下面是一个完整的应用程序json解决方案:

// 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>

确保你的后端API可以解析JSON。

例如，在Express JS中:

import bodyParser from 'body-parser'
app.use(bodyParser.json())

我也遇到过类似的问题，使用相同的帖子和这个链接，我已经解决了我的问题。

 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);

此链接已完成信息。

短而现代

您可以使用FormData捕获表单输入值并通过fetch发送它们

fetch(form.action, {method:'post', body: new FormData(form)});

function send() { let form = document.forms['inputform']; fetch(form.action, {method:'post', body: new FormData(form)}); } <form name="inputform" action="somewhere" method="post"> <input value="person" name="user"> <input type="hidden" value="password" name="pwd"> <input value="place" name="organization"> <input type="hidden" value="key" name="requiredkey"> </form> <!-- I remove type="hidden" for some inputs above only for show them --><br> Look: chrome console>network and click <button onclick="send()">send</button>

不需要插件!

选择下面的代码并将其拖到书签栏中(如果你看不到它，从浏览器设置中启用)，然后编辑该链接:

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);

这是所有!现在你可以访问任何网站，并点击书签栏的按钮!

注意:

上面的方法使用XMLHttpRequest方法发送数据，因此，在触发脚本时必须处于相同的域中。这就是为什么我更喜欢用模拟表单提交发送数据，它可以将代码发送到任何域-这里是为此编写的代码:

 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); 

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

这帮助我，因为我想只使用xmlHttpRequest和post对象作为表单数据:

function sendData(data) {
  var XHR = new XMLHttpRequest();
  var FD  = new FormData();

  // Push our data into our FormData object
  for(name in data) {
    FD.append(name, data[name]);
  }

  // Set up our request
  XHR.open('POST', 'https://example.com/cors.php');

  // Send our FormData object; HTTP headers are set automatically
  XHR.send(FD);
}

https://developer.mozilla.org/en-US/docs/Learn/HTML/Forms/Sending_forms_through_JavaScript