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Re: You have an error in your SQL syntax; [message #174133 is a reply to message #174129] Sun, 22 May 2011 20:35 Go to previous messageGo to previous message
Luuk is currently offline  Luuk
Messages: 329
Registered: September 2010
Karma:
Senior Member
On 22-05-2011 22:11, Co wrote:
> On 22 mei, 19:11, Luuk <L...@invalid.lan> wrote:
>> On 22-05-2011 16:01, Co wrote:
>>
>>> Hi all,
>>
>>> I run a query based on the input in a listbox.
>>> The query looks for users from a certain country.
>>> When no users are from the chosen country I get a error message.
>>> Is there no way to check if there are records in the query before
>>> trying to output
>>> so we only get a message saying: No records for this search.... or
>>> something like it.
>>
>>> Regards
>>> Marco
>>
>> I hope you know that there is a great MANUAL online at:http://www.php.net
>>
>> It has this great info:http://php.net/manual/en/function.mysql-num-rows.php
>>
>> --
>> Luuk
>
> Thanks Luuk,
> I didn't know.
> Anyways I tried an example from the page:
>
> $num_rows = mysql_num_rows($sql2);
> if($num_rows <> 0) {
> while($row = mysql_fetch_array($sql2)) { ......
>
> else {
> print ("<p>No records for this search were found.</p>");
> }
>
> But I still get the error line saying:
> Warning: mysql_num_rows(): supplied argument is not a valid MySQL
> result resource
>
> Marco

The argument to mysql_num_rows() should not be a sql-statement, so
sending a parameter with the name $sql2 seems confusing

The argument you need to pass is the result from mysql_query();

like the example:
$result = mysql_query("SELECT * FROM table1", $link);
$num_rows = mysql_num_rows($result);




--
Luuk
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