如何使用JavaScript进行AJAX调用,而不使用jQuery?


当前回答

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();
}

其他回答

您可以使用以下函数:

function callAjax(url, callback){
    var xmlhttp;
    // compatible with IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp = new XMLHttpRequest();
    xmlhttp.onreadystatechange = function(){
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200){
            callback(xmlhttp.responseText);
        }
    }
    xmlhttp.open("GET", url, true);
    xmlhttp.send();
}

你可以在这些链接上尝试类似的解决方案:

https://www.w3schools.com/xml/tryit.asp?filename=tryajax_first https://www.w3schools.com/xml/tryit.asp?filename=tryajax_callback

在浏览器中使用纯JavaScript:

var xhr = new XMLHttpRequest();

xhr.onreadystatechange = function() {
  if (xhr.readyState == XMLHttpRequest.DONE ) {
    if(xhr.status == 200){
      console.log(xhr.responseText);
    } else if(xhr.status == 400) {
      console.log('There was an error 400');
    } else {
      console.log('something else other than 200 was returned');
    }
  }
}

xhr.open("GET", "mock_data.json", true);

xhr.send();

或者如果你想使用Browserify使用node.js来捆绑你的模块。你可以使用超级代理:

var request = require('superagent');
var url = '/mock_data.json';

 request
   .get(url)
   .end(function(err, res){
     if (res.ok) {
       console.log('yay got ' + JSON.stringify(res.body));
     } else {
       console.log('Oh no! error ' + res.text);
     }
 });

你可以根据浏览器获得正确的对象

function getXmlDoc() {
  var xmlDoc;

  if (window.XMLHttpRequest) {
    // code for IE7+, Firefox, Chrome, Opera, Safari
    xmlDoc = new XMLHttpRequest();
  }
  else {
    // code for IE6, IE5
    xmlDoc = new ActiveXObject("Microsoft.XMLHTTP");
  }

  return xmlDoc;
}

有了正确的对象,GET可以被抽象为:

function myGet(url, callback) {
  var xmlDoc = getXmlDoc();

  xmlDoc.open('GET', url, true);

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

  xmlDoc.send();
}

并将邮件发送到:

function myPost(url, data, callback) {
  var xmlDoc = getXmlDoc();

  xmlDoc.open('POST', url, true);
  xmlDoc.setRequestHeader("Content-type", "application/x-www-form-urlencoded");

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

  xmlDoc.send(data);
}

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.');
    }
  }
}

快速代码获取没有jQuery

async  function product_serach(word) {
            var response = await fetch('<?php echo base_url(); ?>home/product_search?search='+word);
            var json = await response.json();
            for (let [key, value] of Object.entries(json)) 
            {
                console.log(json)
            }                                 
        }