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Difference between sh and Bash
(11 answers)
Closed 5 years ago.
I have a file to be sourced in Centos 7.
It just works fine if I do :
$ source set_puregev_env
however, if I put this in a shell script, it doesn't work..
$ sh xRUN
xRUN: line 3: source: set_puregev_env: file not found
this is my shell script : xRUN
#!/bin/bash
source set_puregev_env
can anyone tell me what I might be doing wrong, or missing?
source is a command implemented in bash, but not in sh.
There are multiple ways to fix your script. Choose either one.
Run the script using bash interpreter
When you are invoking the xRUN script - you are explicitly telling it to be interpreted by sh
$ sh xRUN
To change and interpret the script with bash instead do
$ bash xRUN
This will make bash interpret the source command, and your script will work.
Use dot command to make script bourne compatible
You can also change the source with a dot command which does the same thing but is supported in both bourne and bash.
Change the line:
source set_puregev_env
With:
. set_puregev_env
Now the script will work with either sh or bash.
Make script executable
You should also run the script directly to avoid confusions like these by making it executable chmod +x xRUN, and invoking it like this:
$ ./xRUN
It will then use the command specified in the shebang and use the rest of the script as input. In your case it will use bash - since that is specified in the shebang.
Related
I want to run this script:
#!/bin/bash
echo <(true)
I run it as:
sh file.sh
And I get "Syntax error: "(" unexpected" . I found some similar situations but still can't solve this.
I'm a beginner at shell scripting , but as I understand:
the shebang I use is correct and chooses the bash shell , so the process substitution syntax should work
I try the same from the command line and it works. I checked with echo $0 and it gives me "bash" , so what's the difference from running the command in the command line and from a script that invokes the same shell?
Maybe it's something simple, but I couldn't find an explanation or solution.
You should run your script with bash, i.e. either bash ./script.sh or making use of the shebang by ./script.sh after setting it to executable. Only running it with sh ./script.sh do I get your error, as commented by Cyrus.
See also: role of shebang at unix.SE
Remove export POSIXLY_CORRECT=1 from your ~/.bashrc or ~/.profile (etc.) files.
The issue is that process substitution is an added bash feature that is not part of the posix standards.
sh file.sh
errorsh: 3: Syntax error: "(" unexpected
solution:
bash file.sh
I am trying to run a single command using bash in a sh script. There is no way to use bash for the script, I have to use sh. However, I need to run a bash-only command in sh.
Basically, I want something like the following:
bash --command_in "echo foobar"
Is this possible? I don't want to make a second script file just to run that one command in bash (like bash my_script.bash).
Derp, it's the -c flag. This wasn't easy to Google, and the --help is prety brief.
I want to run this script:
#!/bin/bash
echo <(true)
I run it as:
sh file.sh
And I get "Syntax error: "(" unexpected" . I found some similar situations but still can't solve this.
I'm a beginner at shell scripting , but as I understand:
the shebang I use is correct and chooses the bash shell , so the process substitution syntax should work
I try the same from the command line and it works. I checked with echo $0 and it gives me "bash" , so what's the difference from running the command in the command line and from a script that invokes the same shell?
Maybe it's something simple, but I couldn't find an explanation or solution.
You should run your script with bash, i.e. either bash ./script.sh or making use of the shebang by ./script.sh after setting it to executable. Only running it with sh ./script.sh do I get your error, as commented by Cyrus.
See also: role of shebang at unix.SE
Remove export POSIXLY_CORRECT=1 from your ~/.bashrc or ~/.profile (etc.) files.
The issue is that process substitution is an added bash feature that is not part of the posix standards.
sh file.sh
errorsh: 3: Syntax error: "(" unexpected
solution:
bash file.sh
I can't get the following to work:
c:\cygwin64\bin\bash -c /cygdrive/c/myscript.sh myargument
Specifically, myscript.sh is NOT getting "myargument" passed in as $1.
The script works exactly as I want it to if I hard-code "myargument" inside the script.
It's been 5 years since I have done any shell scripting and rust has accumulated!
The option -c is not used with scripts
Try this instead:
c:\cygwin64\bin\bash /cygdrive/c/myscript.sh myargument
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 $HOME/folder/{folder1,folder2}
and I execute sh ./sample.sh, the first folder will be created then in that a single {folder1,folder2} directory is created. The separate child folders are not created.
My query is
How the script file works when we compared to as terminal command? i.e., why is it not the same?
How to make it work?
bash behaves differently when invoked as sh, to more closely mimic the POSIX standard. One of the things that changes is that brace expansion (which is absent from POSIX) is no longer recognized. You have several options:
Run your script using bash ./sample.sh. This ignores the hashbang and explicitly uses bash to run the script.
Change the hashbang to read #!/bin/bash, which allows you to run the script by itself (assuming you set its execute bit with chmod +x sample.sh).
Note that running it as sh ./sample.sh would still fail, since the hashbang is only used when running the file itself as the executable.
Don't use brace expansion in your script. You could still use as a longer method for avoiding duplicate code:
for d in folder1 folder2; do
mkdir -p "$HOME/folder/$d"
done
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.
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 !.