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Beginner Mistakes

Here are some typical traps:

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.


  • 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:

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:


Brace expansion is not globbing

The following command line is not related to globbing (filename expansion):

# -i1.vob -i2.vob -i3.vob ....

echo -i{*.vob,}

# -i*.vob -i
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:

  • if [ $foo ] …
  • if [-d $dir] …

Please see:

Setting variables

The Dollar-Sign

There is no $ (dollar-sign) when you reference the name of a variable! Bash is not PHP!

$myvar="Hello world!"

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…

="Hello world!"
…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 "The used picture is: $picture"


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

example = Hello

example= Hello

example =Hello

The only valid form is no spaces between the variable name and assigned value:


example=" Hello"

Expanding (using) variables

A typical beginner's trap is quoting.

As noted above, when you want to expand a variable i.e. "get the content", 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:

example="Hello world"

Used formresultnumber of words
$example Hello world2
"$example" Hello world1
\$example $example1
'$example' $example1

If you use parameter expansion, you must use the name (PATH) of the referenced variables/parameters. i.e. not ($PATH):

echo "The first character of PATH is ${$PATH:0:1}"

echo "The first character of PATH is ${PATH:0:1}"

Note that if you are using variables in arithmetic expressions, then the bare name is allowed:

((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.

Please see:


Exporting a variable means to give newly created (child-)processes a copy of that variable. not copy a variable created in a child process 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 ./

$ cat
export hello=world

$ ./
$ echo $hello

Exporting is one-way. The direction is parent process to child process, not the reverse. The above example will work, when you don't execute the script, but include ("source") it:

$ source ./
$ echo $hello
In this case, the export command is of no use.

Please see:

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:

grep ^root: /etc/passwd >/dev/null 2>&1
if [ $? -neq 0 ]; then
  echo "root was not found - check the pub at the corner"

This can be simplified to:

if ! grep ^root: /etc/passwd >/dev/null 2>&1; then
  echo "root was not found - check the pub at the corner"

Or, simpler yet:

grep ^root: /etc/passwd >/dev/null 2>&1 || echo "root was not found - check the pub at the corner"

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

See also:

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:

$(ls -l /tmp)
newvariable=$(printf "foo")

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:

if grep someuser /etc/passwd ; then
    # do something

if ( w | grep someuser | grep sqlplus ) ; then
    # someuser is logged in and running sqlplus

Make sure you're using the form you intended:

if $(grep ERROR /var/log/messages) ; then
    # send alerts

Please see:

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U.Lickert, 2015/09/24 19:37

Reacting to exit codes

You may use the specific value, and do much more by enclosing a group of commands in { }. (and still a 1-liner not pushing the rest of code out of sight)

Note the ';' after the last command, it is necessary

grep ^root: /etc/passwd >/dev/null 2>&1 || { rc=$?; echo "That search returned a '$rc' to me. Maybe time for the pub."; return $rc; }
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  • scripting/newbie_traps.1462972780.txt
  • Last modified: 2016/05/11 13:19
  • by stephenwb