while with "lesser than" not working

[lyngeled]

Hi, I am trying an "while" with "lesser than" or "greater than" but this example I have found do not work, what's wrong?: #!/bin/bash num=1 while [$num -lt 5]; do num=$[$num + 1]; echo $num; done Regards Lars

[unknownPoster]

it's standard practice on this forum to surround your code with the code tags. Just for readability, I assume this is your program/script:



#!/bin/bash
num=1
while [$num -lt 5];
do num=$[$num + 1];
echo $num;
done

[r-senior]

Don't forget Bash is not as lenient with spacing as a typical programming language. The spaces inside the square brackets (and indeed the lack of spaces in the assignment) are significant.



$ cat test.sh
#!/bin/bash
n=1
while [ $n -le 5 ]; do
echo $((n++))
done

$ ./test.sh
1
2
3
4
5


What error message did you get?

[lyngeled]

Ok - that worked. Thanks!
I will continue on my script...

Lars

[psusi]

$[$num + 1] is nonsense, you want $(($num + 1)).

[MadCow108]

$[$num + 1] is nonsense, you want $(($num + 1)).

its not nonsense, its just less portable.
For bash it works fine but it will fail in dash.

[psusi]

its not nonsense, its just less portable.
For bash it works fine but it will fail in dash.

Weird. I can't find any reference to $[ in the bash manual, but it does seem to work.

[hakermania]

Please mark the thread as Solved.