Displaying error message in form

Question

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

Miqdad Ali · Accepted Answer · 2012-06-22 09:31:16Z

2

$msg = "";
if($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; ?>

answered Jun 22, 2012 at 9:31

Miqdad Ali

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2 Comments

user1321271 Over a year ago

Miqdad Ali: thank you very much. This is what i wanted. but how to make the font red in this.

Miqdad Ali Over a year ago

$msg = "<font color='red'>This E-mail id is already registered</font> ";

Anand · Accepted Answer · 2012-06-22 10:05:35Z

0

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 
 }

answered Jun 22, 2012 at 10:05

Anand

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