Try remiving the single quotes around myPage and showRecords.
On 15/06/2011 6:38 PM, "Taco Mathijs Hillenaar-Meerveld" <
tm.hillenaar@xxxxxxxxx> wrote:
> Hello everyone,
>
> I have followed only a short course in PHP and have a book 'PHP for
dummies'
> but i can't find a solution for my problem.
>
> I'm currently working on a photoalbum. most functions work and i started
to
> implement pagination on my photoalbum.
> where the page script works fine when i get a list of users from the db,
it
> does not work the same way for my pictures.
>
> when the page is loading all seem to work fine untill i have more than 5
> photos on the page ( $showRecords)
> also, when i change the second query to just load all the pictures from
the
> db it shows the 5 pictures AND it shows
> how many pages exist.
>
> however, im stuck here, because when i click on page 1, 2 etc it gives me
a
> blank page.
> when the query results are over 5 pictures in my current script (as seen
> below) i get an error message:
>
> *You have an error in your SQL syntax; check the manual that corresponds
to
> your MySQL server version for the right syntax to use near ''0' , '5'' at
> line 1
> *
> for some reason i haven't found a good solution yet. i hope some of you
can
> give me a hand here to understand it better.
> any tips about the script and about the style are welcome, im still
learning
> and could use some constructive critics.
>
> taco mathijs
>
> <?php
>> // this routine checks if a visitor has acces. no session - no acces
>> if(!isset($_SESSION['gbrid'])) {
>> echo "u moet ingelogd zijn als gebruiker";
>> exit();
>> }
>>
>> else {
>>
>> //-- link back to the last page.
>> echo "<a href=\"#\" onclick=\"history.go(-1)\"><strong><h5>Terug
>> naar het overzicht</h5></strong></a>";
>>
>> $pag = $_GET['pag'];
>>
>> // first visit should be 0
>> if($pag == '' || $pag == 1){
>> $pag = 0;
>> }
>> // results to show per page
>> $showRecords= 5;
>>
>>
>> // first query - seems to be okay. the page shows up correct
>> $query = "SELECT * FROM fot WHERE locid= '" .$_GET['locid']."' &&
catnaam=
>> '".$_GET['catnaam']."'";
>>
>> $results = mysql_query($query) or die(mysql_error());
>> $num_rows = mysql_num_rows($results);
>>
>> $pages = $num_rows / $showRecords;
>>
>> if($pages>1){
>> $myPage = $pag * $showRecords;
>>
>> // the second query.
>> $query = "SELECT * FROM fot WHERE locid= '" .$_GET['locid']."' &&
>> catnaam= '".$_GET['catnaam']."' ORDER BY fotid ASC LIMIT '".$myPage."' ,
>> '".$showRecords."'";
>>
>>
>>
>> $results = mysql_query($query) or die(mysql_error());
>> }
>>
>> if($pages > 1){
>> for($i=1; $i<=floor($pages); $i++){
>>
>>
>> echo ' | <a href='.$_SERVER['PHP_SELF'].'?pag='.$i.'>'.$i.'</a>';
>> }
>>
>> echo " |<br/><br/>";
>> }
>>
>>
>>
>>
>> while ($rows = mysql_fetch_array($results))
>> {
>>
>> $foto1 = '<img class="thumb" src="../~taco/'. $rows['path'] . "\" fotid
>> = ".$fotid.">";
>>
>> echo "<a href=\"fotoid.php?fotid=" . $rows['fotid'] .
>> "\">".$foto1."</a>";
>>
>>
>>
>>
>>
>> }
>> }
>>
>> // for debugging purposes
>> echo print_r($query);
>>
>>
>>
>>
>> ?>
>>
