0

我有一个表格，我想用 POST 提交。当我去捕捉 POST 变量时，没有发送任何内容。

<?php require_once("../includes/initialize.php"); ?>
<html><head>
<link href="css/style.css" rel="stylesheet" type="text/css" />
<script src="jquery-1.8.1.min.js" type="text/javascript"></script>
<script type="text/javascript">
$(document).ready(function(){
    $('form').submit(function (e) {
        var $this = $(this);
        e.preventDefault(); // This prevents the form submission.

        $("#messageSent").show("slow");
        $this.closest('#contactForm').slideUp('slow', function () {
           $this[0].submit(); // Actual submission.
        });                
    });

    $("#contactLink").click(function(){
        if ($("#contactForm").is(":hidden")){
            $("#contactForm").slideDown("slow");
        }else{
            $("#contactForm").slideUp("slow");
        }
    });

});
</script></head><body>
<?php 
if(isset($_POST['signupSubmit'])){
echo "Post is set";
echo $_POST['name'], "<br />";
echo $_POST['email'];
}else{
echo "post is not set";
}
if(isset($_POST['signupSubmit'])){
    $signup = new Signup();
    $signup->name = $_POST['name'];
    $signup->email = $_POST['email'];
    if($signup->save()) {
        $session->message("We will contact you with details.");
    } else {
        $session->message("Failed", $signup->errors);
    }
}
echo output_message($message);
?>
    <div id="contactFormContainer">
        <div id="contactLink"></div>
        <div id="contactForm">
            <form action="test2.php" enctype="multipart/form-data" method="post">
            <fieldset>
                <label for="name">Name *</label>
                <input id="name" type="text" name="name" />
                <label for="email">Email address *</label>
                <input id="email" type="text" name="email" />
                <input id="sendMail" type="submit" name="signupSubmit" />
                <span id="messageSent"></span>
            </fieldset>
            </form>
        </div>
    </div>

</body>
</html>

任何帮助，将不胜感激！

4

1 回答 1

2

那是因为当您使用它提交表单时，$this[0].submit();它仍然运行提交处理程序，它无条件地阻止表单提交。设置一些标志，以便表单在动画后提交。

$('form').submit(function (e) {
    var $this = $(this);
    if (!$this.data('afteranimation')){
        $("#messageSent").show("slow");
        $this.closest('#contactForm').slideUp('slow', function () {
            $this.data('afteranimation', true);
            $this.submit(); // Actual submission.
        });                
        e.preventDefault(); // This prevents the form submission.
    }

});
于 2012-09-05T03:19:23.480 回答