justAnoob

Sorry for the confusion, let me try to explain again Page 1: has a list of categories when user clicks on one of the categories, they are taken to a new page, which would look like this http://www.mysite.com/category.php?selecting=Video Games The above URL would be called page 2 Page 2: when page 2 is diplayed, there may be 500 video games to be listed which brings in my pagination. my pagination script is having probs working with my url from above because there is alreay a variable in it. Here is some of my pag script $targetpage = "category.php"; $page = mysql_escape_string($_GET['page']); <a href='$targetpage?page= Of course the page= could be a number of variables within my pagination script. I'm not sure but it almost seems like my $targetpage var should be someting like category.php?selecting=video games ..... this way my pagination script variables come after the complete url of whatever category was chosen. Does any get this?

an undefined index error

I get an error on this line $page = mysql_escape_string($_GET['page']); when my url reads like this already http://www.mysite.com/category.php?selecting=Video Games This is hard to explain

yeh

Ok, The reason I ask is cause the page that loads has a URL like the following http://www.mysite.com/category.php?selecting=Video Games Video Games can differ depending on what category they choose on the previous page. Then, there is a pagination script with page numbers. $targetpage = "mypage.php"; <a href='$targetpage?page=1' So I would have to edit my $targetpage var ?

Can $_GET be used twice within one page? Meaning two different variables to be passed through the URL at the same time?

can't get it to work that way. The example below works, but I'm looking to change the display amount to 10, But I can't figure out what to change the if statement to so it will still work correctyly. <?php $startrow = $_GET['startrow'] ? $_GET['startrow'] : 0; $pages = intval($count / $display); if ($count % $display) { $pages++; } if ($pages > 1) { for ($i=1; $i <= $pages; $i++) { $next = $display * ($i - 1); echo '<a href="testing2.php?startrow=' . $next . '" class="active" > ' . $i . ' </a> '; } } $last = ceil($count / $display); if(! (($last + 6) < $startrow ) && $pages != 1) { $next = $startrow + $display; echo '<a href="testing2.php?startrow=' . $next . '" class="active" >Next >>></a> '; } else { echo '<span class="inactive">Next >>></span>'; } ?> Most of the code is from an example online that I found.

Well I tried that also, but it seems I am right back where I started. It does the same as the code I posted in my frist post. The Next button is always there no matter what page your on. I'm looking to have it there, unless your on the last page. Maybe I confused you.

I tried ceil and it still echos (nothing)

Nothing yet This is what I got <?php if (floor($startrow/$display) < $pages && $pages != 1) { echo ''; } else { $next = $startrow + $display; echo '<a href="testing2.php?startrow=' . $next . '">Next</a> '; } ?> This is my $startrow var below $startrow = $_GET['startrow'] ? $_GET['startrow'] : 0;

having trouble find last page for pagination. if i'm on the last page, i would like it to not show the "Next" button. Here is the echo for the "next button" if (!(($startrow / $display) == $pages) && $pages != 1) { $next = $startrow + $display; echo '<a href="testing2.php?startrow=' . $next . '">Next</a> '; } $count is my total number of rows from mysql and $display is the number of pics to be displayed on each page and this gets my total number of pages pages = intval($count / $display); This seems so simple, but I cannot get it to work.

hey thanks, could you explain what that means? btw, it worked.

Ok, I found an example online, and gave it a little tweak(not much) so I can get the idea of how pagination page numbers work. My page displays with the corrent amount of rows from the database that I want on a page. Here is the prob. When I click on a page number, the same entries are shown. Can anyone throw me a pointer. <?php include 'connection.php'; $cat = "DVDs"; $query = "SELECT id FROM my_table WHERE category = '$cat' "; $result = mysql_query($query); $count = mysql_num_rows($result); $display = 2; if (empty($startrow)) { $startrow=0; } $query2 = "SELECT thumb_1 FROM my_table WHERE category = 'DVDs' LIMIT $startrow, $display"; $result2 = mysql_query($query2); echo "<table border='0' CELLPADDING=5 STYLE='font-size:16px'>"; while ($row = mysql_fetch_array($result2)) { echo "<tr><td align='center'>"; echo '<a href="viewmovie.php?sendto='.$row['id'].'"><img src="' . $row['thumb_1'] . '" width="90" border="0" alt=""></a></td></tr>'; echo '<tr><td>'; echo '<hr width="550">'; echo "</td></tr>"; } echo "</table>"; if ($startrow > 1) { $prevrow = $startrow - $display; echo '<a href="testing2.php?startrow=$prevrow">Previous</a> '; } $pages = intval($count / $display); if ($count % $display) { $pages++; } if ($pages > 1) { for ($i=1; $i <= $pages; $i++) { $next = $display * ($i - 1); echo '<a href="testing2.php?startrow=$next">$i</a> '; } } if (!(($startrow / $display) == $pages) && $pages != 1) { $next = $startrow + $display; echo '<a href="testing2.php?startrow=$next">Next</a> '; } if ($count < 1) { echo "Nothing, sorry. Please try again."; } ?>

move up to a deluxe hosting account if you can, that is what I did. Usually does not cost that much more.

