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I have a CMD Command and it only works after the third time of execution. I want to open a command which gives me a path after that I want to convert the path and in the end I want to show the input of the file.

Here is my command:

@echo off & for /f %A in ('getFilePath') do set string="C:%A" & set newstring=%string:/=\% & type %newstring% & @echo on
Marged
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Gonzo Gonzales
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    What is the `getFilePath` command? Would you consider reformatting this as a .bat script until it is working, then convert it into a one-liner? Please edit the original post. – lit Feb 03 '16 at 14:26

2 Answers2

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Finally i found a solution i am sure it’s not the best way but it works for me.

@echo off & for /f %A in ('getFilePath') do ( set script=%A & call type C:%script:/=\\%) & @ECHO ON
Gonzo Gonzales
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@echo off & for /f %A in ('getFilePath') do set string="C:%A" & set newstring=%string:/=\% & type %newstring% & @echo on

would need delayed expansion (also your answer wouldn't work without delayed expansion), but astonishingly type "c:/path/file.ext" works (type c:/path/file.ext doesn't), so you can just do:

@echo off & @for /f %A in ('getFilePath') do @type "C:%A"  & @echo on

(assuming getFilePath gives you something like /path/file.ext)

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Stephan
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