我有以下jQuery代码:
dataString = 'test'; // array?
$.ajax({
type: "POST",
url: "tokenize.php",
data: {
data: dataString
},
cache: false,
success: function (data) {
returnedvalue = data;
console.log(data); //alert isn't for debugging
}
});
这个jQuery代码工作正常,但我想要一个简单的JavaScript代码,我无法弄清楚该怎么做。我只在Stack Overflow的帮助下编写了这段代码。
我已经看到可以使用XMLHttpRequest
完成此操作:
var http = new XMLHttpRequest();
var url = "tokenize.php";
var params = "lorem=ipsum&name=binny"; // What will be done here in my case?
http.open("POST", url, true);
//Send the proper header information along with the request
http.setRequestHeader("Content-length", params.length);
http.setRequestHeader("Connection", "close");
// Call a function when the state changes.
http.onreadystatechange = function () {
if (http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
http.send(params);
答案 0 :(得分:0)
application/x-www-form-urlencoded
数据的格式为:
key=value&key=value&key=value
通过encodeURIComponent
运行每个键和值,以处理具有特殊含义或编码中不允许的字符。