How to apply %% to a variable in a script with input parameter?

Question

I asked a question yesterday about how to manage to get %% around a variable without getting the evaluation. Well, the thing is, it does not work that way in my case...

I have a .bat file which gets an input parameter. Later on I want to use the value of this input parameter and put %...% around, like:

call script.bat testValue

script.bat:

 set inputPar=%1
 set newValue=%%inputPar%%

Now I get:

 echo %inputPar%
testValue
 echo %newValue%
%inputPar%

But I would like to get:

echo %newValue%
 %testValue%

Is that somehow possible?

paxdiablo · Accepted Answer · 2010-05-21 06:52:48Z

2

The reason you're getting those results is because %%inputPar%% is being treated as %%inputPar%% and %% becomes a literal %.

You need:

first.cmd:
    @echo off
    call script.cmd testValue
script.cmd:
    @echo off
    set inputPar=%1
    set newValue=%%%inputPar%%%
    echo %inputPar% should be testValue
    echo %newValue% should be %%testValue%%

This gives you %%%inputPar%%% and, again, %% becomes a literal % with %inputPar% becoming testValue.

answered May 21, 2010 at 6:52

paxdiablo

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murxx · Accepted Answer · 2010-05-21 06:47:26Z

0

I found the solution:

set inputPar=%1

set inputParDollar=%%%inputPar%%%

:: installDir is an already set environment variable

set installDirDollar=%%%installDir%%%

echo inputPar: %inputPar%

echo inputParDollar: %inputParDollar%

echo installDirDollar: %installDirDollar%

call script.bat testValue

inputPar: testValue

inputParDollar: %testValue%

installDirDollar: %D:\testenv%

answered May 21, 2010 at 6:47

murxx

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