Shell Programming and Scripting

Hello Team,

I have a script in which more than 500 trigger statemetns are written.

i need to filter all lines from that script file which containts "CREATE TRIGGER" word in each line

is it possible using linux commands??

example:

CREATE TRIGGER S1.T1
...............
..............
.......
END

CREATE TRIGGER s2.t2
............
........
.......
...
END
..
..
..
..


out put should

CREATE TRIGGER s1.t1
CREATE TRIGGER s2.t2
.....
...

Have a look at:

Code:

man grep

Thanks for the reply.

i am trying this

grep -i create trigger trig1.sql | awk '{print "DROP TRIGGER\t" $4}'

after $4 i need a semi colon like

DROP TRIGGER X.Y ;

how to add semi colon here??

grep -i create trigger trig1.sql | awk '{print "DROP TRIGGER\t" $4 ";"}'

This is not working for me can any body suggest??

Try this, Its $3 not $4

Code:

grep -i create trigger trig1.sql | awk '{print "DROP TRIGGER\t" $3 ";"}'

Use " ;" instead of ";" to get a space after the value.

when i use this
grep -i create trigger trig1.sql | awk '{print "DROP TRIGGER\t" $4 ";"}'

its giving out put as

;ROP TRIGGER X.Y

how come??

even i tried this
grep -i create trigger trig1.sql | awk '{print "DROP TRIGGER\t" $4 " ;"}'

another kind of out put
;OP TRIGGER x.y

Post your input file trig1.sql