| Re: isset not working with select [message #181422 is a reply to message #181418] |
Wed, 15 May 2013 21:15  |
Tim Streater
Messages: 328
Registered: September 2010
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Senior Member |
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In article <195lftw1ts4is(dot)1wz8p6f1a44my$(dot)dlg(at)40tude(dot)net>,
richard <noreply(at)example(dot)com> wrote:
> 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)
~~~~~~~~
Missing $ sign.
> { 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.
--
Tim
"That excessive bail ought not to be required, nor excessive fines imposed,
nor cruel and unusual punishments inflicted" -- Bill of Rights 1689
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