PHP - Capturing All The Data Send By Post Method

Hi,

I'm trying to setup a quick PHP script that will grab the email from the url (see below) and after inserting into MySQL db - which is working fine - the script will complete two additional tasks:

1. send that same captured email out to a external db as in shown via http://domain1.com/insert.php?email=$lead (example), but then send to a DIFFERENT source - the originator of the lead - a portback acknowledgement using Header (sending the status and email to http://domain2.com/check.php?e=$lead&s=$status for their records).

See the code below:
-------------------------

Code: [Select]

$lead = $_REQUEST['e_mail'];                // will grab email from posted url string and assign to local variable
$result = mysql_query($command);      // this is just to execute the MySQL insert which works just fine but included here to explain validation below

// Create API Call string to insert lead into iContact folder
$requestURL = "http://domain1.com/insert.php?email=$lead";

// Execute API Call to CAKE
$xml = simplexml_load_file($requestURL) or die("feed not loading");


if ($result) {
    $status = 1;       // mark lead as sucess

    // send postback on lead status
    header("Location: http://domain2.com/check.php?e=$lead&s=$status");
}

--------

Problem: I'm getting all sorts of errors with the simplexml_load_file() function and can't figure out why it won't work.
Any input appreciated as this the only way I know how to pass the lead onward and then inform/update the other party of receipt of information.

thanks!

I needed a server side PHP validation script for my form to use on top of client side Javascript as a backup. I found a nice package where I just include the validator and set a few options then it does the rest. However it doesn't seem to work with the action="whateveraction.php" for the form. It will skip all the validation and immediately go to the action script. So I left that blank as you can see in the form below and figured I would send the post data somehow to the script using PHP when the validation is successful. However I am stumped on how to do that can I get some coding help?

This is the small validation script. I will attach the formvalidator.php in this post.
Code: [Select]
<?php
require_once "formvalidator.php";
if(isset($_POST['submit']))
{
    $validator = new FormValidator();
    $validator->addValidation("age","req","Please fill in your Age");
    $validator->addValidation("mcid","req","Please fill in your Minecraft Username");
    $validator->addValidation("description","req","Please fill in a Description");
    if($validator->ValidateForm())
    {

      //need something here to have it send the post data to /mcbuildapp/form_submit.php

    }
    else
    {
        echo "<div class=blockrow><b><font size=5>Form Errors:</font><b></div>";

        $error_hash = $validator->GetErrors();
        foreach($error_hash as $inpname => $inp_err)
        {
          echo "<div class=blockrow><p><font color=red>$inp_err</font></p>\n</div>";
        }
    }
}

Then the form if it matters....
Code: [Select]
<div class="blockrow">
<font size=4><b>Minecraft Building Rights Application</b></font>
<br />
<br />
<b>Failing to fill in any of these fields sufficiently will result in an automatic denial of your application!</b><br />
<b>Think of it this way, being lazy and applying wrong is only wasting your time. Take the time to do it correctly the first time.</b>
 
<br />
<br />

<b><font color=red>To ensure proper user authentication and security you must first join the server at least once before applying for your building rights to work!</font></b>
<br />
<br />
<form action="" id="mcbuilderapp" name="mcbuilderapp" method="post">

<label><b>Your Minecraft Username (required)</b><br />
Use exact capitalization and punctuation that matches your Minecraft.net account!</label>
 
<br />
<br />
 
<input type="text" name="mcid" id="mcid" maxlength="25"/>
 
<br />
<br />
 
<label><b>Your Age (required)</b></label>
 
<br />
<br />
                 
<input type="text" name="age" id="age" maxlength="3"/>
   
<br />
<br />
 
<label><b>Describe Yourself (required)</b><br />
Describe yourself in <b>no less</b> than 3 sentences using <b>correct grammar</b>.</label>
 
<br />
<br />
 
<textarea name="description" id="description" minlength="10" maxlength="1000" cols=60 rows=5 wrap="physical">
</textarea>
 
<br />
<br />
 
<input type="submit" name="submit" value="Submit" />
</form>
</div>
MOD EDIT: PHP manual tags changed to code tags . . .

[attachment deleted by admin]

I'm sure this is a simple question, and I'm not a total newbie, but I keep drawing a blank.

In accordance with existing specifications, a client has created a PHP page on their web server which responds to a query by returning a small block of XML.  I need to use that XML, and I have no problem doing that using XMLReader->open(clientURL).  I can call the client's PHP page and get XML back that I can parse and everything is wonderful.

Except for one thing:  I need to send a product ID when I call the client's PHP page, and I need to send it using the POST method.

I guess that means my clean and neat one-step process is going to have to become a two-step process, and I'm going to have to use some other means besides XMLReader->open() to retrieve the XML.

