ES6 fetch函数与后台交互实现

var request = new XMLHttpRequest();request.onreadystatechange = (e) = >{  if (request.readyState !== 4) {    return;  }  if (request.status === 200) {    console.log('success', request.responseText);  } else {    console.warn('error');  }};request.open('GET', 'https://mywebsite.com/endpoint.php');request.send();

fetch('https://mywebsite.com/endpoint/', {  method: 'POST',  headers: {    'Accept': 'application/json',    'Content-Type': 'application/json',  },  body: JSON.stringify({    firstParam: 'yourValue',    secondParam: 'yourOtherValue',  })}).then(function(res) {　　console.log(res)})

fetch('https://mywebsite.com/endpoint/', {  method: 'POST',  headers: {    'Content-Type': 'application/x-www-form-urlencoded',  },  body: 'key1=value1&key2=value2'}).then(function(res) {  　　console.log(res)})

fetch('https://mywebsite.com/endpoint/').then(function(res) {  if (res.ok) {    res.json().then(function(obj) {      // 这样数据就转换成json格式的了    })  }}, function(ex) {  console.log(ex)})

fetch('doAct.action', {   method: 'post',   headers: {    "Content-type": "application/x-www-form-urlencoded; charset=UTF-8"   },   body: 'foo=bar&lorem=ipsum'  }) .then(json)  .then(function (data) {   console.log('Request succeeded with JSON response', data);  })  .catch(function (error) {   console.log('Request failed', error);  });