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Home/ Questions/Q 9082595
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Editorial Team
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Editorial Team
Asked: June 16, 20262026-06-16T20:35:48+00:00 2026-06-16T20:35:48+00:00

Trying to pass a variable into awk from user input: Have tried variations of

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Trying to pass a variable into awk from user input:

Have tried variations of awk -v with errors stating ‘awk: invalid -v option’, even though the option is listed in man files.

#! /bin/bash
read -p "Enter ClassID:" CLASS
read -p "Enter FacultyName:" FACULTY


awk '/FacultyName/ {print}' data-new.csv > $FACULTY.csv
awk -vclass=${CLASS} '/class/ {print}' data-new.csv >> $FACULTY.csv 


echo Class is $CLASS
echo Faculty member is $FACULTY
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  1. Editorial Team
    Editorial Team
    2026-06-16T20:35:49+00:00Added an answer on June 16, 2026 at 8:35 pm

    Some versions of awk require a space between the -v and the variable assignment. Also, you should put the bash variable in double-quotes to prevent unwanted parsing by the shell (e.g. word splitting, wildcard expansion) before it’s passed to awk. Finally, in awk /.../ is a constant regular expression (i.e. /class/ will search for the string “class”, not the value of the variable “class”). With all of this corrected, here’s the awk command that I think will do what you want:

    awk -v class="${CLASS}" '$0 ~ class {print}' data-new.csv >> $FACULTY.csv
    

    Now, is there any reason you’re using this instead of:

    grep "$CLASS" data-new.csv >> $FACULTY.csv
    
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