Counting letters in a string
Hi,
I am new to Java. What's the most efficient way to count the number of characters in a string. I have a string of text called MyText. I was thinking of a for loop but maybe there is better solution?
Regards,
Rob.
FYI:
public static void main(String[] args)
String word;
System.out.println("Enter word");
word = EasyScanner.nextString();
word.length(); // this is redundant and serves no purpose
System.out.println(word.length());
}Also, when posting your code, please use code tags so that your code will retain its formatting and be readable. To do this, either use the "code" button at the top of the forum Message editor or place the tag [code] at the top of your block of code and the tag [/code] at the bottom, like so:
[code]
// your code block goes here.
[/code]Your posted code will need to be properly formatted to begin with for this to work.
Good luck.
Edited by: Encephalopathic on Jan 6, 2008 10:45 AM
Similar Messages
-
How to find MATCH count for # in a String
Hi all,
How to find MATCH count for # in a String.
Ex: 6170#0400-0002-00#API3PT#AL#AUST#DEVE#KG#100.00#100.00#100.00#0.00##10.20
Regards,
Balavardhan.K>
balavardhan k wrote:
> I have used below syntax to find count but it is not returning the value and SY-SUBRC = 4.
>
>
> Data : l_text type string.
>
> l_text = '6170#0400-0002-00#API3PT#AL#AUST#DEVE#KG#100.00#100.00#100.00#0.00##10.20'.
>
>
> FIND ALL OCCURRENCES OF '#' IN L_TEXT MATCH COUNT mcnt .
Then it's not a # but possible a horizontal tab-sign.
Do a find on cl_abap_char_utilities=>horizontal_tab instead.
Edited by: Maen Anachronos on Nov 23, 2010 2:48 PM -
I am using lingo to count and display letters in a word but my script is stuck on a point. Can any help to solve this?
property
SpriteNum
global
gCurrentCounter, CurrentWord, lettersofCurrentWord, sSoundName
On
MouseDown me
if gCurrentCounter < 10 then
p = "00" & gCurrentCounter
else if gCurrentCounter < 100 then
p = "0" & gCurrentCounter
elsep = gCurrentCounterend ifj = 1
letter_no = spritenum
letter_name = member("PE_" & string(j)).text
letter_uppercase = "A"
-- puppetSound 1, "A sound"
CurrentWord = member("word - " & p)i =
1
If sprite (letter_no).within(sprite(spritenum)) Thenif
charToNum(member(CurrentWord).paragraph[1].char[1]) <= 90 then
member ("pe_space_" & string(i)).text = letter_uppercase
elsemember ("pe_space_" & string(i)).text = letter_name
end if
end if set the loc of sprite letter_no to point(300,330)
-- i = 2
put i
-- repeat while i <= CurrentWord.char.count
if charToNum(member(CurrentWord).paragraph[1].char[i]) >= 200 then
member ("pe_space_" & string(i)).text = member(CurrentWord).paragraph[1].char[i]
sprite (letter_no).within(sprite (10 + i))
elsemember ("pe_space_" & string(i)).text = letter_name
end if
i = i + 1
-- end repeat
end mousedown me
end**** Hello Sean see responses to questions in RED
So what sort of sprite/member holds the completed word?
Sean - A complete set of sprite equal the number of letters in
the word named pe_space_
Presumably the code you posted is attached to individual letters of the
alphabet?
Sean - You assumed right
I have re-posted your code with questions and comments interspersed -
perhaps you could addess some of these:
global gCurrentCounter, CurrentWord, lettersofCurrentWord, sSoundName
property SpriteNum
on mouseDown me
if gCurrentCounter < 10 then
p = "00" & gCurrentCounter
else if gCurrentCounter < 100 then
p = "0" & gCurrentCounter
else
p = gCurrentCounter
end if
j = 1
letter_no = spritenum
-- I don't understand: since you declared j = 1
-- your local var 'letter_name' will always be member("PE_1").text
letter_name = member("PE_" & string(j)).text
-- again, you have hard-coded something with no apparent reason
letter_uppercase = "A" *** Sean - In case the first letter is uppercase
-- puppetSound 1, "A sound"
CurrentWord = member("word - " & p)
i = 1
if sprite (letter_no).within(sprite(spritenum)) then
if charToNum(member(CurrentWord).paragraph[1].char[1]) <= 90 then
member("pe_space_" & string(i)).text = letter_uppercase
else
member("pe_space_" & string(i)).text = letter_name
end if
end if
sprite(letter_no).loc = point(300,330)
-- i = 2
-- i will always equal 1 because that's how you declared it above, and it
isn't changed
Yes Sean, this is the reason for my been stuck on the first sprite
or letter of my word. I was hoping to achive repeat effect by clicking on the mouse
or keydown to get word spelt.
