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Im tring to assign echo value which to a variable but im getting error
Var='(echo $2 | sed -e 's/,/: chararray /g'| sed -e 's/$/: chararray/')'
echo $var
Input : sh load.sh file 1,2,3,4
Error load.sh: line 1: chararray: command not found
885
asked Mar 12 '15 13:03
$ FILES=`sudo find . -type f -print | wc -l` $ echo "There are $FILES in the current working directory." That's it for now, in this article, we explained the methods of assigning the output of a shell command to a variable. You can add your thoughts to this post via the feedback section below.
Two commands, `echo` and `who` are used in this example as the nested command. Here, `who` command will execute first that print the user's information of the currently logged in user. The output of the `who` command will execute by `echo` command and the output of `echo` will store into the variable $var.
We can declare a variable as a variable name followed by assigning operator (=) and the value, which may be a character or string or a number or a special character. There should not be a space between the assignment operator and the variable name, and the corresponding value.
Var=$(echo "$2" | sed -e 's/,/: chararray /g' | sed -e 's/$/: chararray/')
echo "$Var"
OR
Var=`echo "$2" | sed -e 's/,/: chararray /g' | sed -e 's/$/: chararray/'`
echo "$Var"
Use either $(…) or perhaps `…` backtick notation. However, the backtick notation is deprecated and should be avoided. Also, check the comments by mmgross, Etan Reisner and svlasov to your question. They are all correct.
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