CODEWITHSUNDEEP
php
user1321271
user1321271

Reputation: 253

Displaying error message in form

I want to display error message or alert message after checking the condition.

here is my code

    session_start();

    $plan = @$_GET['plan'];
    $plan = +$plan; 
    include('connect.php');

            If (isset($_POST['submit']))
            {
            $CompanyName = $_POST['CompanyName'];

            $CompanyEmail = $_POST['CompanyEmail'];
            $CompanyContact = $_POST['CompanyContact'];
            $CompanyAddress = $_POST['CompanyAddress']; 

            $RegistrationType = $_POST['RegistrationType'];
            $Plans = $_POST['plan'];
            $Status = "Active";
    $query1 = "select count(CompanyEmail) from ApplicationRegister where CompanyEmail = '$CompanyEmail'" ;
    $result1 = mysql_query($query1) or die ("ERROR: " . mysql_error());
    //$result1 = mysql_result($query1, 0, 0);
    //echo $result1;
     while ($row = mysql_fetch_array($result1))

                {
                    //$companyemail = $row['CompanyEmail'];
                    //if($companyemail != '' || $companyemail!= 'NULL')
                    //{
                        if($row['count(CompanyEmail)'] > 0)
                        {


        echo "This E-mail id is already registered ";
    }
    else
    {

    $sql = "INSERT INTO ApplicationRegister(CompanyName, CompanyEmail, CompanyContact, CompanyAddress, RegistrationType, ApplicationPlan, ApplicationStatus, CreatedDate) VALUES ('$CompanyName', '$CompanyEmail', '$CompanyContact', '$CompanyAddress', '$RegistrationType', '$Plans', '$Status', NOW() )";
$result = mysql_query($sql) or die(mysql_error());

$id = mysql_insert_id(); 

$_SESSION['application_id'] = $id;


//if(isset($plan == "trail"))
if($plan == "trail")
{

header("Location: userRegister.php?id=$id");
exit();
}
else
{
    header("Location : PaymentGateway.php");
    exit();
}


    }
    }
                }

?>

and after this i have placed my html form

<!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" />
<link href="style.css" rel="stylesheet" type="text/css" />


<title>Welcome</title>
</head>
<body>
<table width="100%" border="0" cellspacing="0" cellpadding="0">
              <tr>
                <td><h2><br />
                  Application Registration</h2>
                  <p>&nbsp;</p></td>
              </tr>
              <tr>
                <td><form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" enctype="multipart/form-data" name="form1" id="form1"  onsubmit="return Validate();">
                <input type="hidden" name="plan" value="<?php echo $plan ?>"/>
    Company Name: 
        <input type="text" name="CompanyName" style="width:230px; height:20px;" /><br /><br />
    Company E-mail : 
    <input type="text" name="CompanyEmail" style="width:230px; height:20px;" /><br /><br />
     Company Contact <input type="text" name="CompanyContact" style="width:230px; height:20px;" /><br /><br />
     Company Address: <input type="text" name="CompanyAddress" style="width:230px; height:20px;" /><br /><br />
    Select Your Registration Type : <br /><br />
Trail: <input type="radio" name="RegistrationType" value="Trail" /><br />
                                     Paid<input type="radio" name="RegistrationType" value="Paid" /><br /><br />

     <input type="hidden" name="form_submitted" value="1"/> 
     <input type="submit" value="REGISTER" name="submit" />



</form> 


    </td>
              </tr>

            </table>

When the user entered companyemail is already exists it should display the alert message just below the company email field in the form. But now its displaying in the starting of the page. I dont know what to use and how to use. Please suggest

Upvotes: 0

Views: 222

Answers (2)

Anand
Anand

Reputation: 1021

This query will return only one record. So its not need to use while loop

$query1 = "select count(CompanyEmail) from ApplicationRegister where CompanyEmail = '$CompanyEmail'" ;
$result1 = mysql_query($query1) or die ("ERROR: " . mysql_error());

Try this,

    $query1 = "select count(CompanyEmail) from ApplicationRegister where CompanyEmail = '$CompanyEmail'" ;
    $result1 = mysql_query($query1);     
    $row = mysql_ferch_array($result1); 
 if($row['count(CompanyEmail)'] > 0){
     error message 
}else{ 
 insert uery 
 }

Upvotes: 0

Miqdad Ali
Miqdad Ali

Reputation: 6147

$msg = "";
if($row['count(CompanyEmail)'] > 0)
                    {


    $msg = "This E-mail id is already registered ";
}

and HTML

<input type="text" name="CompanyEmail" style="width:230px; height:20px;" /><br />
<?php echo $msg; ?>

Upvotes: 2

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