
int index = Collections.binarySearch(aCd.getTitle(),lc);
if ( index < 0 )
System.out.println("Sorry, CD not avaiable");
else
// System.out.println(aCd.getPrice().get(index));
System.out.println( index + aCd.getTitle().get(index) + "\t\t" + aCd.getArtist().get(index) + "\t\t" + aCd.getPrice().get(index));
this.aCd = new CdList();
public void sort()
System.out.println("\t\t-== Alphabetised CD Titles ==--\n");
Collections.sort(aCd.getTitle());
int size = aCd.getTitle().size();
for ( int i = 0; i < size; i++ )
System.out.println( "\t" + aCd.getTitle().get(i));
this.aCd = new CdList();
public void exit()
System.out.println("\nSystem shutting down...");
System.exit(0);
* Write a description of class CdList here.
* @author (your name)
* @version (a version number or a date)
import java.util.*;
public class CdList
//instance variables
private ArrayList<String> title;
private ArrayList<String> artist;
private ArrayList<Double> price;
public CdList()
//create instances
title = new ArrayList<String>();
artist = new ArrayList<String>();
price = new ArrayList<Double>();
//populate arrays
//add titles
title.add("Boom Boom Pow");
title.add("Summertime");
title.add("Number 1");
title.add("Shake It");
title.add("The Climb");
title.add("Not Fair");
title.add("Love Story");
title.add("Just Dance");
title.add("Poker Face");
title.add("Right Round");
title.add("Dance Wiv Me");
title.add("I'm Not Alone");
title.add("Hot 'n' Cold");
title.add("Viva La Vida");
title.add("Heartless");
//HEX codes
artist.add("Black Eyed Peas");
artist.add("Will Smith");
artist.add("Tinchy Stryder");
artist.add("Metro Station");
artist.add("Miley Cyrus");
artist.add("Lily Allen");
artist.add("Taylor Swift");
artist.add("Lady GaGa");
artist.add("Lady GaGa");
artist.add("Flo Rida");
artist.add("Dizzee Rascal");
artist.add("Calvin Harris");
artist.add("Katy Perry");
artist.add("ColdPlay");
artist.add("Kanye West");
//RGB Co-ods
price.add(0.99);
price.add(0.75);
price.add(1.99);
price.add(2.99);
price.add(2.99);
price.add(0.55);
price.add(2.75);
price.add(1.98);
price.add(1.25);
price.add(1.55);
price.add(0.99);
price.add(2.55);
price.add(0.55);
price.add(1.99);
price.add(0.99);
} //end of constructor
public ArrayList<String> getTitle()
return ( title );
} //end method
public ArrayList<String> getArtist()
return ( artist );
}//end method
public ArrayList<Double> getPrice()
return ( price );
}//end method

It sounds like you are having Scanner woes. Remember that the call:
select = input.nextInt();Reads the number inputted but not the rest of the line (the enter). Then the next call to readLine will read this.
Suggestion: do this, to read a number:
select = input.nextInt();
String restOfLine =input.nextLine(); //discard

How to parse a BigDecimal from a localized number String

I want to be able to build a BigDecimal instance from a localized String.
i.e.: From a string like "2.123,45123" I want to get when I get by doing: new BigDecimal("2123.45123");
The NumberFormat parse(String) method answers doubles, so I lose presition. When I do nf.parse(2,0000000000000000001) y get a Double representing 2. I can't make a BigDecimal from such a string afaik.
The problem is the localized version on the number (in the string) otherwise it works fine (with the US locale).

