#!/bin/bash awk '$1 == "abc" {print}' file # print lines first column matching "abc"
How to print lines when the first column matching members of array(“12” or “34” or “56”)?
#!/bin/bash ARR=("12" "34" "56")
Add
Also, how to print lines when the first column exactly matching members of array(“12” or “34” or “56”)?
Advertisement
Answer
You could use bash
to interpolate the string to a regex pattern used in Awk
, by changing the IFS
value to a |
character and do array expansion as below:
ARR=("12" "34" "56") regex=$( IFS='|'; echo "${ARR[*]}" ) awk -v str="$regex" '$1 ~ str' file
The array expansion converts the list elements to a string delimited with |
, for e.g. 12|34|56
in this case.
The $()
runs in the sub-shell do that the value of IFS
is not reflcted in the parent shell. You could make it in one line as
awk -v str="$( IFS='|'; echo "${ARR[*]}" )" '$1 ~ str' file
OP had also asked for an exact match of the strings from the array in the file, in that case using grep
with its ERE support can do the job
regex=$( IFS='|'; echo "${ARR[*]}" ) egrep -w "$regex" file
(or)
grep -Ew "$regex" file