# 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]{.underline}, and run them with Bash in [native
mode]{.underline}.

**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:

-   [Bash startup mode: SH mode](/scripting/bashbehaviour#sh_mode)
-   [Bash run mode: POSIX mode](/scripting/bashbehaviour#posix_run_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:

    /home/user/bin/test

## Globbing

### Brace expansion is not globbing

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

    # YOU EXPECT
    # -i1.vob -i2.vob -i3.vob ....

    echo -i{*.vob,}

    # YOU GET
    # -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:

-   [brace](/syntax/expansion/brace)

## Test-command

-   `if [ $foo ] ...`
-   `if [-d $dir] ...`
-   \...

Please see:

-   [The classic test command -
    pitfalls](/commands/classictest#pitfalls_summarized)

## Variables

### Setting variables

#### The Dollar-Sign

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

    # THIS IS WRONG!
    $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\"

#### Whitespace

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

    # INCORRECT 1
    example = Hello

    # INCORRECT 2
    example= Hello

    # INCORRECT 3
    example =Hello

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

    # CORRECT 1
    example=Hello

    # CORRECT 2
    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 form      result          number of words
  -------------- --------------- -----------------
  `$example`     `Hello world`   2
  `"$example"`   `Hello world`   1
  `\$example`    `$example`      1
  `'$example'`   `$example`      1

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

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

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

Note that if you are using variables in [arithmetic
expressions](/syntax/arith_expr), 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:

-   [words](/syntax/words)
-   [quoting](/syntax/quoting)
-   [wordsplit](/syntax/expansion/wordsplit)
-   [pe](/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`):

    $ cat script.sh
    export hello=world

    $ ./script.sh
    $ echo $hello
    $

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 (\"source\") it:

    $ source ./script.sh
    $ echo $hello
    world
    $

In this case, the export command is of no use.

Please see:

-   [processtree](/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:

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

if [ $? -ne 0 ]; then
  echo "root was not found - check the pub at the corner"
fi
```

This can be simplified to:

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

Or, simpler yet:

``` bash
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:

-   [Exit codes](/scripting/basics#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:

    $(ls -l /tmp)
    newvariable=$(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
    fi

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

Make sure you\'re using the form you intended:

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

Please see:

-   [intro](/syntax/ccmd/intro)
-   [cmdsubst](/syntax/expansion/cmdsubst)
-   [grouping_subshell](/syntax/ccmd/grouping_subshell)