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我不是jQuery的專家,考慮我更新鮮。這裏是我的代碼,它不負責由Request Body提交jQuery JSON數據。如何在jQuery中通過Request Body提交JSON數據?
<!doctype html>
<html lang="en">
<head>
<title>jQuery Data submitted by JSON Body Request</title>
<script type="text/javascript" src="jquery-1.3.2.js"></script>
<script type="text/javascript">
$.ajax({
url : "/",
type: "POST",
data: [
{id: 1, name: "Shahed"},
{id: 2, name: "Hossain"}
],
contentType: "application/json; charset=utf-8",
dataType : "json",
success : function(){
console.log("Pure jQuery Pure JS object");
}
});
</script>
</head>
<body>
<p>
Example of submission JS Object by JSON Body Request<br/>
Its could submitted mass amount of data by Message body<br/>
It's secured and faster than any data submission .
</p>
</body>
</html>
源後出現了:
Shahed=undefined&Hossain=undefined
但所需的帖子來源是:
[{"id":1,"name":"Shahed"},{"id":2,"name":"Hossain"}]
如何獲取所需的源後爲每個請求的身體?
問題不清 –
首先,您必須實際提交json數據而不是對象。 –
如果您熟悉firebug調試器,那麼您將在控制檯的所有選項卡中看到每個請求。你會在哪裏得到Post Source。 Firebug控制檯負責在'Post Source'上顯示有效的'JSON''請求''Body' –