Problem with posting select box value to php mysql query

Question

I have a page of images populated with information queried with PHP from a MySQL database. I'm attempting to split the query results into pages, and have a select box to allow users to decide how many results per page.

Problem is, I'm stuck! I'm using javascript to post the select box onChange, but don't know how to get that result into the query ($limit). I also have no idea how to get the page number (represented by $curPage in the query).

Hope you can fix this and fill in the blanks:

<form name=imgNum action="new_arrivals_img.php" method=POST>
  <p>
    <select name=ComboName size=1 onChange="imgNum.submit();">
      <option value="12" SELECTED>12</option> 
      <option>16</option> 
      <option>20</option> 
    </select>
  </p>
</form>

<ul class="new_arrivals_gallery">

  <?php
    $mysql_link = mysql_connect("localhost", "root", "root");   
    mysql_select_db("new_arrivals_imgs") or die("Could not select database");

    $query = mysql_query("SELECT `imgURL`,`imgTitle` FROM `images` ".
    "ORDER BY `imgDate` DESC LIMIT " . $limit * $curPage . ", $limit") or die(mysql_error());

    if(!$query) {
      echo "Cannot retrieve information from database.";
    } else { 
      while($row = mysql_fetch_assoc($query)) { 
        echo "<li><a href='new_arrivals_img/".$row['imgURL']."' class='gallery' title='".$row['imgTitle']."'><img src='new_arrivals_img/thumbnails/".$row['imgURL']."'></a></li>";
      }
    }
  ?>
</ul>

Also, is it ok that the the form action points to the current page? Or do I have to have two pages?

<form name="imgNum">, <select name="ComboName"> <option selected="selected"> .... Test your HTML code with validator.w3.org — Julien Lafont
– Julien Lafont, Commented Aug 16, 2011 at 7:18

Mihai Iorga · Accepted Answer · 2011-08-16 07:11:53Z

1

change method from post to get, to pass on the variables ... and check for page:

<?php
     $curPage = 0;
     if(isset($_GET['page'])){
        $curPage = (int) $_GET['page'];
     }
?>
<form name=imgNum action="new_arrivals_img.php" method="get">
  <p>
    <select name=ComboName size=1 onChange="imgNum.submit();">
      <option value="12" SELECTED>12</option> 
      <option>16</option> 
      <option>20</option> 
    </select>
    <input type="hidden" name="page" value="<?php echo $curPage; ?>" />
  </p>
</form>

<ul class="new_arrivals_gallery">

  <?php
    if((int) $_GET['ComboName'] > 0){ 
        $limit = (int) $_GET['ComboName']; 
    } else { 
        $limit = 12; 
    }
    $mysql_link = mysql_connect("localhost", "root", "root");   
    mysql_select_db("new_arrivals_imgs") or die("Could not select database");

    $query = mysql_query("SELECT `imgURL`,`imgTitle` FROM `images` ".
    "ORDER BY `imgDate` DESC LIMIT " . $limit * $curPage . ", $limit") or die(mysql_error());

    if(!$query) {
      echo "Cannot retrieve information from database.";
    } else { 
      while($row = mysql_fetch_assoc($query)) { 
        echo "<li><a href='new_arrivals_img/".$row['imgURL']."' class='gallery' title='".$row['imgTitle']."'><img src='new_arrivals_img/thumbnails/".$row['imgURL']."'></a></li>";
      }
    }
  ?>
</ul>

answered Aug 16, 2011 at 7:11

Mihai Iorga

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4 Comments

stefmikhail Over a year ago

Excellent! Thank you so much. But how do I get the select box to load showing the current setting? For example, if it defaults to 12, then 12, but if it is changed to 16, then when the page reloads, it shows 16?

stefmikhail Over a year ago

hmmm it doesn't seem to be working. When the page reloads, it still shows "12" as the selected number in the combo box.

Mihai Iorga Over a year ago

in combobox you should change for every option <option value="12"<?php if($_GET['page'] == 12 || !isset($_GET['page']){ echo " SELECTED"; } ?>12</option> ... etc

stefmikhail Over a year ago

thanks again for solving this for me. I'm curious how difficult it would be to get Ajax involved by using jQuery. I'm stuck as to how to do it. Any advice?

k102 · Accepted Answer · 2011-08-16 07:17:35Z

1

it's in $_POST["ComboName"] variable:

$limit = (isset($_POST["ComboName"]) ? $_POST["ComboName"] : 16); // 16 is some default value

edited Aug 16, 2011 at 7:17

answered Aug 16, 2011 at 7:08

k102

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