如何通过jquery发布一些数据，并从myxmlfeed.php中获取所有json编码的数据，我从某个网站得到了这个代码。。。在这里，他们将一些数据发布到myxmlfeed.php并进行处理，在json编码后，他们将打印数据。我还想打印id、title、date_start、date_end
<head>
<title>jQuery Test</title>
<script src="js/jquery-1.11.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#submit").click(function(){
$.ajax({
url: "myxmlfeed.php",
type: "POST",
data: {
amount: $("#amount").val(),
firstName: $("#firstName").val(),
lastName: $("#lastName").val(),
email: $("#email").val()
},
dataType: "JSON",
success: function (jsonStr) {
$("#result").text(JSON.stringify(jsonStr));
}
});
});
});
</script>
</head>
<body>
<div id="result"></div>
<form name="contact" id="contact" method="post">
Amount : <input type="text" name="amount" id="amount"/><br/>
firstName : <input type="text" name="firstName" id="firstName"/><br/>
lastName : <input type="text" name="lastName" id="lastName"/><br/>
email : <input type="text" name="email" id="email"/><br/>
<input type="button" value="Get It!" name="submit" id="submit"/>
</form>
</body>
myxmlfeed.php
<?php
$amount = $_POST["amount"];
$firstName = $_POST["firstName"];
$lastName = $_POST["lastName"];
$email = $_POST["email"];
if(isset($amount)){
$data = array(
"amount" => $amount,
"firstName" => $firstName,
"lastName" => $lastName,
"email" => $email
);
echo json_encode($data);
}
?>
<?php
$age=array("id"=>"3","title"=>"first entry","date_start"=>"2014-11-22","date_end"=>"2014-11-30");
echo json_encode($age);
?>
试着这样做，告诉我它是如何为你工作的：
<?php
$amount = $_POST["amount"];
$firstName = $_POST["firstName"];
$lastName = $_POST["lastName"];
$email = $_POST["email"];
if(isset($amount)){
$age=array("id"=>"3","title"=>"first entry","date_start"=>"2014-11-22","date_end"=>"2014-11-30");
$data = array(
"amount" => $amount,
"firstName" => $firstName,
"lastName" => $lastName,
"email" => $email
);
echo json_encode(array_merge($data, $age));
}
?>
如果这是你需要的，不要忘记接受答案！