But on mouseDown me or keyDown does not work with my code
put i - This tell me that I am stuck on i = 1
-- can you explain what this block of code is doing?
-- I have no idea what the various members
-- member("pe_space_" & string(i))
-- are or how they're used
Sean the pe_space_ are the fields where my letter PE_ get attached to,
so when I release my letters on grab and move they are within these
sprites/fields
So, Sean can still help?
repeat while i <= CurrentWord.char.count
if charToNum(member(CurrentWord).paragraph[1].char[i]) >= 200 then
member("pe_space_" & string(i)).text =
member(CurrentWord).paragraph[1].char[i]
-- 'within' is an operator that returns TRUE/FALSE
-- so it isn't clear what you're trying to do here
sprite(letter_no).within(sprite (10 + i))
else
member("pe_space_" & string(i)).text = letter_name
end if
i = i + 1
end repeat
end mouseDown
Message was edited by: efeelo -
How can i get a query to count a number of strings?
I want the query to count for me how many holidays there are to the USA in my table. But im not sure how to get it to retreive the result of a string or varchar2 datatype. This is what i have tried so far.
select count(COUNTRYVIS) <<this column contains all the countries visited
from "IT220_HOLIDAYDETAILS" <<this is the table where the column is
where COUNTRYVIS = "USA" << this is the result i want it to count
ORA-00904: "USA": invalid identifier <<this is the error message im getting, is there a way to count strings?
Thanks in advance!This has nothing to do with APEX. Please post basic SQL questions on the +{forum:id=75}+ forum.
In any OTN forum, alll code should be posted wrapped in <tt>\...\</tt> tags as described in the FAQ:
select count(COUNTRYVIS) -- this column contains all the countries visited
from "IT220_HOLIDAYDETAILS" -- this is the table where the column is
where COUNTRYVIS = "USA" -- this is the result i want it to countYou appear to be struggling with the distinction between identifiers and text literals. Hint: One uses double quotes, the other single quotes.
Note that quoted identifiers are generally not a good idea. -
Counting Characters in a string
HI all!
I am new working with java and i have an application where I have to count specific characters in a string e.g. 'a','.','H','4' and so on! I found an entry in the forum with this script:
String text = "Hello World!";
int count = 0;
for (int x = 0; x < text.length(); x++)
if (text.charAt(x) == 'l')
count++;
This Script count all the 'I', but I could not get CharAt to count different characters at once!
Is there a way?
Thx for your help!
LeBiteString text = "Hello World!";
int countl = 0;
int counto = 0;
for (int x = 0; x < text.length(); x++)
if (text.charAt(x) == 'l') countl++;
if (text.charAt(x) == 'o') counto++;
[/code[ -
Counting character in a string
I'm trying to count how many times a characters occurs in a string, and I'm having a lot of trouble. This is what I have so far:
note: the method call has to be public static int countOccs(char f, String s) and I know that it can be done using only loops, for, charAt, and length () and I want to learn it like that.
[I just chose f, s randomly.]
public class Counting {
public static void main(String [ ] args) {
String s = "Testing";
int countOccs = s.length( );
char [ ] charArray = new new char[counting];
for (int i = 0; i < counting; i++) {
charArray[i]
public static int counting(char f, String s) {
I'm really lost. When I try to define char f = "f" or String s = "Testing" in the method call (public static int counting...) I get messages saying that it has already been declared in the call, but there has been nothing assigned to it. Anyways, I feel really lost. I did find a pattern by looking at my notes from awhile ago, but am having trouble filling out parts of it. (Thanks, memory...) I think the problem comes from having to have the method call be public static int counting (char f, String s). That part confuses me a lot because I want it to print out how many characters of f are in a string, but it doesn't have "void" in it so it doesn't return a value.
