how to print lines with specific column matching members of array in bash

#!/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")

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Also, how to print lines when the first column exactly matching members of array(“12” or “34” or “56”)?

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