How to call another shell script from a shell script in Unix

Question

I am trying to call another shell script from inside a shell script in unix (bash)

1st Shell script(script_1.sh):

VARIABLE_1=<VALUE>
<unix_code> 
<unix_code>
..
..
..
export VARIABLE_1
bash -x script_2.sh
bash -x script_3.sh

my requirement here is to pass the value of VARIABLE_1 to script_2.sh & script_3.sh. I was able to execute script_2.sh pass the argument to script_2.sh and it was working fine. Problem has started when i included script_3.sh. It gives me below error, when i execute the first shell script.

/usr/bin/sh: /usr/bin/sh: cannot execute binary file

Not sure why this happening. Also, Is this recommended way to pass arguments and call other shell scripts ? Please let me know

Does script_3.sh run properly when executed outside of script_1.sh? — Zumo de Vidrio
– Zumo de Vidrio, Commented Apr 19, 2017 at 7:21
Check that you can run /usr/bin/sh from the command-line. It looks like script_3.sh makes a call to /usr/bin/sh somewhere. — cdarke
– cdarke, Commented Apr 19, 2017 at 8:58
Can you show the full error output, in particular the xtrace (-x) output immediately preceding the error? — cdarke
– cdarke, Commented Apr 19, 2017 at 9:13

kontiki · Accepted Answer · 2017-04-19 07:56:56Z

2

cannot execute binary file does not mean the file lacks execute permissions. In fact, running a script as "bash filename", does not need the executable permissions on the filename.

The error message is telling you that the script file is not an actual script file, but a binary file, probably an executable. If that is the case, do not do:

bash +x filename

just do:

filename

Additionally, to pass arguments to the script, you can do so like this:

scriptfile arg1 arg2 arg3

then refer inside your script as $1, $2, $3. Or $*to refer to all of them.

EDIT:

Also note that to execute a script you do not need to use bash scriptfile, you may just run scriptfile. In that case, scriptfile must have the executable permission (i.e., chmod +x scriptfile). Additionally, if the file is in the current directory, but the . is not part of the PATHvariable, the scriptfile will not be found. In that case you should execute your scriptfile as ./scriptfile

Your script would change to:

VARIABLE_1=<VALUE>
...
./script_2.sh "$VARIABLE_1"
./script_3.sh "$VARIABLE_1"

You no longer need to export the VARIABLE_1. The quotes are necessary to make sure the contents of VARIABLE_1 are passed as a single argument. script_2.sh and script_3.sh will see the value as $1. If the VARIABLE_1 has spaces in it, and you omit the quotes, each word will be pass as a separate value. That is, $1 will have the first word, $2the second, etc.

edited Apr 19, 2017 at 7:56

answered Apr 19, 2017 at 7:42

kontiki

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3 Comments

cdarke Over a year ago

I think the error message means that /usr/bin/sh cannot be executed. If it was the script name then it would say script_3.sh: script_3.sh cannot execute binary file

kontiki Over a year ago

I wonder what the script_3.sh contents are. Maybe the error message is coming from something executed inside script_3.sh, not from the actual call of script_3.sh

cdarke Over a year ago

that's my guess, but the OP has not answered my questions

sqlandplsql · Accepted Answer · 2017-04-19 07:27:54Z

-1

script_3.sh may not have permission to execute. Use chmod to provide appropriate permission and try again

answered Apr 19, 2017 at 7:27

sqlandplsql

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