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Beginner Mistakes
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/shshebang- depends on what
/bin/shactually is, for a Bash it means compatiblity mode, not native mode
- depends on what
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:
/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:
Test-command
if [ $foo ] ...if [-d $dir] ...- ...
Please see:
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
readbuiltin 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 ${PATH:0:1}"
Note that if you are using variables in arithmetic expressions, then the bare name is allowed:
((a=$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
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:
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:
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:
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:
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
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: