wiki.bash-hackers.org/scripting/newbie_traps.markup

====== Beginner Mistakes ======

{{keywords>bash shell scripting pitfalls traps beginners}}

Here are some typical traps:

===== Script execution =====

==== Your perfect Bash script executes with syntax errors ====

If you write Bash scripts with Bash specific syntax and features, run them with __Bash__, and run them with Bash in __native mode__.

**Wrong**:
  * no shebang
    * the interpreter used depends on the OS implementation and current shell
    * **can** be run by calling bash with the script name as an argument, e.g. ''bash myscript''
  * ''#!/bin/sh'' shebang
    * depends on what ''/bin/sh'' actually is, for a Bash it means compatiblity mode, **not** native mode

See also:
  * [[scripting:bashbehaviour#sh_mode | Bash startup mode: SH mode]]
  * [[scripting:bashbehaviour#posix_run_mode | Bash run mode: POSIX mode]]

==== Your script named "test" doesn't execute ====

Give it another name. The executable ''test'' already exists.

In Bash it's a builtin. With other shells, it might be an executable file. Either way, it's bad name choice!

Workaround: You can call it using the pathname:
<code>
/home/user/bin/test
</code>

===== Globbing =====

==== Brace expansion is not globbing ====

The following command line is not related to globbing (filename expansion):
<code>
# YOU EXPECT
# -i1.vob -i2.vob -i3.vob ....

echo -i{*.vob,}

# YOU GET
# -i*.vob -i
</code>
**Why?** The brace expansion is simple text substitution. All possible text formed by the prefix, the postfix and the braces themselves are generated. In the example, these are only two: ''-i*.vob'' and ''-i''. The filename expansion happens **after** that, so there is a chance that ''-i*.vob'' is expanded to a filename - if you have files like ''-ihello.vob''. But it definitely doesn't do what you expected.

Please see:
  * [[syntax:expansion:brace]]


===== Test-command =====

  * ''if [ $foo ] ...''
  * ''if [-d $dir] ...''
  * ...
Please see:
  * [[commands:classictest#pitfalls_summarized|The classic test command - pitfalls]]

===== Variables =====

==== Setting variables ====

=== The Dollar-Sign ===

There is no ''$'' (dollar-sign) when you reference the **name** of a variable! Bash is not PHP!
<code>
# THIS IS WRONG!
$myvar=&quot;Hello world!&quot;
</code>

A variable name preceeded with a dollar-sign always means that the variable gets **expanded**. In the example above, it might expand to nothing (because it wasn't set), effectively resulting in...
<code>
=&quot;Hello world!&quot;
</code>
...which **definitely is wrong**!

When you need the **name** of a variable, you write **only the name**, for example
  * (as shown above) to set variables: ''picture=/usr/share/images/foo.png''
  * to name variables to be used by the ''read'' builtin command: ''read picture''
  * to name variables to be unset: ''unset picture''

When you need the **content** of a variable, you prefix its name with **a dollar-sign**, like
  * echo &quot;The used picture is: $picture&quot;

=== Whitespace ===

Putting spaces on either or both sides of the equal-sign (''='') when assigning a value to a variable **will** fail.

<code>
# INCORRECT 1
example = Hello

# INCORRECT 2
example= Hello

# INCORRECT 3
example =Hello
</code>

The only valid form is **no spaces between the variable name and assigned value**:
<code>
# CORRECT 1
example=Hello

# CORRECT 2
example=&quot; Hello&quot;
</code>

==== Expanding (using) variables ====

A typical beginner's trap is quoting.

As noted above, when you want to **expand** a variable i.e. &quot;get the content&quot;, the variable name needs to be prefixed with a dollar-sign. But, since Bash knows various ways to quote and does word-splitting, the result isn't always the same.

Let's define an example variable containing text with spaces:
<code>
example=&quot;Hello world&quot;
</code>

^Used form^result^number of words^
|''$example'' |''Hello world''|2|
|''&quot;$example&quot;'' |''Hello world''|1|
|''\$example'' |''$example''|1|
|''<nowiki>'$example'</nowiki>'' |''$example''|1|

If you use parameter expansion, you **must** use the **name** (''PATH'') of the referenced variables/parameters. i.e. **not** (''$PATH''):
<code>
# WRONG!
echo &quot;The first character of PATH is ${$PATH:0:1}&quot;

# CORRECT
echo &quot;The first character of PATH is ${PATH:0:1}&quot;
</code>

Note that if you are using variables in [[syntax:arith_expr | arithmetic expressions]], then the bare **name** is allowed:
<code>
((a=$a+7))         # Add 7 to a
((a = a + 7))      # Add 7 to a.  Identical to the previous command.
((a += 7))         # Add 7 to a.  Identical to the previous command.

a=$((a+7))         # POSIX-compatible version of previous code.
</code>


Please see:
  * [[syntax:words]]
  * [[syntax:quoting]]
  * [[syntax:expansion:wordsplit]]
  * [[syntax:pe]]

==== Exporting ====

Exporting a variable means giving **newly created** (child-)processes a copy of that variable.  It does **not** copy a variable created in a child process back to the parent process. The following example does **not** work, since the variable ''hello'' is set in a child process (the process you execute to start that script ''./script.sh''):

<code>
$ cat script.sh
export hello=world

$ ./script.sh
$ echo $hello
$
</code>

Exporting is one-way. The direction is from parent process to child process, not the reverse. The above example **will** work, when you don't execute the script, but include (&quot;source&quot;) it:
<code>
$ source ./script.sh
$ echo $hello
world
$
</code>
In this case, the export command is of no use.

Please see:
  * [[scripting:processtree]]

===== Exit codes =====
==== Reacting to exit codes ====

If you just want to react to an exit code, regardless of its specific value, you **don't need** to use ''$?'' in a test command like this:

<code bash>
grep ^root: /etc/passwd >/dev/null 2>&1

if [ $? -ne 0 ]; then
  echo &quot;root was not found - check the pub at the corner&quot;
fi
</code>

This can be simplified to:
<code bash>
if ! grep ^root: /etc/passwd >/dev/null 2>&1; then
  echo &quot;root was not found - check the pub at the corner&quot;
fi
</code>

Or, simpler yet:
<code bash>
grep ^root: /etc/passwd >/dev/null 2>&1 || echo &quot;root was not found - check the pub at the corner&quot;
</code>

If you need the specific value of ''$?'', there's no other choice. But if you need only a &quot;true/false&quot; exit indication, there's no need for ''$?''.

See also:
  * [[scripting:basics#exit_codes | Exit codes]]

==== Output vs. Return Value ====

It's important to remember the different ways to run a child command, and whether you want the output, the return value, or neither.

When you want to run a command (or a pipeline) and save (or print) the **output**, whether as a string or an array, you use Bash's ''$(command)'' syntax:
<code>
$(ls -l /tmp)
newvariable=$(printf &quot;foo&quot;)
</code>

When you want to use the **return value** of a command, just use the command, or add ( ) to run a command or pipeline in a subshell:
<code>
if grep someuser /etc/passwd ; then
    # do something
fi

if ( w | grep someuser | grep sqlplus ) ; then
    # someuser is logged in and running sqlplus
fi
</code>

Make sure you're using the form you intended:
<code>
# WRONG!
if $(grep ERROR /var/log/messages) ; then
    # send alerts
fi
</code>

Please see:
  * [[syntax:ccmd:intro]]
  * [[syntax:expansion:cmdsubst]]