jQuery PHP表單提交-如何顯示表單輸入？

Question

我有一個聯系表> www.bgv.co.za/testspace/contact_3.php，它使用jquery validate和PHP。 我有一些添加/刪除類位來隱藏或顯示“謝謝”標題，還有一些php（如果將isset設置為回顯成功的表單提交輸入）。

我的問題是，在提交時，您會看到“謝謝”頁面，但是div> sadhu不會顯示用戶輸入的數據...實際上，它沒有顯示為div-請幫助

這是我的Jquery：

$(document).ready(function(){
$('#contactform').validate({
showErrors: function(errorMap, errorList) {
   //restore the normal look
   $('#contactform div.xrequired').removeClass('xrequired').addClass('_required');
   //stop if everything is ok
   if (errorList.length == 0) return;
   //Iterate over the errors
   for(var i = 0;i < errorList.length; i++)
   $(errorList[i].element).parent().removeClass('_required').addClass('xrequired');
},
submitHandler: function(form) {             
    $('h1.success_').removeClass('success_').addClass('success_form');
    $("#content").empty();
    $("#content").append('#sadhu');
    $('#contactform').hide();
    var usr = document.getElementById('contactname').value;
    var eml = document.getElementById('email').value;
    var msg = document.getElementById('message').value;
    document.getElementById('out').innerHTML = usr + " " + eml + msg;
    document.getElementById('out').style.display = "block";
    form.submit();
}
});
});

這是我的PHP：

$subject = "Website Contact Form Enquiry";

//If the form is submitted
if(isset($_POST['submit'])) {

//Check to make sure that the name field is not empty
if(trim($_POST['contactname']) == '') {
    $hasError = true;
} else {
    $name = trim($_POST['contactname']);
}

//Check to make sure sure that a valid email address is submitted
if(trim($_POST['email']) == '')  {
    $hasError = true;
} else if (!eregi("^[A-Z0-9._%-]+@[A-Z0-9._%-]+\.[A-Z]{2,4}$", trim($_POST['email']))) {
    $hasError = true;
} else {
    $email = trim($_POST['email']);
}

//Check to make sure comments were entered
if(trim($_POST['message']) == '') {
    $hasError = true;
} else {
    if(function_exists('stripslashes')) {
        $comments = stripslashes(trim($_POST['message']));
    } else {
        $comments = trim($_POST['message']);
    }
}

//If there is no error, send the email
if(!isset($hasError)) {
    $emailTo = 'info@bgv.co.za'; //Put your own email address here
    $body = "Name: $name \n\nEmail: $email \n\nComments:\n $comments";
    $headers = 'From: My Site <'.$emailTo.'>' . "\r\n" . 'Reply-To: ' . $email;

    mail($emailTo, $subject, $body, $headers);
    $emailSent = true;

}
}

這是我的表格：

            <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" id="contactform" >
        <div class="_required"><p class="label_left">Name*</p><input type="text" size="50" name="contactname" id="contactname" value="" class="required" /></div><br/><br/>
        <div class="_required"><p class="label_left">E-mail address*</p><input type="text" size="50" name="email" id="email" value="" class="required email" /></div><br/><br/>
        <p class="label_left">Message</p><textarea rows="5" cols="50" name="message" id="message" class="required"></textarea><br/>
        <input type="submit" value="submit" name="submit" id="submit" />
        </form>

Answer 1

<div id='sadhu'>沒有在任何地方定義。 您正在使用Javascript代碼將其附加到#content ，但是它不存在。

$("#content").append('#sadhu');

您可能想要類似

$("#content").append("<div id='sadhu'>stuff in here...</div>");

您已將表單條目放入ID為out 。 如果這東西也應該在<div id='sadhu'>那么你可以放置.html()的out進去：

document.getElementById('out').innerHTML = usr + " " + eml + msg;
document.getElementById('out').style.display = "block";
$("#content").append("<div id='sadhu'>" + $("#out").html() + "</div>");