The colon command is a null command.
The : construct is also useful in the conditional setting of variables. For example,
: ${var:=value}
Without the :, the shell would try to evaluate $var as a command. <=???
I don't quite understand the last sentence in above statement. Can anyone give me some details?
Thank you
Try
var=badcommand
$var
you will get
bash: badcommand: command not found
Try
var=
${var:=badcommand}
and you will get the same.
The shell (e.g. bash) always tries to run the first word on each command line as a command, even after doing variable expansion.
The only exception to this is
var=value
which the shell treats specially.
The trick in the example you provide is that ${var:=value} works anywhere on a command line, e.g.
# set newvar to somevalue if it isn't already set
echo ${newvar:=somevalue}
# show that newvar has been set by the above command
echo $newvar
But we don't really even want to echo the value, so we want something better than
echo ${newvar:=somevalue}.
The : command lets us do the assignment without any other action.
I suppose what the man page writers meant was
: ${var:=value}
Can be used as a short cut instead of say
if [ -z "$var" ]; then
var=value
fi
${var} on its own executes the command stored in $var. Adding substitution parameters does not change this, so you use : to neutralize this.
Try this:
$ help :
:: :
Null command.
No effect; the command does nothing.
Exit Status:
Always succeeds.
Related
I've seen many examples where $USER and similar commands are used but I could never figure out what it meant.
Whenever I search $ on Google, it doesn't recognise the symbol.
it is used to access system variable.
for example:
if you type in linux and/or unix shell
...$ my_var="some value"
...$ echo my_var
will print "some value"
...$ echo $USER
will print the name of shell user
...$ echo $?
will print the result of the previous command when successfully will print 0
in other word the result of "exit (num)" of you shell.
"$" also can indicate you are logged as no-root user for some shells the root will be indicated "#"
The $ is a special character that tells the shell interpreter to interpret the contents following as a value for a variable. It is also called variable substitution.
In the case of commands or command output, it can be used to call a shell command and store it's output as a variable's value. For example:
VAR_1="$(ip link show)"
Calling the variable VAR_1 will print the output of the command ip link show.
This is called command substitution.
You can find out more information on special characters Here
as well as information on wildcards, keywords and more.
I have a Bash script which generates, stores and modifies values in an array. These values are later used as arguments for a command.
For a MCVE I thought of an arbitrary command bash -c 'echo 0="$0" ; echo 1="$1"' which explains my problem. I will call my command with two arguments -option1=withoutspace and -option2="with space". So it would look like this
> bash -c 'echo 0="$0" ; echo 1="$1"' -option1=withoutspace -option2="with space"
if the call to the command would be typed directly into the shell. It prints
0=-option1=withoutspace
1=-option2=with space
In my Bash script, the arguments are part of an array. However
#!/bin/bash
ARGUMENTS=()
ARGUMENTS+=('-option1=withoutspace')
ARGUMENTS+=('-option2="with space"')
bash -c 'echo 0="$0" ; echo 1="$1"' "${ARGUMENTS[#]}"
prints
0=-option1=withoutspace
1=-option2="with space"
which still shows the double quotes (because they are interpreted literally?). What works is
#!/bin/bash
ARGUMENTS=()
ARGUMENTS+=('-option1=withoutspace')
ARGUMENTS+=('-option2=with space')
bash -c 'echo 0="$0" ; echo 1="$1"' "${ARGUMENTS[#]}"
which prints again
0=-option1=withoutspace
1=-option2=with space
What do I have to change to make ARGUMENTS+=('-option2="with space"') work as well as ARGUMENTS+=('-option2=with space')?
(Maybe it's even entirely wrong to store arguments for a command in an array? I'm open for suggestions.)
Get rid of the single quotes. Write the options exactly as you would on the command line.
ARGUMENTS+=(-option1=withoutspace)
ARGUMENTS+=(-option2="with space")
Note that this is exactly equivalent to your second option:
ARGUMENTS+=('-option1=withoutspace')
ARGUMENTS+=('-option2=with space')
-option2="with space" and '-option2=with space' both evaluate to the same string. They're two ways of writing the same thing.
(Maybe it's even entirely wrong to store arguments for a command in an array? I'm open for suggestions.)
