Ways to find records based on their first letter awk Unix

Question

I would like to take out only persons with name starting with A using awk. How do i do it

REGNO   NAME            DEPT            AVG             STATUS
100     Tony            ECE             30              FAIL
200     Jimmy           ECE             67              PASS
300     Sanjay          CSE             89              PASS
400     Nethra          IT              90              PASS
500     Ramesh          MECH            45              FAIL
600     Ranjani         CSE             34              FAIL
700     Aarthi          IT              56              PASS
800     Ram             MECH            98              PASS

awk '{if($2=="A*" && $3=="IT" )  print $2,$3}' output.txt

I used this. What is wrong?

Answer 1

awk '$2 ~ "^A" && $3 == "IT" {print $2, $3}'

没有通配符匹配，例如== "A*"

Answer 2

This will work for you:

kent$  awk '$2~/^A/&&$3=="IT"{print $2,$3}' file
Aarthi IT

Your problem is, confused by glob and regex.

In awk, == won't do glob check, it compares two values, check if they are equal. ~ is a regex matching check, it expect a static or dynamic regex expression.