Using gawk to extract rows with string in a column - regex

Question

I was trying to extract rows from a tab separated file, if it contained a certain word in the 4th column. For example, if input file test.txt is:
chr 8 1234 abc ; xyz
chr 8 1255 abc
chr 8 987 xyz
chr 8 5467 jxyzm
The following code correctly outputs only the 1st and 3rd line:
gawk -F"\t" ' { if($4 ~ /\<xyz\>/) print $0 } ' test.txt >> test.out
However, when I try to run this in a loop, in a bash script, my output file is blank. the code I am using is:
while read id
do
OFILE=${ODIR}/${id}.txt
gawk -v id="$id" -F"\t" ' { if($4 ~ /\<id\>/) print $0 } ' ${IFILE} >> ${OFILE}
done < ${GFILE}
The file ${GFILE} has one word per line, e.g.:
xyz
fg45
tre2y
What am I doing wrong?
thanks!
Edited to:
Add fourth row in input file
Added -v id="$id" to command...script still doesn't work!

Answer 1

You can very well use awk to read search patterns from one file and find matches in other like this:
awk -F '\t' '
NR == FNR {
words[$1]
next
}
{
for (w in words)
if (index($4, w)) {
print > w ".txt"
break
}
}' "$GFILE" "$IFILE"
Then check output:
cat xyz.txt
chr 8 1234 abc ; xyz
chr 8 987 xyz
If you really-really want to fix your shell script then here it is:
while read id; do
awk -F '\t' -v id="$id" '$4 ~ id' "$IFILE" > "$id.txt"
done < "$GFILE"