我想在JavaScript中使用XMLHttpRequest发送一些数据。
假设我在HTML中有以下表单:
<form name="inputform" action="somewhere" method="post">
<input type="hidden" value="person" name="user">
<input type="hidden" value="password" name="pwd">
<input type="hidden" value="place" name="organization">
<input type="hidden" value="key" name="requiredkey">
</form>
如何在JavaScript中使用XMLHttpRequest编写等效内容?
当前回答
下面的代码演示了如何做到这一点。
var http = new XMLHttpRequest();
var url = 'get_data.php';
var params = 'orem=ipsum&name=binny';
http.open('POST', url, true);
//Send the proper header information along with the request
http.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
http.onreadystatechange = function() {//Call a function when the state changes.
if(http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
http.send(params);
如果你有/创建一个对象，你可以用下面的代码把它转换成参数，即:
var params = new Object();
params.myparam1 = myval1;
params.myparam2 = myval2;
// Turn the data object into an array of URL-encoded key/value pairs.
let urlEncodedData = "", urlEncodedDataPairs = [], name;
for( name in params ) {
urlEncodedDataPairs.push(encodeURIComponent(name)+'='+encodeURIComponent(params[name]));
}
其他回答
下面的代码演示了如何做到这一点。
var http = new XMLHttpRequest();
var url = 'get_data.php';
var params = 'orem=ipsum&name=binny';
http.open('POST', url, true);
//Send the proper header information along with the request
http.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
http.onreadystatechange = function() {//Call a function when the state changes.
if(http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
http.send(params);
如果你有/创建一个对象，你可以用下面的代码把它转换成参数，即:
var params = new Object();
params.myparam1 = myval1;
params.myparam2 = myval2;
// Turn the data object into an array of URL-encoded key/value pairs.
let urlEncodedData = "", urlEncodedDataPairs = [], name;
for( name in params ) {
urlEncodedDataPairs.push(encodeURIComponent(name)+'='+encodeURIComponent(params[name]));
}
var xhr = new XMLHttpRequest();
xhr.open('POST', 'somewhere', true);
xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
xhr.onload = function () {
// do something to response
console.log(this.responseText);
};
xhr.send('user=person&pwd=password&organization=place&requiredkey=key');
或者如果你可以指望浏览器的支持，你可以使用FormData:
var data = new FormData();
data.append('user', 'person');
data.append('pwd', 'password');
data.append('organization', 'place');
data.append('requiredkey', 'key');
var xhr = new XMLHttpRequest();
xhr.open('POST', 'somewhere', true);
xhr.onload = function () {
// do something to response
console.log(this.responseText);
};
xhr.send(data);
只是为了让专题读者发现这个问题。我发现，只要你有一个给定的路径，接受的答案就可以工作，但如果你让它为空，它将在IE中失败。以下是我想到的:
function post(path, data, callback) {
"use strict";
var request = new XMLHttpRequest();
if (path === "") {
path = "/";
}
request.open('POST', path, true);
request.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded; charset=UTF-8');
request.onload = function (d) {
callback(d.currentTarget.response);
};
request.send(serialize(data));
}
你可以这样写:
post("", {orem: ipsum, name: binny}, function (response) {
console.log(respone);
})
最小限度地使用FormData来提交AJAX请求
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="X-UA-Compatible" content="IE=Edge, chrome=1"/>
<script>
"use strict";
function submitForm(oFormElement)
{
var xhr = new XMLHttpRequest();
xhr.onload = function(){ alert (xhr.responseText); } // success case
xhr.onerror = function(){ alert (xhr.responseText); } // failure case
xhr.open (oFormElement.method, oFormElement.action, true);
xhr.send (new FormData (oFormElement));
return false;
}
</script>
</head>
<body>
<form method="post" action="somewhere" onsubmit="return submitForm(this);">
<input type="hidden" value="person" name="user" />
<input type="hidden" value="password" name="pwd" />
<input type="hidden" value="place" name="organization" />
<input type="hidden" value="key" name="requiredkey" />
<input type="submit" value="post request"/>
</form>
</body>
</html>
讲话
This does not fully answer the OP question because it requires the user to click in order to submit the request. But this may be useful to people searching for this kind of simple solution. This example is very simple and does not support the GET method. If you are interesting by more sophisticated examples, please have a look at the excellent MDN documentation. See also similar answer about XMLHttpRequest to Post HTML Form. Limitation of this solution: As pointed out by Justin Blank and Thomas Munk (see their comments), FormData is not supported by IE9 and lower, and default browser on Android 2.3.
有一些重复的作品涉及到这一点，但没有人真正阐述它。我将借用公认答案的例子来说明
http.open('POST', url, true);
http.send('lorem=ipsum&name=binny');
为了说明，我过度简化了这一点(我使用http.onload(function(){})而不是那个答案的旧方法)。如果你按原样使用，你会发现你的服务器可能会将POST正文解释为字符串，而不是实际的key=value参数(即PHP不会显示任何$_POST变量)。你必须在http.send()之前传递表单头文件来获取它。
http.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
如果您使用的是JSON而不是url编码的数据，则应该传递application/ JSON
