Question

我想在检查条件后显示错误消息或警告消息。

这是我的代码

    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();
}


    }
    }
                }

?>

之后我放了我的html表格

<!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>

当用户输入companyemail已经存在时，它应该在表单中的公司电子邮件字段正下方显示警告消息。但现在它显示在页面的开头。我不知道该使用什么以及如何使用。请建议

Answer 1

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


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

和HTML

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

Answer 2

此查询只返回一条记录。所以它不需要使用while循环

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

试试这个，

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

在表单中显示错误消息

2 个答案: