Understanding script language

Question

I'm a newbie to scripting languages trying to learn bash programming.

I have very basic question. Suppose I want to create three folders like $HOME/folder/ with two child folders folder1 and folder2.

If I execute command in shell like
```
mkdir -p $HOME/folder/{folder1,folder2}
```
folder will be created along with child folder.
If the same thing is executed through script I'm not able get expected result. If sample.sh contains
```
#!/bin/sh

mkdir -p $HOME/folder/{folder1,folder2}
```
and I execute sh ./sample.sh, the first folder will be created then in that a single {folder1,folder2} directory is created. The separate child folders are not created.

My query is

How the script file works when we compared to as terminal command? i.e., why is it not the same?
How to make it work?

before voting down please pass the comment why you voting down — vinay hunachyal, Oct 03 '13 at 11:25
http://stackoverflow.com/questions/19112031/difference-between-korn-and-bash-shell/19112167#19112167 — Srini V, Oct 03 '13 at 12:34

score 2 · Answer 1 · answered Oct 03 '13 at 12:05

bash behaves differently when invoked as sh, to more closely mimic the POSIX standard. One of the things that changes is that brace expansion (which is absent from POSIX) is no longer recognized. You have several options:

Run your script using bash ./sample.sh. This ignores the hashbang and explicitly uses bash to run the script.
Change the hashbang to read #!/bin/bash, which allows you to run the script by itself (assuming you set its execute bit with chmod +x sample.sh).

Note that running it as sh ./sample.sh would still fail, since the hashbang is only used when running the file itself as the executable.
Don't use brace expansion in your script. You could still use as a longer method for avoiding duplicate code:
```
for d in folder1 folder2; do
    mkdir -p "$HOME/folder/$d"
done
```

score 1 · Accepted Answer · answered Oct 03 '13 at 11:19

1

Brace expansion doesn't happen in sh.

In sh:

$ echo {1,2}

produces

{1,2}

In bash:

$ echo {1,2}

produces

1 2

Execute your script using bash instead of using sh and you should see expected results.

answered Oct 03 '13 at 11:19

devnull

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It depends on what `sh` is. Sometimes it actually will be Bash. :) – John Zwinck Oct 03 '13 at 11:20
@vinayhunachyal note that it is `#!/bin/bash` <--- there is a `!`. – fedorqui Oct 03 '13 at 11:26
Sorry while commenting i missed but in my code i have included `!`still im getting same output – vinay hunachyal Oct 03 '13 at 11:27
Is `/bin/bash` the path of your `bash`? Check it with `which bash`. – fedorqui Oct 03 '13 at 11:28
1

@vinayhunachyal If you're executing your script by saying `sh ./sample.sh`, it implies that you're using `sh`. Execute it by saying `bash ./sample.sh` or even `./sample.sh` **if** you're running `bash`. – devnull Oct 03 '13 at 11:29
Ya i checked its in `/bin/bash` and which bash `/bin/bash` – vinay hunachyal Oct 03 '13 at 11:30
@devnull Thank you its worked .. How to make it work with sh since if i have only sh – vinay hunachyal Oct 03 '13 at 11:31
1

@vinayhunachyal As mentioned above, *brace expansion* wouldn't happen in `sh`. – devnull Oct 03 '13 at 11:32
@devnull then its not possible to create child folder using sh shell? – vinay hunachyal Oct 03 '13 at 11:34
@vinayhunachyal you can do `mkdir -p foo/bar/baz` in sh, but not `mkdir -p foo/{bar,baz}`, i.e. the latter wouldn't *expand* to subdirectories `bar` and `baz`. – devnull Oct 03 '13 at 11:36
@devnull but it wont create two subdirectoy – vinay hunachyal Oct 03 '13 at 11:37
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/38542/discussion-between-devnull-and-vinay-hunachyal) – devnull Oct 03 '13 at 11:46

score 1 · Answer 3 · answered Oct 03 '13 at 11:19

1

This is probably happening because while your tags indicate you think you are using Bash, you may not be. This is because of the very first line:

#/bin/sh

That says "use the system default shell." That may not be bash. Try this instead:

#!/usr/bin/env bash

Oh, and note that you were missing the ! after #. I'm not sure if that's just a copy-paste error here, but you need the !.

answered Oct 03 '13 at 11:19

John Zwinck

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while writing here i missed but its there in my script file ! – vinay hunachyal Oct 03 '13 at 11:26
I didn't downvote, but changing the hashbang will have no effect when the script is run using `sh ./sample.sh`. – chepner Oct 03 '13 at 11:59

Understanding script language

3 Answers3