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.
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="$(ip link show)"
Calling the variable VAR_1 will print the output of the command ip link show.
This is called command substitution.
You can find out more information on special characters Here
as well as information on wildcards, keywords and more.
My python script can take a series of bitwise operators as one of its arguments. They all work fine except for "=<<" which is roll left, and "=>>" which is roll right. I run my script like ./script.py -b +4,-4,=>>10,=<<1, where anything after -b can be any combination of similar operations. As soon as the terminal sees "<<" though, it just drops the cursor to a new line after the command and asks for more input instead of running the script. When it sees ">>", my script doesn't process the arguments correctly. I know it's because bash uses these characters for a specific purpose, but I'd like to get around it while still using "=>>" and "=<<" in my arguments for my script. Is there any way to do it without enclosing the argument in quotation marks?
Thank you for your help.
You should enclose the parameters that contain special symbols into single quotation marks (here, echo represents your script):
> echo '+4,-4,=>>10,=<<1'
+4,-4,=>>10,=<<1
Alternatively, save the parameters to a file (say, params.txt) and read them from the file onto the command line using the backticks:
> echo `cat params.txt`
+4,-4,=>>10,=<<1
Lastly, you can escape some offending symbols:
> echo +4,-4,=\>\>10,=\<\<1
+4,-4,=>>10,=<<1
I saw the line data=$(cat) in a bash script (just declaring an empty variable) and am mystified as to what that could possibly do.
I read the man pages, but it doesn't have an example or explanation of this. Does this capture stdin or something? Any documentation on this?
EDIT: Specifically how the heck does doing data=$(cat) allow for it to run this hook script?
#!/bin/bash
# Runs all executable pre-commit-* hooks and exits after,
# if any of them was not successful.
#
# Based on
# http://osdir.com/ml/git/2009-01/msg00308.html
data=$(cat)
exitcodes=()
hookname=`basename $0`
# Run each hook, passing through STDIN and storing the exit code.
# We don't want to bail at the first failure, as the user might
# then bypass the hooks without knowing about additional issues.
for hook in $GIT_DIR/hooks/$hookname-*; do
test -x "$hook" || continue
echo "$data" | "$hook"
exitcodes+=($?)
done
https://github.com/henrik/dotfiles/blob/master/git_template/hooks/pre-commit
cat will catenate its input to its output.
In the context of the variable capture you posted, the effect is to assign the statement's (or containing script's) standard input to the variable.
The command substitution $(command) will return the command's output; the assignment will assign the substituted string to the variable; and in the absence of a file name argument, cat will read and print standard input.
The Git hook script you found this in captures the commit data from standard input so that it can be repeatedly piped to each hook script separately. You only get one copy of standard input, so if you need it multiple times, you need to capture it somehow. (I would use a temporary file, and quote all file name variables properly; but keeping the data in a variable is certainly okay, especially if you only expect fairly small amounts of input.)
Doing:
t#t:~# temp=$(cat)
hello how
are you?
t#t:~# echo $temp
hello how are you?
(A single Controld on the line by itself following "are you?" terminates the input.)
As manual says
cat - concatenate files and print on the standard output
Also
cat Copy standard input to standard output.
here, cat will concatenate your STDIN into a single string and assign it to variable temp.
Say your bash script script.sh is:
#!/bin/bash
data=$(cat)
Then, the following commands will store the string STR in the variable data:
echo STR | bash script.sh
bash script.sh < <(echo STR)
bash script.sh <<< STR
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
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.