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); 

 });