| Re: isset not working with select [message #181346 is a reply to message #181342] |
Mon, 13 May 2013 12:43   |
Peter H. Coffin
Messages: 245
Registered: September 2010
Karma:
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Senior Member |
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On Mon, 13 May 2013 07:01:46 -0400, richard wrote:
> http://mroldies.net/showtable.php?year=1960
>
>
>
> $result = mysql_query("SELECT acover,bcover FROM A$year WHERE id =
> $number");
> if (!$result) { echo 'Could not run query: ' . mysql_error(); exit; }
> $cov = mysql_fetch_row($result);
>
>
> if (isset($cov[0]))
> {echo "<img src='http://mroldies.net/covers/$year/".$cov[0]."'>";}
> echo "<img src='http://mroldies.net/covers/$year/".$cov[1]."'>";
>
> On item #1 in the table, the panel appears with the desired images as
> wanted.
> On item #2 image place holders appear because I do not have them online
> yet.
> I do not want to see the place holders if there is not an image.
> What to use to do that?
If the first one works and the second one doesn't, maybe you should just
do them both like the first one?
> Also, if you click on 1961, or any other year, and then any item in the
> table, you will get a message below the videos stating the column does not
> exist so it can't continue processing.
> How can I bypass this when that database does not have that column?
Again, check for zero rows being returned in your program and build code
to handle that circumstance. die() is NOT your friend.
--
I wish there was a knob on the TV to turn up the intelligence. There's a
knob called "brightness", but it doesn't work.
-- Gallagher
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