Why does my bash code change when I write it to a file?

Question

I have been trying to write this for loop to a file:

echo "for i in 1 2 3 4; do echo $i; done" >> test.sh

It should make a file that will print the numbers 1 to 4. But whenever I run this the output is 4 4 4 4 each appearing on a new line.

When I look at the sh file the do echo $i I entered has changed to do echo 4. The rest of the code is the same.

Is there some syntax that I'm missing?

Between double quotes, `$i` expands before `echo` is run. You need to use single quotes to prevent that. — oguz ismail, Mar 20 '21 at 14:33
@oguzismail Both this solution and John's work. Is there one that I should use over the other in the future? — CyberSkull311, Mar 20 '21 at 14:54
Depends. Like, if you have too many variable references in the string, wrapping it in single quotes is preferable. — oguz ismail, Mar 20 '21 at 14:58

John Kugelman · Accepted Answer · 2021-03-20T18:59:37.700

2

$i is being evaluated by the outer shell. Escape the dollar sign to ensure a literal dollar sign is echoed.

echo "for i in 1 2 3 4; do echo \$i; done" >> test.sh

Or use single quotes to prevent any expansion within the string.

echo 'for i in 1 2 3 4; do echo $i; done' >> test.sh

edited Mar 20 '21 at 18:59

answered Mar 20 '21 at 14:33

John Kugelman

1 Answers1