for (int i=0; i < s.length(); i++)
char f = s.charAt(i);
if (something that looks at the characters)
do something like count
Edited by: MAresJonson on Feb 6, 2009 6:50 PMh1. The Ubiquitous Newbie Tips
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(d) read the relevant section of the [API Docs|http://java.sun.com/javase/6/docs/api/index-files/index-1.html] and maybe even
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We the New To Java regulars do hereby pledge to refrain from flaming anybody, no matter how gumbyish the question, if the OP has demonstrably tried to cover these bases. The rest are fair game. -
Counting SPACES in a string field.
Hi ,
can any one plz send me the code for counting the number of spaces in a particular string.
like if we have the value as 100 00 0000. we must get the out put as 2 .try this...
DATA : str TYPE string VALUE '100 00 0000',
str_len TYPE i,
str_len2 TYPE i,
space_count TYPE i.
str_len = STRLEN( str ).
CONDENSE str NO-GAPS.
str_len2 = STRLEN( str ).
space_count = ( str_len - str_len2 ).
WRITE :/ space_count. -
Read Text file and count occurences of certain string
Hello,
I have a text file with lines of data in it. I would like to read this text file and count how many lines match a certain string of text.
For example my text file has this data in it.
dog,blue,big
dog,red,small
dog,blue,big
cat,blue,big
If the certain string of text is "dog,blue,big" then the count would return "2".
Thanks for your helpHello,
Thank you for your post.
I am afraid that the issue is out of support range of VS General Question forum which mainly discusses the usage issue of Visual Studio IDE such as
WPF & SL designer, Visual Studio Guidance Automation Toolkit, Developer Documentation and Help System
and Visual Studio Editor.
I am moving your question to the moderator forum ("Where is the forum for..?"). The owner of the forum will direct you to a right forum.
In addition, if you are working with Windows Forms app. please consult on Windows Forms Forum:http://social.msdn.microsoft.com/Forums/windows/en-US/home?category=windowsforms
If you are working with WPF app, please consult on WPF forum:
http://social.msdn.microsoft.com/Forums/vstudio/en-US/home?forum=wpf
If you are working with ASP.NET Web Application, I suggest that you can consult your issue on ASP.NET forum:
http://forums.asp.net/
for better solution and support.
Visual Studio Language Forums:
http://social.msdn.microsoft.com/Forums/vstudio/en-US/home?category=vslanguages
Best regards,
We are trying to better understand customer views on social support experience, so your participation in this interview project would be greatly appreciated if you have time. Thanks for helping make community forums a great place.
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HERE to participate the survey. -
Replacing the letters in a string
I have an odd set of fields for properties for my client.
Some have multiple
letters which correspond to a code table. How can I loop
through the string
and replace the letters with words
For example
A = central air
B = ceiling fan
C = A/C Unit
If the field looks like 'ABC', it would be outputted to the
page as 'central
air ceiling fan A/C Unit'
Wally Kolcz
MyNextPet.org
Founder / Developer
586.871.4126You don't say how many unique letters you have, but I presume
it is 26 or less. If that's the case, you could index an array that
has been initialized with the words. See concept code below. It
assumes only upper case letters. -
Extract specific letters from a string of text
Hello,
I have an idea that I would like to implement into Numbers but I'm not quite sure how to do it, or even if it is possible. What I would like to do is replace a multitude of checkboxes with one cell. In that one cell I will have a string of text, with a letter representing a condition. From this one cell, I would like to be able to extract a certain letter into its own cell and then hide the multitude of cells.
My idea is to eventually just be able to type the conditions into a single cell, extract the letter of the certain condition into another cell, and then extract that information into another table, all the while hiding all of the auxiliary columns in the original table. Because it would be much easier and prettier to simply type a bunch of letters into one box instead of having to check a bunch of boxes, no?
For example Left Header Row 1 would contain the letters BMRT. Column A would return a "FALSE" value, while Column B would return "B" or "TRUE." Columns C through L return "FALSE," etc.
In Left Header Row 2 we would have the string AJQTVZ, for instance. Column A would return "A" (or "TRUE"), and Columns B through I would return "FALSE."
Is it possible to write such a formula? Or is there an easier way to be thinking about this problem?Let's start with the easy part.
Cells on a Numbers table can contain data entered directly, or can contain a formula. They can't contain both. That means you cannot 'type the conditions into a single cell' in column A ("the left header" cell) AND have a formula which sets that cell to TRUE if the typed in data contains an "A".
There's no problem doing this using column A as the key holder and columns B:Z to hold the TRUE/FALSE results, staring in both cases on row 2.
Here's an example
The column header cells (row 1) contain the letter corresponding to that column.
The row header cells (starting at row 2) contain the 'bunch of letters' you describe. Note (A4) that the letters do not have to be entered in any particular order, and that extraneous characters (eg. a space) are ignored.
The formula shown is entered B2, and filled down and right from there.
=IFERROR(FIND(B$1,$A2)>0,FALSE)
FIND returns the position of the first occurrence of the target string (in this case, the single letter at the top of the column) in the search string, then compares that with the value zero. For any letter that is included in the search string, the find value will be at least 1, so the comparison will return TRUE. If the target letter is not found, FIND returns an error. IFERROR traps this and returns FALSE.
Since the target depends on the letter at the top of the column, all that's needed to extend the range of possible letters to the full alphabet is to enter an A in cell B1, then run through the alphabet A to Z, with Z in cell AA1.
Depending what you want to do with the TRUE or FALSE values in these 26 columns, it may be possible to skip the auxiliary column step and use a formula similar to the one above as the condition argument of an IF(condition,do-if-true,do-if-false) statement.
Regards,
Barry -
Counting specific characters in string
// method that counts and returns the number of times a specific character
// appears in the data string
public int numChar(char _charData)
for(int i = 0; i < dataString.length(); i++)
if(_charData == dataString.indexOf(_charData))
int count = 1;
int countIt = 0;
countIt = countIt + count;
return countIt;
int nothing = 0;
return nothing;
}And your question is....?
I have one: Where does dataString come from? You don't pass it in. Seems ridiculous to make it a class member.
This is wrong:
if(_charData == dataString.indexOf(_charData))Should be:
if(_charData == dataString.charAt(i))The logic inside the if is incorrect, too. You don't want to return after the first time you find a character. You want to loop all the way through and count all the instances.
You might have been trying to stop the loop once the indexOf returned -1, but you should be doing it on the remaining substring in that case.
The style is poor. If I were reviewing your code, I'd say that your "nothing = 0" doesn't add any value.
I'd write it like this:
public class CharCounter
public static void main(String [] args)
if (args.length > 1)
String str = args[0];
char ch = args[1].charAt(0);
System.out.println("str : " + str);
System.out.println("ch : " + ch);
System.out.println("count: " + getCharCount(ch, str));
public static int getCharCount(char ch, String str)
int charCount = 0;
if (str != null)
int numChars = str.length();
for (int i = 0; i < numChars; ++i)
if (ch == str.charAt(i))
++charCount;
return charCount;
}% -
Counting occurences of a string (in a cell)
This is a little like the topic "Counting occurences of a word," except I want to be able to count how many times a specified search string occurs in a cell, not count how many cells contain the string. I would also like a formula that gave the offset of the starting character of the search string into the cell, such that a zero signified that the search string did not occur.
Basically, I want a general set of tools to parse a cell's string data, enabling me to determine where in the cell's string data each occurrence of another string is located & to extract parts of the string data relative to where the search string occurs.
For example, if the cell contains the string, "1ab2cd2e" I want to be able to extract "cd" & "e" based on the characteristic that they follow "2."Here it is:
--(SCRIPT decoupeur.app]
Save the script as a Script, an Application or an Application Bundle named decoupeur
Put the file in the folder
<startup Volume>:Users:<yourAccount>:Library:Scripts:Applications:Numbers:
Maybe you would have to create the folder Numbers and even the folder Applications by yourself.
To use it,
Select the group of cells to convert
Copy to the clipboard
Select the destination cell
Goto Scripts > Numbers > decoupeur
You will be asked to define the separator to use.
Given the separator "2", the strings of the kind "1zer6sd2az2rt8aa2er7kkk5uu2a"
would be parsed and return az TAB rt TAB er TAB a
The converted values will be pasted at the cursor.
