CODEWITHSUNDEEP
windowsbatch-fileif-statementcmd
Dusan
Dusan

Reputation: 831

string comparison in batch file not working

I have this code:

SETLOCAL enabledelayedexpansion enableextensions

SET OutofService=Weblog_Doc(file:///C:/Bla/BlaKon.htm?send=2000)


REM - Declare and set the Array

FOR /f "delims=" %%a IN (C:\Folder1\log.dat) DO (
    SET /a c+=1
    SET x[!c!]=%%a          
)   

REM - Read Array from end to start
FOR /l %%I IN (!c! -1 1) DO (                   

    SET _result=!x[%%I]:~-47%!      

    ECHO !_result!
    ECHO !OutofService!

    IF "!_result!"=="!OutofService!" (              
        ECHO yahooo!                
    )           
)

The string in file log.dat is as follows:

09/17/15 15:18:52:577 Container:
Weblog_Doc(file:///C:/Bla/BlaKon.htm?send=2000)

My ECHO statement is outputting the result:

Weblog_Doc(file:///C:/Bla/BlaKon.htm?send=2000)
Weblog_Doc(file:///C:/Bla/BlaKon.htm?send=2000)

The if statement is returning false and not outputting yahooo! as it should, given the fact that these two strings should be equal. What am I doing wrong?

Upvotes: 1

Views: 562

Answers (1)

MC ND
MC ND

Reputation: 70961

SET x[!c!]=%%a          
              ^^^^^^^^^^ Spaces included in value
....
SET _result=!x[%%I]:~-47%!      
                          ^^^^^^ Spaces included in value

Try with (percent sign in the third line is not needed)

SET "OutofService=Weblog_Doc(file:///C:/Bla/BlaKon.htm?send=2000)"
....
SET "x[!c!]=%%a"
....
SET "_result=!x[%%I]:~-47!"

This way, while the assignment operation is quoted, the quotes are not included in the value, nor the ending spaces if present.

Upvotes: 1

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