如何使用JavaScript进行AJAX调用,而不使用jQuery?
当前回答
尝试使用Fetch Api (Fetch Api)
fetch('http://example.com/movies.json').then(response => response.json()).then(data => console.log(data));
非常清澈,100%香草味。
其他回答
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();
}
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);
}
});
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)
}
}
这是一个没有JQuery的JSFiffle
http://jsfiddle.net/rimian/jurwre07/
function loadXMLDoc() {
var xmlhttp = new XMLHttpRequest();
var url = 'http://echo.jsontest.com/key/value/one/two';
xmlhttp.onreadystatechange = function () {
if (xmlhttp.readyState == XMLHttpRequest.DONE) {
if (xmlhttp.status == 200) {
document.getElementById("myDiv").innerHTML = xmlhttp.responseText;
} else if (xmlhttp.status == 400) {
console.log('There was an error 400');
} else {
console.log('something else other than 200 was returned');
}
}
};
xmlhttp.open("GET", url, true);
xmlhttp.send();
};
loadXMLDoc();
使用“vanilla”(普通)JavaScript:
function loadXMLDoc() {
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == XMLHttpRequest.DONE) { // XMLHttpRequest.DONE == 4
if (xmlhttp.status == 200) {
document.getElementById("myDiv").innerHTML = xmlhttp.responseText;
}
else if (xmlhttp.status == 400) {
alert('There was an error 400');
}
else {
alert('something else other than 200 was returned');
}
}
};
xmlhttp.open("GET", "ajax_info.txt", true);
xmlhttp.send();
}
jQuery:
$.ajax({
url: "test.html",
context: document.body,
success: function() {
$(this).addClass("done");
}
});