我需要在JavaScript中做一个HTTP GET请求。最好的方法是什么?
我需要在Mac OS X的dashcode小部件中做到这一点。
我需要在JavaScript中做一个HTTP GET请求。最好的方法是什么?
我需要在Mac OS X的dashcode小部件中做到这一点。
当前回答
原型让它变得非常简单
new Ajax.Request( '/myurl', {
method: 'get',
parameters: { 'param1': 'value1'},
onSuccess: function(response){
alert(response.responseText);
},
onFailure: function(){
alert('ERROR');
}
});
其他回答
对于那些使用AngularJs的人来说,它是$http.get:
$http.get('/someUrl').
success(function(data, status, headers, config) {
// this callback will be called asynchronously
// when the response is available
}).
error(function(data, status, headers, config) {
// called asynchronously if an error occurs
// or server returns response with an error status.
});
我不熟悉Mac OS的Dashcode小部件,但如果他们让你使用JavaScript库和支持xmlhttprequest,我会使用jQuery,做这样的事情:
var page_content;
$.get( "somepage.php", function(data){
page_content = data;
});
一种支持旧浏览器的解决方案:
function httpRequest() {
var ajax = null,
response = null,
self = this;
this.method = null;
this.url = null;
this.async = true;
this.data = null;
this.send = function() {
ajax.open(this.method, this.url, this.asnyc);
ajax.send(this.data);
};
if(window.XMLHttpRequest) {
ajax = new XMLHttpRequest();
}
else if(window.ActiveXObject) {
try {
ajax = new ActiveXObject("Msxml2.XMLHTTP.6.0");
}
catch(e) {
try {
ajax = new ActiveXObject("Msxml2.XMLHTTP.3.0");
}
catch(error) {
self.fail("not supported");
}
}
}
if(ajax == null) {
return false;
}
ajax.onreadystatechange = function() {
if(this.readyState == 4) {
if(this.status == 200) {
self.success(this.responseText);
}
else {
self.fail(this.status + " - " + this.statusText);
}
}
};
}
这段代码可能有点过分,但绝对是安全的。
用法:
//create request with its porperties
var request = new httpRequest();
request.method = "GET";
request.url = "https://example.com/api?parameter=value";
//create callback for success containing the response
request.success = function(response) {
console.log(response);
};
//and a fail callback containing the error
request.fail = function(error) {
console.log(error);
};
//and finally send it away
request.send();
短的、干净的:
const http = new XMLHttpRequest() http。打开(“得到”,“https://api.lyrics.ovh/v1/toto/africa”) http.send () http。onload = () => console.log(http.response)
现代、干净、简洁
fetch('https://baconipsum.com/api/?type=1')
让url = 'https://baconipsum.com/api/?type=all-meat¶s=1&start-with-lorem=2'; //只发送GET请求而不等待响应 fetch (url); //使用then来等待结果 获取(url)。然后(r = > r.json()。then(j=> console.log('\nREQUEST 2',j))); //或async/await (异步()= > console.log('\nREQUEST 3', await(await fetch(url)).json()) ) (); 打开Chrome控制台网络选项卡查看请求