PHP - Moved: Need Help With A Submit Form

Hello, first time poster..  I've looked the web over for a long time and can't figure this one out.

- Below is basic code that successfully checks MySQL for a match and displays result.  I was debugging and forced the "height" and "width" to be 24 and 36 to make sure that wasn't the problem.  That's good.. 
- I'd like to give the user ability to select width and height from a form.. and have it do an onchange this.form.submit so the form can be changing as fields are altered (thus the onchange interaction)
- In a normal coding environment I've done this numerous times with no "Page cannot be displayed" problems.  It would simply change one select-option value at a time til they get down the form and click submit...  but in WordPress I'm having trouble making even ONE single onchange work!
- I've implemented the plugins they offer which allows you to "copy+paste" your php code directly into their wysiwyg editor.  That works with basic tests like my first bullet point above.
- I've copied and pasted the wordpress url (including the little ?page_id=123) into the form "action" url... that didn't work...  tried forcing it into an <option value=""> tag.. didn't work.  I'm just not sure. 

I've obviously put xx's in place of private info..

Why does this form give me Page Cannot Be Displayed in WordPress every time?  It won't do anything no matter how simple.. using onchange..

Code..

$con = mysql_connect("xxxx.xxxxxxx.com","xxxxxx","xxxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("xxxxxx", $con);
$myprodwidth=24;
$myprodheight=36;
$result = mysql_query("SELECT * FROM product_sizes WHERE prodwidth='$myprodwidth' and prodheight='$myprodheight'");
while($row = mysql_fetch_array($result))
  {
  echo $row['prodprice'];
  }
mysql_close($con);


<form method="post" action="">
<select name="myheight" onchange="this.form.submit();">
<option selected="selected" value="">select height</option>
<option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">36</option>
<option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">48</option>
</select>

Hi-
the code below lets me upload a CSV file to my database if I have 1 field in my database and 1 column in my CSV.
I need to add to my db "player_id" from the CVS file and "event_name" and "event_type" from the form... any ideas???

here's the code:
Code: [Select]
<?php
$hoststring =""; 
$database = "";
$username = "";
$password = "";
$makeconnection = mysql_pconnect($hoststring, $username, $password);
?>
<?php
ob_start();
mysql_select_db($database, $makeconnection);
$sql_get_players="
SELECT * 
FROM tabel 
ORDER BY player_id ASC";
//
$get_players = mysql_query($sql_get_players, $makeconnection) or die(mysql_error());
$row_get_players = mysql_fetch_assoc($get_players);
//
$message = null;
$allowed_extensions = array('csv');
$upload_path = '.'; //same directory

if (!empty($_FILES['file'])) {

if ($_FILES['file']['error'] == 0) {
// check extension
$file = explode(".", $_FILES['file']['name']);
$extension = array_pop($file);

if (in_array($extension, $allowed_extensions)) {

if (move_uploaded_file($_FILES['file']['tmp_name'], $upload_path.'/'.$_FILES['file']['name'])) {

if (($handle = fopen($upload_path.'/'.$_FILES['file']['name'], "r")) !== false) {
$keys = array();
$out = array();
$insert = array();
$line = 1;

while (($row = fgetcsv($handle, 0, ',', '"')) !== FALSE) {

foreach($row as $key => $value) {    

if ($line === 1) {
$keys[$key] = $value;       
} else {       
$out[$line][$key] = $value;              
}       
}                
$line++;          
}        

fclose($handle);            

if (!empty($keys) && !empty($out)) {        
$db = new PDO(
'mysql:host=host;dbname=db', 
'user', 
'pw');   

$db->exec("SET CHARACTER SET utf8");        

foreach($out as $key => $value) {        
$sql  = "INSERT INTO `table` (`";
$sql .= implode("`player_id`", $keys);    
$sql .= "`) VALUES (";    
$sql .= implode(", ", array_fill(0, count($keys), "?"));    
$sql .= ")";    
$statement = $db->prepare($sql);    
$statement->execute($value);       
}      

$message = '<span>File has been uploaded successfully</span>';      
}
}
}
} else {
$message = '<span>Only .csv file format is allowed</span>';
}
} else {
$message = '<span>There was a problem with your file</span>';
}
}
ob_flush();?>
<!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>CSV File Upload</title>
</head>
<body>

<form class="form" action="" method="post" enctype="multipart/form-data">
<h3>Select Your File</h3>
<p><?php echo $message; ?></p>
<input type="file" name="file" id="file" size="30" />
<br/>
<label>Event Name:</label><input name="event_name" type="text" value="" />
<br/>
<label>Event Type:</label><input name="event_type" type="text" value="" />
<br/>
<input type="submit" id="btn" class="button" value="Submit" />
</form>

<br/>
<h3>Results:</h3>
<?php do { ?>
<p><?php echo $row_get_players['player_id'];?></p>
<?php } while ($row_get_players = mysql_fetch_assoc($get_players)); ?>
</body>
</html>

Hi all,

What I am trying to achieve is, I thought quite simple!

Basically, a user signs up and chooses a package, form is submitted, details added to the database, email sent to customer, then I want to direct them to a paypal payment screen, this is where I am having issues!

Is their any way in php to submit a form without user interaction? Here is my code for the form process page
Code: [Select]
<?php
include('config.php');
require('scripts/class.phpmailer.php');

$package = $_POST['select1'];
$name = $_POST['name'];
$email = $_POST['email'];
$password = md5($_POST['password']);

$domain = $_POST['domain'];
$a_username = $_POST['a_username'];
$a_password = $_POST['a_password'];


$query=mysql_query("INSERT INTO orders (package, name, email, password, domain, a_username, a_password)
VALUES
('$package', '$name', '$email', '$password', '$domain', '$a_username', '$a_password')");

if (!$query)
{
echo "fail<br>";
echo mysql_error();
}

else
{
$id = mysql_insert_id();
$query1=mysql_query("INSERT INTO customers (id, name, email, password)
values
('$id', '$name', '$email', '$password')");

if (!$query1)
{
echo "fail<br>";
echo mysql_error();
}


if($package=="Reseller Hosting")
{

//email stuff here - all works - just cutting it to keep the code short


if(!$mail->Send())
{
   echo "Message could not be sent. <p>";
   echo "Mailer Error: " . $mail->ErrorInfo;
   exit;
}
?>


<form name="_xclick" action="https://www.paypal.com/cgi-bin/webscr" method="post">
<input type="hidden" name="cmd" value="_xclick-subscriptions">
<input type="hidden" name="business" value="subscription@jollyhosting.com">
<input type="hidden" name="currency_code" value="USD">
<input type="hidden" name="item_name" value="Jolly Hosting Reseller Packages">
<input type="hidden" name="no_shipping" value="1">

<!--1st month -->
<input type="hidden" name="currency_code" value="USD"> 
<input type="hidden" name="a3" value="3.00">
<input type="hidden" name="p3" value="1">
<input type="hidden" name="t3" value="M">
<input type="hidden" name="src" value="1">
<input type="hidden" name="sra" value="1">
</form>';


<?php
}

//last
}
//end
?>

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I can't submit php form using IE 8.  This works fine on Firefox very well.
I tried following code but none of them is working.   
1.  <button type="submit">Submit</button>

2.  <input type="submit" value="Submit" name="submit">


 I would appreciate if any one help me in this.

Hi, I have setup a basic enquiry form with a Captcha - once the code has been inserted and is correct I want to action the form (submit) in PHP.

<?php

session_start();
if($_SERVER['REQUEST_METHOD'] == 'POST'){
$vResult = '';
if(strtolower($_SESSION['security_code']) != strtolower($_POST['security_code'])){
$vResult = 'Invalid code!';
}
else{
"/enquiry.php"
}
}
?>

I want to submit the form to /enquiry.php - I tried header(Location.. and realised that just redirects, and doesn't submit the form.

Any suggestions or tips would be great.

Cheers,

Paul

hello everyone, I have a php form, everything works, I do not carry any header() at the end of the form because I have to stay on the same page and because I feel it wipes out the message you sent the form successfully, but for this though if reloading  the page shows me the popup asking me to resend the form.  how can i solve?  I found a function in js with replacestate but I saw that it doesn't work with mobile Safari.

if ( window.history.replaceState ) { window.history.replaceState( null, null, window.location.href ); }

 

Can anybody see as to why, if any, this form takes forever to execute?
Code: [Select]
<?php
if(isset($_POST['submitted'])){
    $z = $_POST['zipcode'];
    $r = $_POST['radius'];
    
    $sql = mysql_query("SELECT DISTINCT m.LocAddZip, m.MktName,m.LocAddSt,m.LocAddCity,m.LocAddState,m.x1,m.y1,z1.lat,z2.long
    FROM mrk m, zip z1, zip z2
    WHERE m.LocAddZip = z1.zipcode
    AND z2.zipcode = $z
    AND ( 3963 * acos( truncate( sin( z2.lat / 57.2958 ) * sin( z1.lat / 57.2958 ) + cos( z2.lat / 57.2958 ) * cos( z1.lat / 57.2958 ) * cos( z1.long / 57.2958 - z2.long / 57.2958 ) , 8 ) ) ) <= $r ")
    
    or die(mysql_error());
        while($row = mysql_fetch_array( $sql )) {
            $store = $row['MktName']."<br />";
            $store .= $row['LocAddSt']."<br />";
            $store .= $row['LocAddCity'].", ".$row['LocAddState']." ".$row['LocAddZip'];
            $lat1 = $row['lat'];
            $lon1 = $row['long'];
            $lat2 = $row['y1'];
            $lon2 = $row['x1'];
            $dis = distance($lat1, $lon1, $lat2, $lon2);
     
            echo "<p>".$store."</p>";        
            echo ceil($dis) . " mile(s) away";
            echo "<hr/>";
        }
}
?>

I get a timeout error sometimes, and sometimes I don't.
Is it the form, or would it be on the server?

Thanks in advance

Hi,
I want to submit a form using ajax ans jquery with two fields input text and file.

Code: [Select]
<script>
$(document).ready(
function()
{


$('#basicinfofrm').ajaxForm(function() {
url: ''+PN+'.php',
type:'POST',
data:'action=yes&'+frmFeilds,
success: function(html)
{ 
 alert(html);

}

    }).submit();
</script>
<form action='' method='post'>
    <label>name</label>
    <input type='text' name='txtname' />
    <input type='file' name  = 'txtfile' />
   <input type='submit' value='save' />
</form>

but is is not working...
can anyone tell me how to?
i don't want to refresh my page
Thanks