I suppose without seeing the class, I will not be able to learn much from that example, right?

I hope I'm allowed to post links http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/Q_23060532.html?eeSearch=true If you look at the very last reply all the way at the bottom of the page, this is what I'm looking to do.

The $_GET['p'] , is that just a session variable that was set? I found this on google and wanted to use something similar for my pagination <?php $pager = new pager($_GET['p'], $records_per_page, $total_count, $number_of_links); ?> When I try to use it with my code, it is saying that "class pager not found"

alright, i threw some info into the query to make sure it works and it does. So I'll go back and check my variables.

if ($result = mysql_query($sql)) = nothing displayed and if (mysql_num_rows($sql) > 0) = nothing displayed

Here is my table the is suppose to be displayed. But is not. No table is displayed and also in the error console of FF I get an "unterminated string literal" which points to this below. image0.src ="<?php echo $row['imgpath']; ?>" image1.src ="<?php echo $row['imgpath2']; ?>" image2.src ="<?php echo $row['imgpath3']; ?>" image3.src ="<?php echo $row['imgpath4']; ?>" image4.src ="<?php echo $row['imgpath5']; ?>" <?php include "connection.php"; $sql = mysql_query("SELECT id, imgpath, imgpath2, imgpath3, imgpath4, imgpath5, thumb_1, thumb_2, thumb_3, thumb_4, thumb_5 FROM member_pictures WHERE thumb_1 = " . $_SESSION['picposted'] . ""); $result=mysql_query($sql); if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { $row = mysql_fetch_array($result); ?> <table width="705" bordercolor="#000000" bgcolor="#9AA4AD" id="table"> <tr> <td> <table width="400" border="1" cellpadding="0" cellspacing="0" bgcolor="#FFFFFF"> <tr> <td colspan="4"><div align="center"><img src="<?php echo $row['imgpath']; ?>" align="middle" border="0" width="380 height="225" name="large"></div></td> </tr> <tr> <td colspan="4"> <div align="center"> Mouse over thumbnails to view larger image. </div> </td> </tr> <tr> <td bordercolor="#FFFFFF"> <div align="center"> <a onmouseover="javascript:image_click(0)"><img src="<?php echo $row['thumb_1']; ?>" width="90" height="80" name="pimage0" border=0></a> </div> </td> <td bordercolor="#FFFFFF"> <div align="center"> <a onmouseover="javascript:image_click(1)"><img src="<?php echo $row['thumb_2']; ?>" width="90" height="80" name="pimage1" border=0></a> </div> </td> <td bordercolor="#FFFFFF"> <div align="center"> <a onmouseover="javascript:image_click(2)"><img src="<?php echo $row['thumb_3']; ?>" width="90" height="80" name="pimage2" border=0></a> </div> </td> <td bordercolor="#FFFFFF"> <div align="center"> <a onmouseover="javascript:image_click(3)"><img src="<?php echo $row['thumb_4']; ?>" width="90" height="80" name="pimage3" border=0></a> </div> </td> <td bordercolor="#FFFFFF"> <div align="center"> <a onmouseover="javascript:image_click(4)"><img src="<?php echo $row['thumb_5']; ?>" width="90" height="80" name="pimage4" border=0></a> </div> </td> </tr> </table> <?php } } ?> </td> </tr> </table> </div> </div> </div> </td> </tr> </table> I'm pretty sure that I am displaying the table correctly. I have done this before. Maybe something is wrong though. I'm stumped.

Nope, totally confused now. And wondering if the general way I have the code will even work at all, or if I should change everything.

got it. I think. I try a few things.

Actually, just the part where you say that the rest of the code goes here. I'm sorry, I'm not just looking for someone to give me the answer. I would still like to understand it.

wo, what? sorry.

just changed it whle you were probably typing changed $result=mysql_query($sql); to this $row = mysql_fetch_array($sql) Now this. Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource One day I think i understand it, next day I'm confused