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Removing Spaces from a string

Posted on 2003-10-29
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Last Modified: 2010-04-17
Easy, so I have a string, this string could have any number of spaces at its end, and then I have another string, which could have any number of spaces at the beginning.  My question is how can I remove the unwanted spaces so that I have just the data.  Here is an example of what I need...  The number of spaces that are at the end and beginning of the string varies and are not constant...  Thanks!

--------  Example  --------

(String 1)
Before:
String = "John Doe                "

After
String = "John Doe"

(String 2)
Before:
String = "     100"

After:
String = "100"
------- Break -------
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Question by:gideonn
7 Comments
 
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Expert Comment

by:bobbit31
ID: 9642760
what language?

e.g.,

vb: Trim("John Doe        ");

java:
String str = "    100";
str = str.trim();

c++: depends, are you using AnsiString, string, CString, char[] ???

   
0
 
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Author Comment

by:gideonn
ID: 9642782
Ops My bad I thought I was in the VB forum..  VB Thanks Bobbit!
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Accepted Solution

by:
MattWare earned 250 total points
ID: 9643086
strSpacelessString = RTrim(strString1)+LTrim(strString2)
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LVL 18

Expert Comment

by:bobbit31
ID: 9643287
lol... why didn't i get the points?

strSpacelessString = Trim(str1)

will remove leading and trailing spaces (no need for RTrim, LTrim)

In addition,

In VB the string concat operator is & not +

strSpacelessString = RTrim(strString1) & LTrim(strString2)
0
 
LVL 1

Author Comment

by:gideonn
ID: 9643428
Sorry Bobbit I just looked at your post again and the first time I looked at it I didn't see the VB part, I just saw Java and disregarded it, I am sorry please see my next post for your points.
0
 

Expert Comment

by:vkgonline
ID: 9656900
It's easy.
use Trim command to remove extra spaces from the string.
Syntax:

newStr=trim(oldStr)

Vinod
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Expert Comment

by:danessin
ID: 9664956
re: In VB the string concat operator is & not +

Either will concatenate strings. The difference is that & is early-bound (i.e. the compiler knows that the reference is a string). + is late bound because the target might be a number, in which case math is required or a string, in which case a concatenation is required. Therefore & is considerably faster than +. Otherwise either will work.

btw - Neither of these solutions will remove multiple spaces between words within the string - in case that's an issue
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