Jump to content

Recommended Posts

Hi at All

 

I have the problem, that I do not know, how to get the value from the selected option in my dropdown list. Hope someone can help me. I have the follow code:

 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>File Uploader</title>
<link href="css/reset.css" rel="stylesheet" type="text/css" />

<style type="text/css">
body {
	margin: 10px;
	font: 62% Tahoma, Arial, sans-serif;
}
#main_container{
	font-size: 1.4em;
}
h2 {
	font-size: 2em;
	padding-bottom: 20px;
}
</style>

<link href="css/ui-lightness/jquery-ui-1.8.14.custom.css" rel="stylesheet" type="text/css" />
<link href="css/fileUploader.css" rel="stylesheet" type="text/css" />

<script src="js/jquery-1.6.2.min.js" type="text/javascript"></script>
<script src="js/jquery-ui-1.8.14.custom.min.js" type="text/javascript"></script>
<script src="js/jquery.fileUploader.js" type="text/javascript"></script>

</head>

<body>
<div id="main_container">
	<h2>Bilder Girls Uploader</h2>
	<p> </p>
    <p>
      <?php
		mysql_connect("localhost:8888", "root", "root") or die("Connection Failed");
		mysql_select_db("mydb")or die("Connection Failed");
		$query = "SELECT idgirls,name FROM girls";
		$result = mysql_query($query);
	?>
      <form name="form1" method="POST" action="<?=$PHP_SELF?>">
      <select name="select1" onchange="this.form1.submit();" method="post">
        <?php
	while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
	?>
        <option value="<?php echo $line['idgirls'];?>"> <?php echo $line['name'];?> </option>
        
        <?php
	}
	?>
      </select>
      </form>   
    </p>
    <p> </p>
    <p> </p>
	<form action="upload.php" method="post" enctype="multipart/form-data">
    <input type="file" name="userfile" class="fileUpload" multiple>
    <input type="hidden" name='girl' value='<?php echo $_POST['select1'];?>' />
		
		<button id="px-submit" type="submit">Upload</button>
		<button id="px-clear" type="reset">Clear</button>
</form>
	<script type="text/javascript">
		jQuery(function($){
			$('.fileUpload').fileUploader();
		});
	</script>
</div>
</body>
</html>

 

 

Link to comment
https://forums.phpfreaks.com/topic/276461-get-value-from-dropdown-menulist/
Share on other sites


               echo "<option value = ''>Select Category</option>";
               echo "<select name = 'cat_name'>";
	while($s = mysql_fetch_array($cat_class->get_the_result))
		  {
		echo "<option value = $s[cat_id]> $s[cat_name]</option>";
		   }
		echo "</select>";
		

 

a simple way, to get the record from database using select option, hope its help

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.