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  1. #1
    Join Date
    Sep 2005
    Posts
    73

    PHP pagination problem

    Hi guys,

    well I'm trying to formulate pagination view of my results from a Mysql query in my database. The thing is that the query is not known beforehand, since the user of the database can enter several search terms in text fields of the webpage (text_search.php) and when he hits the "Submit" button, the new page (retrieve_text.php) looks about which text fields have been filled and appends the values in the mysql query like this:

    Code:
    $text_query = ...... WHERE [user-defined conditions]
    Everything is OK until now. But I wanted to show the results more clear to the user, since it can have e.g 600 results so it's not so good to view them in a loooooooooong page. So I tried to paginate them.

    Now comes the problem: the first page looks Ok, and brings only 25 results, which I want to be the limit of results to show per page. BUT, when I click on the "Next button", I see nothing. The reason? In pages 2,3...N, the $text_query is empty since there are no values submitted through the $_POST variable..., So, in the 1st page, if I print the SQL query I have something like that:

    Code:
    SELECT DISTINCT protein.protein_id, protein.protein_name FROM protein LEFT JOIN protein_reference ON protein_reference.prot_ref_protein_id = protein.protein_id LEFT JOIN reference ON reference.reference_id = protein_reference.prot_ref_reference_id LEFT JOIN families ON protein.protein_families_id = families.families_id WHERE protein.protein_name LIKE '%protein%' ORDER BY protein.protein_id LIMIT 0, 25
    while in the 2nd page for example I get:

    Code:
    SELECT DISTINCT protein.protein_id, protein.protein_name, protein.protein_seq_len, protein.protein_organism, protein_ncbi FROM protein LEFT JOIN protein_reference ON protein_reference.prot_ref_protein_id = protein.protein_id LEFT JOIN reference ON reference.reference_id = protein_reference.prot_ref_reference_id LEFT JOIN families ON protein.protein_families_id = families.families_id WHERE ORDER BY protein.protein_id LIMIT 25, 25
    The reason is, as I wrote before, that the where clause in the sql statement is written on-the-fly so the sql search is dynamic and not static. If it was static, like

    Code:
    SELECT * from protein WHERE name LIKE '%protein%';
    things would be straight-forward.

    What can I do? Should/Could I somehow "send" the sql query from page to page? I suppose this is BAD practice, but I don't really know how to proceed...

  2. #2
    Join Date
    May 2002
    Posts
    10,943

    Re: PHP pagination problem

    Quote Originally Posted by ktsirig View Post
    The thing is that the query is not known beforehand
    Um...I've never known a PHP page that knows the query before a page is submitted. This should have absolutely no affect on pagination. Just add the LIMIT on the end as any pagination example does.

    PHP Code:
    <?php
    $show 
    10;
    $page = (isset($_GET['page'])) ? $_GET['page'] : 1;
    $page--; // because databases start at zero, humans start at one
    $start $page $show;

    $query "SELECT * FROM blog ORDER BY date DESC LIMIT $start$show";
    $sql mysql_query($query);
    $rows mysql_fetch_object(mysql_query("SELECT FOUND_ROWS() AS `rows`"))->rows;

    while (
    $row mysql_fetch_object($sql)) {
      
    // do your outputting from the database here
    }

    if (
    $page 0) {echo '<a href="?page=' $page '">&lt;&lt; Newer Posts</a>';}
    $total ceil(mysql_num_rows(mysql_query("SELECT id FROM blog")) / $show);
    if (
    $page $total 1) {echo '<a href="?page=' . ($page 2) . '">Older Posts &gt;&gt;</a>';}
    ?>
    If the post was helpful...Rate it! Remember to use [code] or [php] tags.

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