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Re: Dynamic dropdown: where am I wrong?

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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@xxxxxxxxx> wrote in message 
news:BANLkTim47cjrUsXYz=t-Yfhaoz_iUntesA@xxxxxxxxxxxxxxxxx
> 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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