Re: Re: Dynamic dropdown: where am I wrong?

You could try using Ajax to get the field values from your database. And you wouldn't have to refresh the page. jQuery ajax does a nice job of this. HTH, Karl On May 24, 2011, at 10:27 PM, Jim Giner wrote:

not an expert but from some things that I've done... When you handle the selection use JS and set a hidden field on your form wiht the choice. Do the same with your second dropdown and then let them click on a 'submit' <input> "Fahim Mohammad" <fahim.md@gmail.com> wrote in message news:BANLkTim47cjrUsXYz=t-Yfhaoz_iUntesA@mail.gmail.com...

Hi Php users, I have a form that has two (has more but for sake of brevity I am assuming it to be two) dropdown menu. menu 1: select organism. menu 2: select genome. This menu depends on the selected value in the first menu. It query into database and populate the result in this menu. Once both are selected, I want a submit button that takes me to another page where I can process these values to perform some specific task on the database. I am using the following form tag <form name="theForm" method="get" enctype="multipart/form-data" action = 'processForm.php' > I have the following problem now: 1. If I keep the action value in the form tag above, after selecting the first option it direct me to the second page and I am not able to select the seconf option in the form. 2. If I remove the action value above then I can not submit the form as intended. 3. If I remove the "onChange="autoSubmit();"" from select tag then I am not able to pass the selected value in first option to mysql to get the result. I think I reached a place from where it is not possible for me to proceed ahead. I may need to change something which I am not aware of. Help needed. Thanks [php] <?php include'login.php'; $conn = mysql_connect($db_hostname, $db_username, $db_password ); $db = mysql_select_db('rugit',$conn); if (!$db) die("Unable to connect to MySQL: " . mysql_error()); ?> <?php $organism = $genomeVer = null; //declare vars if(isset($_GET["organism"]) && is_string($_GET["organism"])) { $organism = $_GET["organism"]; } if(isset($_GET["genomeVer"]) && is_string($_GET["genomeVer"])) { $genomeVer = $_GET["genomeVer"]; } ?> <script language="JavaScript"> function autoSubmit() { var formObject = document.forms['theForm']; formObject.submit(); } </script> <!-- Form tag --> <form name="theForm" method="get" enctype="multipart/form-data" > <!-- Option 1: Select the organism --> <select name="organism" onChange="autoSubmit();">

     <option value="null">Select Organism</option>
     <option value= "human" <?php if(strcmp($organism, "human") == 0) echo

" selected"; ?>>human</option>

     <option value= "mouse" <?php if(strcmp($organism, "mouse") == 0) echo

" selected"; ?>>mouse</option>

     <option value= "rat" <?php if(strcmp($organism,"rat") == 0) echo "

selected"; ?>>rat</option> </select> <br></br> <!--Option 2: select the genome --> <?php

       //POPULATE DROP DOWN MENU FOR Genome version corresponding to a

given organism

      $sql = "SELECT genomeVer FROM organismGenomeVer WHERE organism =

\"$organism\"";

      $result = mysql_query($sql,$conn);

?>

      <br>
      <select name="genomeVer" onChange="autoSubmit();">
      <option value="null">genomeVer</option>

<?php

      while($row = mysql_fetch_array($result))
               {
                echo ("<option value=\"$row[0]\" " . ($genomeVer ==

$row["0"] ? " selected" : "") . ">$row[0]</option>");

                 }
               mysql_free_result($result);
      echo"</select>";

?> <!-- Submit button --> <br></br> <input type="submit" value="submit" class="html-text-box"> </form> [/php] --Fahim

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