如何使用JavaScript进行AJAX调用,而不使用jQuery?


当前回答

xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
        alert(this.responseText);
    }
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();

其他回答

var load_process = false;
function ajaxCall(param, response) {

 if (load_process == true) {
     return;
 }
 else
 { 
  if (param.async == undefined) {
     param.async = true;
 }
 if (param.async == false) {
         load_process = true;
     }
 var xhr;

 xhr = new XMLHttpRequest();

 if (param.type != "GET") {
     xhr.open(param.type, param.url, true);

     if (param.processData != undefined && param.processData == false && param.contentType != undefined && param.contentType == false) {
     }
     else if (param.contentType != undefined || param.contentType == true) {
         xhr.setRequestHeader('Content-Type', param.contentType);
     }
     else {
         xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
     }


 }
 else {
     xhr.open(param.type, param.url + "?" + obj_param(param.data));
 }

 xhr.onprogress = function (loadTime) {
     if (param.progress != undefined) {
         param.progress({ loaded: loadTime.loaded }, "success");
     }
 }
 xhr.ontimeout = function () {
     this.abort();
     param.success("timeout", "timeout");
     load_process = false;
 };

 xhr.onerror = function () {
     param.error(xhr.responseText, "error");
     load_process = false;
 };

 xhr.onload = function () {
    if (xhr.status === 200) {
         if (param.dataType != undefined && param.dataType == "json") {

             param.success(JSON.parse(xhr.responseText), "success");
         }
         else {
             param.success(JSON.stringify(xhr.responseText), "success");
         }
     }
     else if (xhr.status !== 200) {
         param.error(xhr.responseText, "error");

     }
     load_process = false;
 };
 if (param.data != null || param.data != undefined) {
     if (param.processData != undefined && param.processData == false && param.contentType != undefined && param.contentType == false) {
             xhr.send(param.data);

     }
     else {
             xhr.send(obj_param(param.data));

     }
 }
 else {
         xhr.send();

 }
 if (param.timeout != undefined) {
     xhr.timeout = param.timeout;
 }
 else
{
 xhr.timeout = 20000;
}
 this.abort = function (response) {

     if (XMLHttpRequest != null) {
         xhr.abort();
         load_process = false;
         if (response != undefined) {
             response({ status: "success" });
         }
     }

 }
 }
 }

function obj_param(obj) {
var parts = [];
for (var key in obj) {
    if (obj.hasOwnProperty(key)) {
        parts.push(encodeURIComponent(key) + '=' + encodeURIComponent(obj[key]));
    }
}
return parts.join('&');
}

我的ajax调用

  var my_ajax_call=ajaxCall({
    url: url,
    type: method,
    data: {data:value},
    dataType: 'json',
    async:false,//synchronous request. Default value is true 
    timeout:10000,//default timeout 20000
    progress:function(loadTime,status)
    {
    console.log(loadTime);
     },
    success: function (result, status) {
      console.log(result);
    },
      error :function(result,status)
    {
    console.log(result);
     }
      });

对于中止先前的请求

      my_ajax_call.abort(function(result){
       console.log(result);
       });

使用XMLHttpRequest。

简单的GET请求

httpRequest = new XMLHttpRequest()
httpRequest.open('GET', 'http://www.example.org/some.file')
httpRequest.send()

简单的POST请求

httpRequest = new XMLHttpRequest()
httpRequest.open('POST', 'http://www.example.org/some/endpoint')
httpRequest.send('some data')

我们可以通过可选的第三个参数指定请求应该是异步(true)(默认值)或同步(false)。

// Make a synchronous GET request
httpRequest.open('GET', 'http://www.example.org/some.file', false)

我们可以在调用httpRequest.send()之前设置头信息

httpRequest.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');

我们可以通过设置httpRequest来处理响应。在调用httpRequest.send()之前,onreadystatechange函数

httpRequest.onreadystatechange = function(){
  // Process the server response here.
  if (httpRequest.readyState === XMLHttpRequest.DONE) {
    if (httpRequest.status === 200) {
      alert(httpRequest.responseText);
    } else {
      alert('There was a problem with the request.');
    }
  }
}

这可能会有帮助:

function doAjax(url, callback) {
    var xmlhttp = window.XMLHttpRequest ? new XMLHttpRequest() : new ActiveXObject("Microsoft.XMLHTTP");

    xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
            callback(xmlhttp.responseText);
        }
    }

    xmlhttp.open("GET", url, true);
    xmlhttp.send();
}
xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
        alert(this.responseText);
    }
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();

我正在寻找一种方法,包括承诺与ajax和排除jQuery。HTML5 Rocks上有一篇文章谈到了ES6的承诺。(您可以使用像Q这样的承诺库填充)您可以使用我从文章中复制的代码片段。

function get(url) {
  // Return a new promise.
  return new Promise(function(resolve, reject) {
    // Do the usual XHR stuff
    var req = new XMLHttpRequest();
    req.open('GET', url);

    req.onload = function() {
      // This is called even on 404 etc
      // so check the status
      if (req.status == 200) {
        // Resolve the promise with the response text
        resolve(req.response);
      }
      else {
        // Otherwise reject with the status text
        // which will hopefully be a meaningful error
        reject(Error(req.statusText));
      }
    };

    // Handle network errors
    req.onerror = function() {
      reject(Error("Network Error"));
    };

    // Make the request
    req.send();
  });
}

注意:我还写了一篇关于这方面的文章。