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Easy One!

Posted on 2004-10-12
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Last Modified: 2013-12-25
I have been doing this quite a while, perl, this one is just bugging me.

I have a fixed variable called
$A = "0001";
that resides in a delimited database.
I have 2 pics for each data piece, one called 0001 and the other 0001L

Now when I call up this:
$pic = "$A";
print "http://www.mydomain.com/$A.jpg";
pulls pic just fine....

But when I try this:
print "http://www.mydomain.com/$AL.jpg";
to call up the other pic, it does not work!
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Question by:precision
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3 Comments
 
LVL 3

Accepted Solution

by:
WhisperUK earned 1600 total points
ID: 12292900
I dont know PERL but i do a bit of unix scripting and I can see straight away that in the secong example you are actually asking for a variable named $AL  and you haven't got one of them. Curly braces might be the answer here so
try this instead:

print "http://www.mydomain.com/${A}L.jpg";

Hope that helps :)
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LVL 84

Assisted Solution

by:ozo
ozo earned 400 total points
ID: 12294685
I do know Perl, and  WhisperUK is correct.
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Author Comment

by:precision
ID: 12296836
Thanks guys!
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