-
Hi,
I have a file that conatins
blah blah
start
##############
12/12
blah blah
But brthe code below never finds "/"
can any 1 help me pls..
thank you
inrow is each line xtracted
IF k=0 THEN
IF in_row LIKE '%start%' THEN
dbms_output.put_line(in_row);
i:=1;
k:=1;
END IF;
END IF;
IF j=1 THEN
i:=2;
END IF;
IF i=1 THEN
IF in_row LIKE '%/% ' THEN
dbms_output.put_line(in_row);
j:=j+1;
END IF;
END IF;
-
normally that character is reserved.
try it as:
IF in_row LIKE '%''/''%' THEN
maybe that will work.
- magnus
-
Hi
i even tried to find '#' but it dint....in a huge line as
################################
Me really confused.....
thank you
-
How about trying out:
IF INSTR(in_row,"/") > 0 THEN
..............
Maybe this will work??
-amar
-
Thanx a lot.....
it worked.....tough a little late......thanx again..
Posting Permissions
- You may not post new threads
- You may not post replies
- You may not post attachments
- You may not edit your posts
-
Forum Rules
|
Click Here to Expand Forum to Full Width
|