>>
So you only can't convert a string to a BigDoublethat
uses the ',' for the decimal place and the '.' forthe
real number separators? If that's the case, whycan't
you use a String buffer then to strip out all the
periods and replace the comma with a period so it
looks like a US decimal string then use that?Thank you very much for your answer. That would have
been a workaround. I found a BigDecimal parser that's
absolutely perfect for the task.
http://users.belgacombusiness.net/arci/math/
Gives to BigDecimals what java standard gives to all
the other Number subclasses. Good! Frankly, I'm usually just lazy enough to hack a solution.... because I like to say that I hack. evil laugh

Reading a binary file to string variable does not populate correctly

Hi, I am new to Adobe Air/Flex and I'm trying to read a
binary file which also contains text in "cleartext". The problem
I'm having is that when I call FileStream.readUTFBytes method, only
the first 6 characters are showing up in my string variable
"contents" when I debug it in FlexBuilder or use a trace command
and debug it. I have also tried with other types of files but I
have a similar problem unless it's a non-binary file.
Am I doing something incorrectly or should I be reading a
binary file differently than the way I'm reading it currently?
The source code is shown below.
TIA,
Magnus
<?xml version="1.0" encoding="utf-8"?>
<mx:WindowedApplication xmlns:mx="
http://www.adobe.com/2006/mxml"
layout="absolute">
<mx:Script>
<![CDATA[
import flash.filesystem.File;
private function readFiles():void{
var feedFile:File = File.userDirectory.resolvePath( "Local
Settings/Application Data/Microsoft/Feeds/AppScout~.feed-ms" );
var stream:FileStream = new FileStream();
stream.open( feedFile, FileMode.READ );
var contents:String =
stream.readUTFBytes(stream.bytesAvailable);
trace(contents);
stream.close();
]]>
</mx:Script>
<mx:Button x="121" y="66" label="Button"
click="readFiles()" />
</mx:WindowedApplication>

It's difficult to tell what it is, it looks like a binary
pipe symbol but I can copy it from TextPad for example. Some of the
characters following it cannot be copied from TextPad which I
assume is because it's null. I can read the whole file in C#/.Net
and assign it to a string variable without any problems but perhaps
Air is somewhat limited to binary content.

Binary form of string

Hi All,
I have written a small program which prints the addresses of the line. But, I don't know how to get the binary form of the string. How the binary form of a string? Will it be 0s and 1s or someother format?
How to get the binary form of a string?
        try
            // Open or create the output file
            FileReader fr = new FileReader("./Sample.txt");
            BufferedReader br = new BufferedReader(fr);
            String inLine;
            while((inLine = br.readLine())!=null)
                byte[] data = inLine.getBytes();
                BinaryRefAddr binary = new BinaryRefAddr("01",data);
                System.out.println(binary.getContent());
                System.out.println(binary.toString());
        catch (IOException e)
        }

Arivu,
Could you give an example of what a binary form is?
For example
What is the binary form of "2"?
Is it a String of 0s an 1s such as "00110010"?
or is an int with value of 50?
Or what?

Binary Tree: Number of Nodes at a Particular Level

Hi, I'm trying to teach myself about binary tree but am having trouble writing an algorithm. The algorithm that I'm trying to write is one that will take a binary tree and output the number of nodes at each level of the tree (maybe in an array).
I have no trouble writing an algorithm that does a breadth-first traversal, but I have lots of trouble trying to determine where each level ends.
Thanks for the help

Try something like this:
class BTree {
    BTreeNode root;
    public int numberOfNodesAtLevel(int level) {
        if(root == null) return -1;
        return numberOfNodesAtLevel(root, level);
    private int numberOfNodesAtLevel(BTreeNode node, int level) {
        if(level == 0) {
            return 1;
        } else {
            return (node.left == null ? 0 : numberOfNodesAtLevel(node.left, level-1)) +
                   (node.right == null ? 0 : numberOfNodesAtLevel(node.right,level-1));
}

Number - string question

Hello,
I have the next problem:
I have cursor that contains the next thing:
WHERE LKP_INDUSTRY1_ID IN ( p_sector )
p_sector is being delivered in varchar: '23;22;36'
My question is: is it possible to manipulate p_sector in a way that I can paste it in my cursor? I can replace the ';' with ',' but I don't know how I can make it read the numbers as numbers and not as strings.
Thanks in advance.
Oli

