使用 XMLHttpRequest 发送 POST 数据
问题描述：
I'd like to send some data using an XMLHttpRequest in JavaScript.
Say I have the following form in HTML:
<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>
How can I write the equivalent using an XMLHttpRequest in JavaScript?
转载于:https://stackoverflow.com/questions/9713058/send-post-data-using-xmlhttprequest
答
The code below demonstrates on how to do this.
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);
答
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');
Or if you can count on browser support you could use 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);
答
Minimal use of
Minimal use of FormData to submit an AJAX request
<!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>
Remarks
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
GETmethod. 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),
FormDatais not supported by IE9 and lower, and default browser on Android 2.3.
答
var util = {
getAttribute: function (dom, attr) {
if (dom.getAttribute !== undefined) {
return dom.getAttribute(attr);
} else if (dom[attr] !== undefined) {
return dom[attr];
} else {
return null;
}
},
addEvent: function (obj, evtName, func) {
//Primero revisar attributos si existe o no.
if (obj.addEventListener) {
obj.addEventListener(evtName, func, false);
} else if (obj.attachEvent) {
obj.attachEvent(evtName, func);
} else {
if (this.getAttribute("on" + evtName) !== undefined) {
obj["on" + evtName] = func;
} else {
obj[evtName] = func;
}
}
},
removeEvent: function (obj, evtName, func) {
if (obj.removeEventListener) {
obj.removeEventListener(evtName, func, false);
} else if (obj.detachEvent) {
obj.detachEvent(evtName, func);
} else {
if (this.getAttribute("on" + evtName) !== undefined) {
obj["on" + evtName] = null;
} else {
obj[evtName] = null;
}
}
},
getAjaxObject: function () {
var xhttp = null;
//XDomainRequest
if ("XMLHttpRequest" in window) {
xhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
return xhttp;
}
};
//START CODE HERE.
var xhr = util.getAjaxObject();
var isUpload = (xhr && ('upload' in xhr) && ('onprogress' in xhr.upload));
if (isUpload) {
util.addEvent(xhr, "progress", xhrEvt.onProgress());
util.addEvent(xhr, "loadstart", xhrEvt.onLoadStart);
util.addEvent(xhr, "abort", xhrEvt.onAbort);
}
util.addEvent(xhr, "readystatechange", xhrEvt.ajaxOnReadyState);
var xhrEvt = {
onProgress: function (e) {
if (e.lengthComputable) {
//Loaded bytes.
var cLoaded = e.loaded;
}
},
onLoadStart: function () {
},
onAbort: function () {
},
onReadyState: function () {
var state = xhr.readyState;
var httpStatus = xhr.status;
if (state === 4 && httpStatus === 200) {
//Completed success.
var data = xhr.responseText;
}
}
};
//CONTINUE YOUR CODE HERE.
xhr.open('POST', 'mypage.php', true);
xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
if ('FormData' in window) {
var formData = new FormData();
formData.append("user", "aaaaa");
formData.append("pass", "bbbbb");
xhr.send(formData);
} else {
xhr.send("?user=aaaaa&pass=bbbbb");
}
答
NO PLUGINS NEEDED!
Just drag any link (i.e. THIS LINK) in BOOKMARK BAR (if you dont see it, enable from Browser Settings), then EDIT that link :
and insert javascript code:
javascript:var my_params = prompt("Enter your parameters", "var1=aaaa&var2=bbbbb"); var Target_LINK = prompt("Enter destination", location.href); function post(path, params) { var xForm = document.createElement("form"); xForm.setAttribute("method", "post"); xForm.setAttribute("action", path); for (var key in params) { if (params.hasOwnProperty(key)) { var hiddenField = document.createElement("input"); hiddenField.setAttribute("name", key); hiddenField.setAttribute("value", params[key]); xForm.appendChild(hiddenField); } } var xhr = new XMLHttpRequest(); xhr.onload = function () { alert(xhr.responseText); }; xhr.open(xForm.method, xForm.action, true); xhr.send(new FormData(xForm)); return false; } parsed_params = {}; my_params.split("&").forEach(function (item) { var s = item.split("="), k = s[0], v = s[1]; parsed_params[k] = v; }); post(Target_LINK, parsed_params); void(0);
That's all! Now you can visit any website, and click that button in BOOKMARK BAR!
NOTE:
The above method sends data using XMLHttpRequest method, so, you have to be on the same domain while triggering the script. That's why I prefer sending data with a simulated FORM SUBMITTING, which can send the code to any domain - here is code for that:
javascript:var my_params=prompt("Enter your parameters","var1=aaaa&var2=bbbbb"); var Target_LINK=prompt("Enter destination", location.href); function post(path, params) { var xForm= document.createElement("form"); xForm.setAttribute("method", "post"); xForm.setAttribute("action", path); xForm.setAttribute("target", "_blank"); for(var key in params) { if(params.hasOwnProperty(key)) { var hiddenField = document.createElement("input"); hiddenField.setAttribute("name", key); hiddenField.setAttribute("value", params[key]); xForm.appendChild(hiddenField); } } document.body.appendChild(xForm); xForm.submit(); } parsed_params={}; my_params.split("&").forEach(function(item) {var s = item.split("="), k=s[0], v=s[1]; parsed_params[k] = v;}); post(Target_LINK, parsed_params); void(0);
答
I have faced similar problem, using the same post and and this link I have resolved my issue.
var http = new XMLHttpRequest();
var url = "MY_URL.Com/login.aspx";
var params = 'eid=' +userEmailId+'&pwd='+userPwd
http.open("POST", url, true);
// Send the proper header information along with the request
//http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
//http.setRequestHeader("Content-Length", params.length);// all browser wont support Refused to set unsafe header "Content-Length"
//http.setRequestHeader("Connection", "close");//Refused to set unsafe header "Connection"
// Call a function when the state
http.onreadystatechange = function() {
if(http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
http.send(params);
This link has completed information.
答
Use modern JavaScript!
I'd suggest looking into fetch. It is the ES5 equivalent and uses Promises. It is much more readable and easily customizable.
const url = "http://example.com";
fetch(url, {
method : "POST",
body: new FormData(document.getElementById("inputform")),
// -- or --
// body : JSON.stringify({
// user : document.getElementById('user').value,
// ...
// })
}).then(
response => response.text() // .json(), etc.
// same as function(response) {return response.text();}
).then(
html => console.log(html)
);More Info:
</div>
答
Here is a complete solution with
application-json:
// Input values will be grabbed by ID
<input id="loginEmail" type="text" name="email" placeholder="Email">
<input id="loginPassword" type="password" name="password" placeholder="Password">
// return stops normal action and runs login()
<button onclick="return login()">Submit</button>
<script>
function login() {
// Form fields, see IDs above
const params = {
email: document.querySelector('#loginEmail').value,
password: document.querySelector('#loginPassword').value
}
const http = new XMLHttpRequest()
http.open('POST', '/login')
http.setRequestHeader('Content-type', 'application/json')
http.send(JSON.stringify(params)) // Make sure to stringify
http.onload = function() {
// Do whatever with response
alert(http.responseText)
}
}
</script>
Ensure that your Backend API can parse JSON.
For example, in Express JS:
import bodyParser from 'body-parser'
app.use(bodyParser.json())