It's the exact right thing to do. Arrays are perfect for this. Using a flat string would be a mistake.
I am trying to do string substitution in bash, want to understand it better.
I crafted a success case like this:
a=abc_de_f
var=$a
echo ${var//_/-}
outout is abc-de-f. This works.
However, the following script fails:
a=abc_de_f
echo ${$a//_/-}
The error message is ${$a//_/-}: bad substitution.
It seems like related to how we can use a variable in substitution. Why this fails? How bash handles variables in this case?
Also, what is the best practice to handle escape characters in bash string substitution?
In the second case, you don't need the second $ as a is the string.
a=abc_de_f
echo ${a//_/-}
If you wanted to add a level of indirection, you can use ! before the variable as in
a=abc_de_f
b=a
echo ${b//_/-}
will output a, while
echo ${!b//_/-}
will output abc-de-f.
See here for a discussion on the art of escaping in BASH
I've just discovered a strange behaviour in bash that I don't understand. The expression
${variable:=default}
sets variable to the value default if it isn't already set. Consider the following examples:
#!/bin/bash
file ${foo:=$1}
echo "foo >$foo<"
file ${bar:=$1} | cat
echo "bar >$bar<"
The output is:
$ ./test myfile.txt
myfile.txt: ASCII text
foo >myfile.txt<
myfile.txt: ASCII text
bar ><
You will notice that the variable foo is assigned the value of $1 but the variable bar is not, even though the result of its defaulting is presented to the file command.
If you remove the innocuous pipe into cat from line 4 and re-run it, then it both foo and bar get set to the value of $1
Am I missing somehting here, or is this potentially a bash bug?
(GNU bash, version 4.3.30)
In second case file is a pipe member and runs as every pipe member in its own shell. When file with its subshell ends, $b with its new value from $1 no longer exists.
Workaround:
#!/bin/bash
file ${foo:=$1}
echo "foo >$foo<"
: "${bar:=$1}" # Parameter Expansion before subshell
file $bar | cat
echo "bar >$bar<"
It's not a bug. Parameter expansion happens when the command is evaluated, not parsed, but a command that is part of a pipeline is not evaluated until the new process has been started. Changing this, aside from likely breaking some existing code, would require extra level of expansion before evaluation occurs.
A hypothetical bash session:
> foo=5
> bar='$foo'
> echo "$bar"
$foo
# $bar expands to '$foo' before the subshell is created, but then `$foo` expands to 5
# during the "normal" round of parameter expansion.
> echo "$bar" | cat
5
To avoid that, bash would need some way of marking pieces of text that result from the new first round of pre-evaluation parameter expansion, so that they do not undergo a second
round of evaluation. This type of bookkeeping would quickly lead to unmaintainable code as more corner cases are found to be handled. Far simpler is to just accept that parameter expansions will be deferred until after the subshell starts.
The other alternative is to allow each component to run in the current shell, something that is allowed by the POSIX standard, but is not required, either. bash made the choice long ago to execute each component in a subshell, and reversing that would break too much existing code that relies on the current behavior. (bash 4.2 did introduce the lastpipe option, allowing the last component of a pipeline to execute in the current shell if explicitly enabled.)
status=0
$status=1
echo $status
Can anyone tell my what i am doing wrong with this?
It gives me the following error:
0=1: command not found
This line is OK - it assigns the value 0 to the variable status:
status=0
This is wrong:
$status=1
By putting a $ in front of the variable name you are dereferencing it, i.e. getting its value, which in this case is 0. In other words, bash is expanding what you wrote to:
0=1
Which makes no sense, hence the error.
If your intent is to reassign a new value 1 to the status variable, then just do it the same as the original assignment:
status=1
Bash assignments can't have a dollar in front. Variable replacements in bash are like macro expansions in C; they occur before any parsing. For example, this horrible thing works:
foof="[ -f"
if $foof .bashrc ] ; then echo "hey"; fi
Only use the $ when actually using the variable in bash. Omit it when assigning or re-assining.
e.g.
status=0
status2=1
status="$status2"
also this ugly thing works too :
status='a'
eval $status=1
echo $a
1