Awk column with pattern array - awk

Question

Is it possible to do this but use an actual array of strings where it says "array"
array=(cat
dog
mouse
fish
...)
awk -F "," '{ if ( $5!="array" ) { print $0; } }' file
I would like to use spaces in some of the strings in my array.
I would also like to be able to match partial matches, so "snow" in my array would match "snowman"
It should be case sensitive.
Example csv
s,dog,34
3,cat,4
1,african elephant,gd
A,African Elephant,33
H,snowman,8
8,indian elephant,3k
7,Fish,94
...
Example array
snow
dog
african elephant
Expected output
s,dog,34
H,snowman,8
1,african elephant,gd
Cyrus posted this which works well, but it doesn't allow spaces in the array strings and wont match partial matches.
echo "${array[#]}" | awk 'FNR==NR{len=split($0,a," "); next} {for(i=1;i<=len;i++) {if(a[i]==$2){next}} print}' FS=',' - file

Answer 1

The brief approach using a single regexp for all array contents:
$ array=('snow' 'dog' 'african elephant')
$ printf '%s\n' "${array[#]}" | awk -F, 'NR==FNR{r=r s $0; s="|"; next} $2~r' - example.csv
s,dog,34
1,african elephant,gd
H,snowman,8
Or if you prefer string comparisons:
$ cat tst.sh
#!/bin/env bash
array=('snow' 'dog' 'african elephant')
printf '%s\n' "${array[#]}" |
awk -F',' '
NR==FNR {
array[$0]
next
}
{
for (val in array) {
if ( index($2,val) ) { # or $2 ~ val for a regexp match
print
next
}
}
}
' - example.csv
$ ./tst.sh
s,dog,34
1,african elephant,gd
H,snowman,8

Answer 2

This prints no line from csv file which contains an element from array in column 5:
echo "${array[#]}" | awk 'FNR==NR{len=split($0,a," "); next} {for(i=1;i<=len;i++) {if(a[i]==$5){next}} print}' FS=',' - file