php 使用 jquery 将数据发布给自己
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Posting data to self using jquery
提问by Gandalf
I am trying to post some form data to self with no success.My code is as follows
我正在尝试向自己发布一些表单数据，但没有成功。我的代码如下
<html>
<title>Post to self</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js">
</script>
<script type="text/javascript">
$(document).ready(function () {
$('.x-button').live("click", function () {
$.ajax({
type: "POST",
url: "self.php",
data: $(".aj").serialize(),
success: function (data) {
alert("Data Loaded:");
}
});
});
});
</script>
</head>
<body>
<?php
if(isset($_POST['submit']))
{
echo $_POST['firstname'];
}
?>
<form name="input" action="" class="aj" method="post">
<article>
<label>Firstname</label>
<input type="text" name="firstname" value="Ed" class="x-input"
/>
</article>
<article>
<label>Lastname</label>
<input type="text" name="lastname" value="Doe" class="x-input"
/>
</article>
<article>
<label>City</label>
<input type="text" name="city" value="London" class="x-input"
/>
</article>
<input type="submit" value="Update Options" class="x-button" />
</form>
</body>
</html>
When using <form name="test" action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post">in plain php and html it works but i can't get it to work with jquery.
当<form name="test" action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post">在普通的 php 和 html 中使用时，它可以工作，但我无法让它与 jquery 一起使用。
采纳答案by Johndave Decano
Of course you do not want to include the whole page to the response text so you need a statement if ajax is requested
当然，您不想将整个页面包含在响应文本中，因此如果请求 ajax，您需要一个声明
<?php
/* AJAX check */
if(!empty($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest') {
if(isset($_POST))
{
print_r($_POST);
}
}else{
?>
<html>
<title>Post to self</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js">
</script>
<script type="text/javascript">
$(document).ready(function () {
$('.x-button').live("click", function () {
$.ajax({
type: "POST",
url: "self.php",
data: $(".aj").serialize(),
success: function (data) {
alert("Data Loaded:");
}
});
});
});
</script>
</head>
<body>
<form name="input" action="" class="aj" method="post">
<article>
<label>Firstname</label>
<input type="text" name="firstname" value="Ed" class="x-input"
/>
</article>
<article>
<label>Lastname</label>
<input type="text" name="lastname" value="Doe" class="x-input"
/>
</article>
<article>
<label>City</label>
<input type="text" name="city" value="London" class="x-input"
/>
</article>
<input type="submit" value="Update Options" class="x-button" />
</form>
</body>
</html>
<?php }?>
回答by diEcho
you are mixing two approch altogether.
你完全混合了两种方法。
<form id="myform" action="" >
.....
</form>
To send ajax request to the same page you can keep url parameter empty/removed
要将 ajax 请求发送到同一页面，您可以将 url 参数保留为空/已删除
TRY
尝试
<script type="text/javascript">
$(document).ready(function () {
$('.x-button').live("click", function () {
$.post({
data: $('form#myform').serialize(),
success: function (data) {
alert("Data Loaded:");
}
});
});
});
</script>
回答by Andreas Linden
add return false;at the end of the inline javascript function to avoid the form gets submitted the "normal" way
添加return false;在内联javascript函数的末尾以避免表单以“正常”方式提交
