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Re: isset not working with select [message #181418 is a reply to message #181363] Wed, 15 May 2013 20:23 Go to previous messageGo to previous message
richard is currently offline  richard   
Messages: 213
Registered: June 2013
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On Mon, 13 May 2013 19:54:53 +0000 (UTC), Denis McMahon wrote:

> On Mon, 13 May 2013 07:01:46 -0400, richard wrote:
>
>> What to use to do that?
>
> Programming skills and logic.
>
> Perhaps you could check that the returned elements contain some viable
> data after checking if they're set, and if they don't contain such
> expected viable data, then you could output something else?
>
> Given:
>
> $cov = mysql_fetch_row($result);
>
> Then:
>
> if ( isset( $cov[0], $cov[1] ) && is_string( $cov[0] ) && is_string( $cov
> [1] ) && strlen( $cov[0] ) > 4 && strlen( $cov[1] ) > 4 ) {
>
> // here you use $cov[0] and $cov[1] in the output
>
> } else {
>
> // here you have an alternative output
>
> }
>
> Now please note carefully:
>
> The order of multiple tests in an if statement as shown above can be
> critical to prevent errors. Don't change it around!

if (isset($cov[0]) && is_string($cov[0]) && strlen(cov[0])>4)
{ echo "<img src='http://mroldies.net/covers/$year/".$cov[0]."'>"; }

Returns a parse error of unexpected '['.
where did I screw up at?

I want to check both images seperately as I may have one set, but not both.
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