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Search for a string in file and get the next value in it

Posted on 2011-02-22
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Last Modified: 2012-05-11
Example text file contains the below values. I need a batch file to get the value 'sm10' when given 203 as input.
The below master file is one time prepared and values can be seperated other than comma if required.

203,sm10
304,sm20
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Question by:sk7815
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Expert Comment

by:AriMc
ID: 34956178
Is that really for MS-DOS like specified in your question zone, or a command prompt of some Windows version?



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Accepted Solution

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sjklein42 earned 2000 total points
ID: 34956226
Assuming your data is in file "foo.dat" and this command line is in batch file "foo.bat"

@FOR /F "tokens=1,2 delims=," %%i IN (foo.dat) DO @IF "%%i" == "203" echo %%j

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c:\temp>foo.bat
sm10

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You didn't say what you wanted to do with the "sm10" once you got it, but substitute any command for the "echo".

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Author Closing Comment

by:sk7815
ID: 34956325
Excellent!  Thank you so much!  Actually I wanted to get the email ID of the person searching for his ID in a file and this meets the need!
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Expert Comment

by:Steve Knight
ID: 34956540
I know this is answered and working fine, but if you use a FIND command it will be quicker than a for loop if there are lots of entries as FIND will bring in only the one correct line:

@echo off
for /f "tokens=2 delims=," %%a in ('find "203" ^<yourfile.txt') do set email=%%a
echo The address is %email%
Steve
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Expert Comment

by:sjklein42
ID: 34956579
nice trick
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Author Comment

by:sk7815
ID: 34956640
thanks Dragon.  I use %1 or the input from the user and getting FIND parameter is invalid.  

@echo off
for /f "tokens=2 delims=," %%a in ('find %1 ^<foo.dat') do set email=%%a
echo The address is %email%

Output:
C:\junk>foo 203
FIND: Parameter format not correct
The address is
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Expert Comment

by:Steve Knight
ID: 34956661
You probably just need to put " " around it, try "%~1" - the ~ removes any existing " " first.

Steve
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Expert Comment

by:Steve Knight
ID: 34956666
You can get more complicated with findstr too, e.g. /b to say beginning a line.
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Author Comment

by:sk7815
ID: 34956886
Thank you, it worked with " " .
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