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

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)

其他回答

集功能,食谱容易和简单

我准备了一组函数,它们在某种程度上是相似的,但如果你知道如何利用它,就会展示新的功能以及Javascript已经达到的简单性。


让一些基本常数


let data;
const URLAPI = "https://gorest.co.in/public/v1/users";
function setData(dt) {
    data = dt;
}

最简单的


// MOST SIMPLE ONE 
function makeRequest1() {       
    fetch(URLAPI)
        .then(response => response.json()).then( json => setData(json))
        .catch(error => console.error(error))
        .finally(() => {
            console.log("Data received 1 --> ", data);
            data = null;
    });
}

使用承诺和异步工具的变化


// ASYNC FUNCTIONS 
function makeRequest2() { 
    fetch(URLAPI)
        .then(async response => await response.json()).then(async json => await setData(json))
        .catch(error => console.error(error))
        .finally(() => {
            console.log("Data received 2 --> ", data);
            data = null;            
        });
}

function makeRequest3() {    
    fetch(URLAPI)
        .then(async response => await response.json()).then(json => setData(json))
        .catch(error => console.error(error))
        .finally(() => {
            console.log("Data received 3 --> ", data);
            data = null;
        });
}

// Better Promise usages
function makeRequest4() {
    const response = Promise.resolve(fetch(URLAPI).then(response => response.json())).then(json => setData(json) ).finally(()=> {
        console.log("Data received 4 --> ", data);

    })
}

一个线性函数的演示!!


// ONE LINER STRIKE ASYNC WRAPPER FUNCTION 
async function makeRequest5() {
    console.log("Data received 5 -->", await Promise.resolve(fetch(URLAPI).then(response => response.json().then(json => json ))) );
}

值得一提的是——> @Daniel De León可能是最干净的函数*

(async () =>
    console.log(
            (await (await fetch( URLAPI )).json())
            )
)();

上面的答案-> By @tggagne显示了HttpClient API的功能。

使用Fetch也可以实现同样的效果。根据此使用MDN获取展示了如何将INIT作为第二个参数传递,基本上打开了使用经典方法(get, post…)轻松配置API的可能性。


// Example POST method implementation:
async function postData(url = '', data = {}) {
  // Default options are marked with *
  const response = await fetch(url, {
    method: 'POST', // *GET, POST, PUT, DELETE, etc.
    mode: 'cors', // no-cors, *cors, same-origin
    cache: 'no-cache', // *default, no-cache, reload, force-cache, only-if-cached
    credentials: 'same-origin', // include, *same-origin, omit
    headers: {
      'Content-Type': 'application/json'
      // 'Content-Type': 'application/x-www-form-urlencoded',
    },
    redirect: 'follow', // manual, *follow, error
    referrerPolicy: 'no-referrer', // no-referrer, *no-referrer-when-downgrade, origin, origin-when-cross-origin, same-origin, strict-origin, strict-origin-when-cross-origin, unsafe-url
    body: JSON.stringify(data) // body data type must match "Content-Type" header
  });
  return response.json(); // parses JSON response into native JavaScript objects
}

postData('https://example.com/answer', { answer: 42 })
  .then(data => {
    console.log(data); // JSON data parsed by `data.json()` call
  });

Node

在节点(服务器端)上无法获取

最简单的解决方案(2021年底)是使用Axios。

$ npm install axios

然后运行:

const axios = require('axios');
const request = async (url) => await (await axios.get( url ));
let response = request(URL).then(resp => console.log(resp.data));

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.
  });

在你的小部件的信息。plist文件,不要忘记设置你的AllowNetworkAccess键为true。

function get(path) {
    var form = document.createElement("form");
    form.setAttribute("method", "get");
    form.setAttribute("action", path);
    document.body.appendChild(form);
    form.submit();
}


get('/my/url/')

同样的事情也可以为post request做。 看看这个链接JavaScript post请求像一个表单提交

没有回调的版本

var i = document.createElement("img");
i.src = "/your/GET/url?params=here";