[英]Displaying error message in form
我想在檢查條件后顯示錯誤消息或警告消息。
這是我的代碼
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();
}
}
}
}
?>
在此之后我已經放置了我的html表單
<!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> </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>
當用戶輸入companyemail已經存在時,它應該在表單中的公司電子郵件字段下方顯示警告消息。 但現在它顯示在頁面的開頭。 我不知道該使用什么以及如何使用。 請建議
$msg = "";
if($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; ?>
此查詢將僅返回一條記錄。 所以它不需要使用while循環
$query1 = "select count(CompanyEmail) from ApplicationRegister where CompanyEmail = '$CompanyEmail'" ;
$result1 = mysql_query($query1) or die ("ERROR: " . mysql_error());
嘗試這個,
$query1 = "select count(CompanyEmail) from ApplicationRegister where CompanyEmail = '$CompanyEmail'" ;
$result1 = mysql_query($query1);
$row = mysql_ferch_array($result1);
if($row['count(CompanyEmail)'] > 0){
error message
}else{
insert uery
}
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