Can sombody recommend the best way to do this?  I need to send POST data first to get the correct XML response, and then I need to capture the XML in a way that allows me to parse it.

Dear All,

I am trying to check which button is clicked inside the same php file and from there i will do some coding, I tried to echo some text but not working.

PHP:

<?php
 
include('dbcon.php');
include('session.php'); 

$result=mysqli_query($con, "select * from staff where OracleID='$session_id'")or die('Error In Session');
$row=mysqli_fetch_array($result);

if (isset($_POST["inside"])){
		
	echo 'inside';
		
}
	
else{
		
	echo 'outside';
		
}

?>

HTML:

<html>
<head>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<div class="form-wrapper"> 
    <center>
	
		<form action="home.php" method="POST">
			
		<h3>Welcome: <?php echo $row['StaffName']; ?> </h3>
		
		<?php if ($row['ClockedIn'] == True and $row['ClockedOut'] == False) {echo "You already clockedIn today, please clockOut";
											  echo '<button hidden> type="submit" class="buttonstyle" >Clock Time IN!</button>'; 
											  echo '<button type="submit" name = "out" class="buttonstyle">Clock Time OUT!</button>';} ?>
											  
		<?php if ($row['ClockedOut'] == False and $row['ClockedIn'] == False) {echo '<button type="submit" name = "inside" class="buttonstyle" >Clock Time IN!</button>'; 
											  echo '<button type="submit" name = "out" class="buttonstyle" >Clock Time OUT!</button>';} ?>						  
											  
		<?php if ($row['ClockedIn'] == True and $row['ClockedOut'] == True) {echo "You already clockedIn and ClockedOut today!";
											  echo '<button hidden> type="submit" class="buttonstyle" >Clock Time IN!</button>'; 
											  echo '<button hidden> type="submit" class="buttonstyle">Clock Time OUT!</button>';} ?>
	
	
		</form>
	
	</center>
	
</div>
</body>
</html>

Please help on this please.

Hi I am having this error while running my php script using ajax. Everything seems to be fine but I do not know why it is giving this error.
Code: [Select]
<br />
<b>Notice</b>:  Undefined index:  amount in <b>C:\wamp\www\...\add_bid.php</b> on line <b>3</b><br />
{"error":"yes","fieldErrors":null}
Here is my code

here is .php code which contains the form
Code: [Select]
<?php
session_start();
$auctionid = $_POST['auction_id'];
include "design/header.php";
include "DB/db.php";
?>
<div class="contents" id="bid_contents">
    <?php
        $q = "SELECT * FROM bids WHERE productid='$auctionid' ORDER BY bidid DESC LIMIT 0,1";
        $query = mysql_query($q) or mysql_error();
        //echo $q;
        $row = mysql_fetch_assoc($query) or mysql_error();
        $current_bid = $row['current_bid'];
        
        $thumbnail = mysql_query("SELECT * FROM products WHERE productid='$auctionid'");
        $row_thumb = mysql_fetch_assoc($thumbnail);
        $thumb = $row_thumb['thumbnail'];
        $title = $row_thumb['product_title'];
        
    ?>
    <div id="thumb">
        <img src='<?php echo $thumb ?>' border='none' /><br />
        <?php 
        echo $title."<br />";
        echo "<b>Current Bid:</b> Eur ".$current_bid;
        ?>
        <form name="bid_form" id="bid_form">
            <input type="text" name="amount" size="5" />
            <input type="button" name="place_bid_btn" id="bid_btn" value="Place Bid" />
            <div id="message"></div>
        </form>
    </div><!-- thumb -->
</div>
<?php include "design/footer.php" ?>


here is .js file for ajax

Code: [Select]
$(function() {
   $('#bid_btn').click(function() {
     var url = 'add_bid.php';
     var query = $('#bid_form').serialize();
     alert(query);
     $.ajax({
        type: 'POST',
        url: url,
        query: query,
        dataType: 'json',
        success: function(data) {
            if(data.error == 'no') {
                $('#message').css('display','none');
            } else {
                $('#message').html(data.fieldErrors).css('display','block');
            }
        }
     });
   });
});

and here is where i get data from previous .php file and the problem seems to be in this page. I guess. Name of this page is add_bid.php
Code: [Select]
<?php
session_start();
$bid = $_POST['amount'];
$error = 'yes';
$msg = $bid;

$JSON_array = array('error' => $error, 'fieldErrors' => $msg);
$JSON_response = json_encode($JSON_array);

header('Content-type: application/json');
echo $JSON_response;
?>

I am adding jquery library and javascript file in my header.php file so it is fine.

Please help