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I am trying to echo a loop function to another file and I need to echo two percent signs to it

What I have:

echo for /l %%x in (1, 1, 100) do echo %%x > main.bat
echo pause >> main.bat

Output:

for /l %x in (1, 1, 100) do echo %x 
pause
Kep13r
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    Try `%%%%` please. – Yunnosch Jul 13 '21 at 06:31
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    Does this answer your question? [How do I escape %% in a batch file](https://stackoverflow.com/questions/44977892/how-do-i-escape-in-a-batch-file) – Yunnosch Jul 13 '21 at 09:41
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    Sorry, just found a decent duplicate, after getting lost in not-quite answering ones.... That is why I answered AND THEN close-voted for dupe. – Yunnosch Jul 13 '21 at 09:42
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    duplicates: [Escape percent in bat file](https://stackoverflow.com/q/13551829/995714), [Ignore percent sign in batch file](https://stackoverflow.com/q/1907057/995714) – phuclv Jul 13 '21 at 09:56

1 Answers1

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You found out that, because of escaping mechanisms, echo %% gets you one %.
So if you need %% then do echo %%%%.

Yunnosch
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