Delay in a 'for' loop in a batch file

Question

I have this code, and I want to make a delay in the `for` loop to measure one time and then make a delay and continue after that. I tried: ``` timout / t 10/ nobreak ``` As it is shown in the code, but it didn't work. ``` : @echo off set Looping_number=10 or anything else FOR /L %%A IN (1,1,%Looping_number%) DO call :doit %%A goto :eof :doit set pad=00%1 set num=%pad:~-2% @set var1=var1.exe @set var2=C:\...\...\... .txt @set output=C:\....\output\%num% Mkdir %output% %var1% %var2% %Results% timeout / t 10 / nobreak goto :eof ``` But it says you can not delay? How can I fix this problem?

Accepted Answer

You have extra spaces in your command:

```
timeout / t 10 / nobreak
```

You need to say:

```
timeout /t 10 /nobreak
```

Delay in a 'for' loop in a batch file

Solution

Problem