redhat - Bash script to remove lines containing any of a list of words -

i have large config file use define variables script pull it, each defined on single line. looks this:

var   val foo   bar foo1  bar1 foo2  bar2 

i have gathered list of out of date variables want remove list. go through manually, script, @ least more stimulating. file contains vlaues may contain multiple instances. idea find value, , if it's found, remove entire line.

does know if possible? know sed not know how make use file input.

#!/bin/bash shopt -s extglob remove=(foo1 foo2) ifs='|' eval 'pattern="@(${remove[*]})"' while read -r line;     read b <<< "$line"     [[ $a != $pattern ]] && echo "$line" done < input_file.txt > output_file.txt 

or (use copy first)

#!/bin/bash shopt -s extglob file=$1 remove=("${@:2}") ifs='|' eval 'pattern="@(${remove[*]})"' save=() while read -r line;     read b <<< "$line"     [[ $a != $pattern ]] && save+=("$line") done < "$file" printf '%s\n' "${save[@]}" > "$file" 

running with

bash script.sh your_config_file pattern1 pattern2 ... 

or

#!/bin/bash shopt -s extglob file=$1 patterns_file=$2 readarray -t remove < "$patterns_file" ifs='|' eval 'pattern="@(${remove[*]})"' save=() while read -r line;     read b <<< "$line"     [[ $a != $pattern ]] && save+=("$line") done < "$file" printf '%s\n' "${save[@]}" > "$file" 

running with

bash script.sh your_config_file patterns_file