Ubuntu – bash script: different results when called with or without sudo

bashscriptssudo

In Ubuntu 16.04.3, I have a very simple bash script:

test.sh

[[ 0 == 0 ]] && result="true" || result="false"
echo $result
echo $USER $SHELL $0

When I call it as the non-root user me or as root, it works as expected. If I use sudo ./test.sh, it complains about a syntax error:

$ ./test.sh
true
me /bin/bash ./test.sh

$ sudo su
# ./test.sh 
true
root /bin/bash ./test.sh

# exit
$ sudo ./test.sh
./test.sh: 1: ./test.sh: [[: not found
false
root /bin/bash ./test.sh

What could be causing this? How can I fix it so that me can use this script both normally and with sudo?

Best Answer

Every script begins with a Shebang, without it the shell starting your script doesn't know which interpreter should run your script¹ and might – as in the case of sudo ./script.sh here – run it with sh, which in Ubuntu 16.04 is linked to dash. The conditional expression [[ is a bash compound command, so dash doesn't know how to handle it and throws the error you encountered.

The solution here is to add

#!/bin/bash

as the first line of your script. You may get the same result when you call it explicitly with sudo bash ./script.sh, but a shebang is the way to go.
To check which shell runs your script, add echo $0 to it. That's not the same as echo $SHELL, citing wiki.archlinux.org:

SHELL contains the path to the user's preferred shell. Note that this is not necessarily the shell that is currently running, although Bash sets this variable on startup.

1: As you started ./test.sh with bash it just assumed bash, the same goes for the sudo su subshell.

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Perhaps all you want to do is:

exec ./test.sh

This replaces your current shell process by the script, so that when it finishes there is nothing left to run.

Ubuntu – Call and run php script from shell script

The problem here is that you are quoting the entiore command you are trying to run as a single variable. As a result, you're not running php with foo.php as an argument but instead are attempting to execute a file called php foo.php. Here's a simpler example to show you what I mean:

$ var1="echo "
$ var2="foo"
$ set -x ## debugging info
$ "$var1$var2"
+ 'echo foo'      ### The shell tries to execute both vars as a single command
bash: echo foo: command not found

$ "$var1" "$var2"
+ 'echo ' foo     ### The shell tries to execute 'echo ' (echo with a space)
bash: echo foo: command not found

So, the right way is to remove the space and quote each variable separately:

$ var1="echo"
$ var2="foo"
$ "$var1" "$var2"

If you do that though, you'll hit the next error. The . is the source command. That tries to read a shell script and execute it in the current session. You are giving it a php script instead. That won't work, you need to execute it, not source it.

Finally, always avoid using CAPITAL variable names. The shell's reserved variables are capitalized so it's a good idea to always use lower case variable names for your scripts.

Putting all this together (with a few other minor improvements), what you want is something like:

#!/bin/sh

list="/path/to/my/site/dir"
config="/usr/bin/php"

for i in "$list"
do
    "$config" "$i"/test.php
done

Best Answer

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