Yvan KOENIG 30juin 2008
property theApp : "Numbers"
property menuFenetre : 10
property premierNom : 6
property nomDuDocActif : ""
property theDelim : ""
property listeLignes : {}
property listeTemp : {}
property msg90 : ""
property msg91 : ""
property msg92 : ""
--=============
on run
tell application "System Events" to if not (UI elements enabled) then set (UI elements enabled) to true (*
Active le GUI scripting
• Enable GUI scripting *)
my nettoie()
my controleVersion()
set nomDuDocActif to my getFrontDoc()
if nomDuDocActif = "" then return
my prepareMessages()
try
set txtDatas to the clipboard as Unicode text
on error (*
The clipboard was empty *)
return
end try
set my listeLignes to every paragraph of txtDatas
set srcNbRows to count of my listeLignes
if txtDatas contains tab then (*
several columns *)
error "Can't treat several columns" number 8001
else (*
single column *)
tell application theApp
set choix to choose from list {"1", "2", "3", "4", "5", "6", "7", "8", "9", "0"} with prompt msg90 default items {"2"} OK button name msg91 cancel button name msg92
end tell
if choix is false then error -128
set theDelim to item 1 of choix
repeat with i from 1 to srcNbRows
if item i of my listeLignes is not "" then set item i of my listeLignes to my decipher(item i of my listeLignes)
end repeat
end if
set the clipboard to my recolle(listeLignes, return)
my pasteIt()
my nettoie()
end run
--=============
on pasteIt()
tell application theApp to activate
tell application "System Events" to tell (first process whose title is theApp)
click menu item nomDuDocActif of menu 1 of menu bar item menuFenetre of menu bar 1
keystroke "v" using {command down}
end tell
end pasteIt
--=============
on decoupeur(n)
local d, ms, m
set d to n div 1
set ms to (n - d) * 60
set m to round (ms)
return (d as text) & "°" & m & "’" & (round ((ms - m) * 60)) & "”"
end decoupeur
--=============
on decipher(source)
local i, itm, shortimem, j
set my listeTemp to my decoupe(source, theDelim)
set my listeTemp to items 2 thru -1 of my listeTemp
repeat with i from 1 to count of my listeTemp
set itm to item i of my listeTemp
set shortItem to ""
repeat with j from 1 to length of itm
if character j of itm is not in "1234567890" then
set shortItem to shortItem & character j of itm
else
exit repeat
end if
end repeat
set item i of my listeTemp to shortItem
end repeat
return my recolle(my listeTemp, tab)
end decipher
--=============
on getMinutesSeconds(t, d)
local ll
set ll to my decoupe(t, d)
return {(item 1 of ll) as integer, (item 2 of ll) as text}
end getMinutesSeconds
--=============
on decoupe(t, d)
local l
set AppleScript's text item delimiters to d
set l to text items of t
set AppleScript's text item delimiters to ""
return l
end decoupe
--=============
on recolle(l, d)
local t
set AppleScript's text item delimiters to d
set t to l as text
set AppleScript's text item delimiters to ""
return t
end recolle
--=============
on nettoie()
set AppleScript's text item delimiters to ""
set my listeLignes to {}
set my listeTemp to {}
set nomDuDocActif to ""
end nettoie
--=============
Get the name of the active document
on getFrontDoc()
local nw, mm
tell application theApp to activate
tell application "System Events" to tell (first process whose title is theApp)
set nw to name of every menu item of menu 1 of menu bar item menuFenetre of menu bar 1
if (count of nw) < premierNom then
set mm to ""
else
repeat with i from premierNom to count of nw
set mm to item i of nw
if (value of attribute "AXMenuItemMarkChar" of menu item mm of menu 1 of menu bar item menuFenetre of menu bar 1) is not in {"", "•"} then exit repeat
end repeat
end if -- (count of nw)…
end tell
return mm
end getFrontDoc
--=============
on controleVersion()
local v
try
set v to version of application theApp
set menuFenetre to 10 (* index of the Windows menu *)
set premierNom to 6 (* index of the first docName in the list of menu names
The list contains one more item than the displayed menu *)
on error (*
• We are here if Numbers ignores the instruction get version *)
tell application "System Events" to set v to get version of (get (application file of (get first process whose title is theApp)))
if v starts with "1" then
set menuFenetre to 10
set premierNom to 6
else (* ready for a Numbers v2 ignoring AppleScript *)
set menuFenetre to 10
set premierNom to 6
end if
end try
end controleVersion
--=============
on parleAnglais()
local z
try
tell application theApp to set z to localized string "Cancel"
on error
set z to "Cancel"
end try
return (z = "Cancel")
end parleAnglais
--=============
on prepareMessages()
if my parleAnglais() is false then
set msg90 to "Choisir le séparateur"
set msg91 to " OK "
set msg92 to "Annuler"
else
set msg90 to "Choose a separator"
set msg91 to " OK "
set msg92 to "Cancel"
end if
end prepareMessages
--=============
--[/SCRIPT]
Yvan KOENIG (from FRANCE lundi 30 juin 2008 11:16:52) -
Good Afternoon all,
Very new to Labview 2011 and taking it as a class in college and need some help. I have tried many different ways but none are working so I will post the question and if anyone can help me with picture view I would very much appreciate it.
