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

Question

#!/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")?

Answer 1

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 eg 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

Answer 2

awk单线

awk -v var="${ARR[*]}" 'BEGIN{split(var,array," "); for(i in array) a[array[i]] } ($1 in a){print $0}' file

Answer 3

The following code does the trick:

awk 'BEGIN{myarray [0]="aaa";myarray [1]="bbb"; test=0 }{

test=0;
for ( x in myarray ) {
    if($1 == myarray[x]){
    test=1;
    break;
    }

}
if(test==0) print}'

If you need to pass a variable to awk use the -v option, however for array it is a bit tricker but the following syntax should work.

A=( $( ls -1p ) ) #example of list to be passed to awk (to be adapted to your needs)

awk -v var="$A" 'BEGIN{split(var,list,"\n")}END{ for (i in list) print i}'

Answer 4

与印第安人相同

ARR=("34" "56" "12");regex=" ${ARR[*]} ";regex="'^${regex// /\\|^}'";grep -w $regex infile