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。
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