当使用变量进行比较时，awk无法在bash脚本中工作

Question

Following is my bash script. If I use varible oid to compare in awk, it doesnt show matching line.

oid="3586302804992"
SYMBOL_CSV_FILE="symbol/BAC"
awk -F, '$5 == $oid' "$SYMBOL_CSV_FILE"
echo "2nd"
awk -F, '$5 == "3586302804992"' "$SYMBOL_CSV_FILE"

O/P is

2nd
BAC,1,O,1,3586302804992

symbol/BAK file contents are

BAC,1,O,1,3586302804992o

Putting "" around $oid , on 3rd line, doesnt make any difference.

Answer 1

Instead of:

awk -F, '$5 == $oid' "$SYMBOL_CSV_FILE"

use it like this:

awk -F "," -v oid="$oid" '$5 == oid' "$SYMBOL_CSV_FILE"

Answer 2

For bash to interpret your variables, you have to use the double quotes. Single quotes will send $oid as is to your program.

Then, as the $5 will also be interpreted, and you don't want to! You have to escape the $ .

In the end, you have:

awk -F, "\$5 == $oid" "$SYMBOL_CSV_FILE"
        ^^          ^