Does this solve your problem ?
SQL> create or replace type tab_n is table of number;
2 /
Type created.
SQL> declare
2   p_sector varchar2(20) := '7369;7499;7521';
3   t tab_n := tab_n();
4   element number;
5   pos integer := 0;
6   cursor a is select ename from emp where empno in (select * from table(t));
7 begin
8   for i in 1..length(translate(p_sector || ';',';0123456789',';')) loop
9    t.extend;
10    t(t.count) := substr(p_sector,pos+1,instr(p_sector || ';',';',1,i)-pos-1);
11    pos := instr(p_sector,';',1,i);
12   end loop;
13   for v in a loop
14    dbms_output.put_line(v.ename);
15   end loop;
16 end;
17 /
SMITH
ALLEN
WARD
PL/SQL procedure successfully completed.Rgds.

Regular expression - parse version number string

Hi,
I try to parse a string using regular expressions but id did not work correctly in all cases.
The string that i want to parse can have one of the following layout:
3.4.5.v20090305 or
3.4.5The first three parts have to be integer values, the last part can be a free string (which is optional).
I use the following code to parse the version number and extract the parts:
Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
Matcher m = versionPattern.matcher(versionString);
if (!m.find()) {
    throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// assert that we matched every part
// three groups (without qualifier) or four parts (with qualifier)
int groups = m.groupCount();
if (groups != 4) {
     throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// extract the parts
major = parseInt(m.group(1));
minor = parseInt(m.group(2));
micro = parseInt(m.group(3));
qualifier = m.group(4);The above regular expression works in all cases that i tested, except one.
The follwoing string is accepted as correct input: (but it shouldn't)
3.4.5a.v20090305And i get the result:
major = 3
minor = 4
micro = 5
qualifier = a.v20090305
Thanks for help or suggestions :)
Best Regards,
Michael

heissm wrote:
Hi,
I try to parse a string using regular expressions but id did not work correctly in all cases.
The string that i want to parse can have one of the following layout:
3.4.5.v20090305 or
3.4.5The first three parts have to be integer values, the last part can be a free string (which is optional).
I use the following code to parse the version number and extract the parts:
Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
Matcher m = versionPattern.matcher(versionString);
if (!m.find()) {
throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// assert that we matched every part
// three groups (without qualifier) or four parts (with qualifier)
int groups = m.groupCount();
if (groups != 4) {
throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// extract the parts
major = parseInt(m.group(1));
minor = parseInt(m.group(2));
micro = parseInt(m.group(3));
qualifier = m.group(4);The above regular expression works in all cases that i tested, except one.
The follwoing string is accepted as correct input: (but it shouldn't)
3.4.5a.v20090305And i get the result:
major = 3
minor = 4
micro = 5
qualifier = a.v20090305No, that can't be the output. Perhaps you have some old class files you're executing, because with the code you now posted, that can't be the result.
To verify this, execute this:
import java.util.regex.*;
public class Main {
    public static void main(String[] args) {
        String versionString = "3.4.5a.v20090305";
        Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
        Matcher m = versionPattern.matcher(versionString);
        if (!m.find()) {
            throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
        // assert that we matched every part
        // three groups (without qualifier) or four parts (with qualifier)
        int groups = m.groupCount();
        if (groups != 4) {
             throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
        // extract the parts
        System.out.println("1 -> "+m.group(1));
        System.out.println("2 -> "+m.group(2));
        System.out.println("3 -> "+m.group(3));
        System.out.println("4 -> "+m.group(4));
}and you'll see that the following is the result:
1 -> 3
2 -> 4
3 -> 5
4 -> nullSome remarks about your pattern:
"(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?"All the "{1}" can be omitted they don't add anything to your regex and are only cluttering it up. And you'll probably also want to "anchor" your regex using the start- and end-of-string-meta-characters, like this:
"^(\\d+)\\.(\\d+)\\.(\\d+)(?:\\.(\\w+))?$"

Binary or number string

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