Why this code says echo is off?

Question

What is wrong with this code? It says ECHO is off.

@ECHO off set /p pattern=Enter id: findstr %pattern% .\a.txt > result if %errorlevel%==0 ( set var2= <result echo %var2% set var1=%var2:~5,3% echo %var1% > test.txt echo %var1% ) else ( echo error ) del result pause 

Any help is appreciated.

Answer 1

If your variable is empty somewhere, it will be the same as having the command "echo" on its own, which will just print the status of echo.

To avoid this, you should replace all your echo commands with something like this:

echo var2: %var2% 

That way, if %var2% is empty it will just print "echo var2:" instead of "echo off".

Answer 2

As Laurent stated, it's not a problem of the ECHO, it's a problem of your code.

In batch files, blocks are completely parsed before they are executed.
While parsing, all percent expansion will be done, so it seems that your variables can't be changed inside a block.

But for this exists the delayed expansion, the delayed expansion will be evaluated in the moment of execution not while parsing the block.

It must be enabled, as per default the delayed expansion is disabled.

@ECHO off setlocal EnableDelayedExpansion set /p pattern=Enter id: findstr %pattern% .\a.txt > result if %errorlevel%==0 (   set var2= <result   echo(!var2!   set var1=!var2:~5,3!   echo(!var1! > test.txt   echo(!var1! ) else (   echo error ) del result 

I used here the construct echo( instead of echo as this will ensure echoing an empty line even if the variable is empty.