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Showing results for tags 'option'.
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I am trying to use a <select> <option> to display text on a web page. The data comes from a database. This is what I am attempting: When an option is selected and submit button clicked them some text will be displayed. At the moment I do have text on the web page, but none of the selected options change this. I have four pieces of text: Page 1, Page 2 etc. At the moment Page 4 text is displayed. I would like each piece of text to be displayed. This is the php: <?php // Connect to the database $pdo = new PDO("mysql:host=localhost;dbname=###", "###", ""); $sql = "SELECT * FROM testdb ORDER BY id"; try { $stmt = $pdo->prepare($sql); $stmt->execute(); $data = $stmt->fetchAll(); } catch(Exception $ex){ echo ($ex -> getMessage()); } ?> This is the html: <form name="###" method="post" action="#"> <p></p> <select onchange="reload(this.form)"> <option>test one</option> <?php foreach ($data as $output) { ?> <option value=''><?php echo $output['header']; ?></option> <?php } ?> </select> <br> <button type="submit" value="submit">Submit</button> </form> <?php echo $output['pages']; ?> I would appreciate help with this. However, from previous attempts at adding data to a page I used isset() and I think $_POST(), possibly together. If these are the things I need to use, then please could you include their usage within any examples you feel would help. Thank you.
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Hi all, I was hoping you could point me in the right direction, I have 2 questions. 1. I want to have an option drop down which uses the array of a returned sql query. I have the following code which uses onchange to submit the selection, how do I use the mysql_fetch_array($result1) to give a list of the two selected columns in the $Site=array() bit. 2. The selected result will be used in more SQL select statements, so once a site from the list is selected the relevant data from other tables will be displayed using e.g. select person from people [some inner join statement] where site="site 1"; Therefore the $_POST value must be available to be passed to other queries once selected. Thanks for any help, I still feel like a noob, but I'm getting there. Gary <?php include 'header.php'; ?> <div class='container'> <?php include 'menu.php'; ?> <?php include 'connect.php'; ?> <?php $sql1="SELECT Sites.Site_ID, Sites.Site_name_1 FROM `Sites`"; $result1=mysql_query($sql1); ?> <?php function get_options() { $site=array('Site 1'=>'Site 1', 'Site 2'=>'Site 2', 'Site 3'=>'Site 3'); $options=''; while(list($k,$v)=each($site)) { $options.='<option value="'.$v.'">'.$k.'</option>'; } return $options; } if(isset($_POST['site'])) { echo $_POST['site']; } ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST"> <select name="site" onchange="this.form.submit();"> <?php echo get_options(); ?> </select>