Question:
Provide a string control called Name. Provide a string indicator called Greeting. Provide a numerical control (U16 representation) called Last Name Length. The VI should examine the data entered in Name, find the firt name only and produce a Greeting that says "Hello Joe!" The VI should also find the last name, and determine the number of letters in it., and write this value to Last Name Length. Provide an appropriate connector pane, and connect up all three front panel items so that the VI can be used as a subVI.
Please help!!!
The picture is how it is supposed to look.
Attachments:
labview.PNG 64 KBnewstudent wrote:
I have tried many different ways but none are working so I will post the question and if anyone can help me with picture view I would very much appreciate it.
Please show us all your failed attempts and explain in what way they did not work (unresponsive, wrong result, no result, clipped result, computer crashes, computer blows up). Maybe you can see a pattern yourself after doing some tutorials or by looking at the lecture notes you took during the course.
There are no programming environments called "picture view" or anything similar (PictureVIEW, etc.).
Since you posted in a LabVIEW forum and judging from your picture, I assume you mean LabVIEW. Can you confirm that?
LabVIEW Champion . Do more with less code and in less time . -
Hi,
Is it possible to add on one letter at a time to a string, to create a word, like when you go through a loop.
Loop 1: add a - String = a
Loop 2: add b - String = ab
Loop 3: add c - String = abc
ThanksString getit = "";
for(int x = 0;x<6; x++) {
getit += (x == 1) ? "h" : "" ;
getit += (x ==2) ? "e" :"" ;
getit += (x == 3) ? "l" :"" ;
getit += (x == 4) ? "l" :"" ;
getit += (x == 5) ? "o" :"" ;
System.out.println(getit);or
String getit = "";
for(int x = 0;x<6; x++) {
getit += (x == 1) ? "h" : "" ;
getit += (x ==2) ? "e" :"" ;
getit += (x == 3) ? "l" :"" ;
getit += (x == 4) ? "l" :"" ;
getit += (x == 5) ? "o" :"" ;
System.out.println(getit);
} -
Word counting program - new to strings
Im trying to write a program that will tell me how many times the word "rabbit" appears in a text file. This is what I ahve so far and I am not sure if it is correct. I get a cannot find symbol
symbol : method indexOf(java.lang.String) error when I compile.
import java.util.Scanner;
import java.io.File;
import java.lang.String;
public class Rabbitcount
public static void main( String[] args ) throws Exception // new: throws Exception, copy and paste for now
File f = new File("rabbit.txt");
Scanner input = new Scanner(f);
while (true) {
int x = indexOf("rabbit");
System.out.println("There are " + x + " occurences of rabbit in the text.");
// Close the file
input.close();
} Any help is appreciated
Edited by: euchresucks on Nov 3, 2008 2:23 PMindexOf() is a String method. If you were going to use that, you would have to load the file's contents into a String and call indexOf() on that String. You would also need to use the two-argument form of indexOf() so you could start each search at the point where the last match ended.
But you don't need to do any of that. You've already a perfectly good Scanner there; use it to do the searching. Hint: the name of method you need does not start with the word "next".
Maybe you are looking for
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