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Re: You have an error in your SQL syntax; [message #174134 is a reply to message #174133]

Sun, 22 May 2011 20:56

Co
Messages: 75
Registered: May 2011

Karma:

Member

On 22 mei, 22:35, Luuk <L...@invalid.lan> wrote:
> 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

Guys,

I solved it.
I put this in the beginning:
$nr = mysql_num_rows($sql); // Get total of Num rows from the database
query
if ($nr){ // if we found any records we will proceed

If the query returns 0 records we pass all the code and just say "No
records found".

The only problem I still have is when I load the page first time I get
an error saying:
Notice: Undefined index
Somehow on this line:

if (($_POST['listByq'] == "newest_members")) {

listByq has not been defined.
Can I declare it at the top of the page?

Marco

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[Message index]

		You have an error in your SQL syntax; By: Co on Sun, 22 May 2011 14:01
		Re: You have an error in your SQL syntax; By: Luuk on Sun, 22 May 2011 17:11
		Re: You have an error in your SQL syntax; By: Co on Sun, 22 May 2011 20:11
		Re: You have an error in your SQL syntax; By: The Natural Philosoph on Sun, 22 May 2011 20:21
		Re: You have an error in your SQL syntax; By: Jerry Stuckle on Sun, 22 May 2011 20:21
		Re: You have an error in your SQL syntax; By: Luuk on Sun, 22 May 2011 20:35
		Re: You have an error in your SQL syntax; By: Co on Sun, 22 May 2011 20:56
		Re: You have an error in your SQL syntax; By: Jerry Stuckle on Sun, 22 May 2011 21:16
		Re: You have an error in your SQL syntax; By: Co on Sun, 22 May 2011 21:22

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