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="$(ip link show)"
Calling the variable VAR_1 will print the output of the command ip link show.
This is called command substitution.
You can find out more information on special characters Here
as well as information on wildcards, keywords and more.
Related
If I am in a Linux terminal and I start setting variables such as export AGE=45.
Then I have a script to read user data from terminal variables and process it, is this possible to do?
IE:
user#linux$ export AGE=45
user#linux$ ./age.sh
#script asks for input
read -p "what is your age?" scriptAGE
#user inputs variable set in terminal
$AGE
#echo output
echo "your age is: " $scriptAGE"
#should say your age is: 45
There is no such thing as a terminal variable. read just assigns a string to your variable scriptAGE.
If this string contains some $NAME you want to expand, you could apply eval to it, but this is of course extremely dangerous because of possible code injection.
A safer way to do this is using envsubst, but this requires that the variables to be substituted must be environment variables. In your case, AGE is in the environment, so this condition is met.
In your case, you would have to do therefore a
envsubst <<<"$scriptAGE"
which would print on stdout the content of scriptAGE with all environment variables in it substituted.
Variables are not expanded in input, only in the script itself.
You could use eval to force it to process the variable value as shell syntax.
eval "echo 'your age is:' $scriptAGE"
But this will also process other shell syntax. If they enter $AGE; rm * it will say their age is 45 and then delete all their files.
you could just do
age=$1
echo "Your age is $1"
where $1, $2, $3, .., $N are the passed arguments by order
And then run your script
bash script sh Noureldin
For more Info read this:
passing names args
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 read the value of a bash variable defined earlier , but this variable name derived dynamically.
this is the bash script i am trying to do
$ mythreshold=10
$ table=my
$ threshold="$table"threshold
$ echo $("$threshold")
mythreshold
but when i try to read this variable value like
$ echo $("$threshold")
-bash: mythreshold: command not found
however i was expecting it to print
$ echo $("$threshold")
10
is there a way i can get this work, it should have printed the value of mythreshold variable defined above
$() is Command Substitution. It runs the command inside and returns the output. A variable name is not a command.
You can $(echo "$threshold") but that will only get the mythreshold back.
You need indirection for what you want. Specifically Evaluating indirect/reference variables.
As an example, for this specific case:
echo "${!threshold}"
Use eval command :
eval echo \${$threshold}
More details about this command can be found here:
eval command in Bash and its typical uses
I am new to shell scripting just started off.
I have written this script
#!/bin/sh
profile_type= cat /www/data/profile.conf
echo $profile_type
the o/p of this script is
. /tmp/S_panicA1.txt
. /tmp/S_panicA0.txt
away_Def="panicA1 panicA0"
away_Byp=0
away_Sts=$((panicA1+panicA0-away_Byp))
In this i want to get panicA1 panicA0 and 0 and store it in other variable how to do this?
When you want to assign the output of a command to a variable, you use the dollar parenthesis syntax.
foo=$(cat /my/file)
You can also use the backticks syntax.
foo=`cat /my/file`
In your script, you simply run the command cat and assign its result, nothing, to your variable. Hence the output consisting of the content of your file, result of cat, followed by an empty line, result of echo with an empty variable.
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.