从搜索引擎，我结束了这个主题的非json发布数据与fetch，所以我认为我将添加这个。

对于非json，你不必使用表单数据。你可以简单地设置Content-Type头为application/x-www-form-urlencoded，并使用一个字符串:

fetch('url here', {
    method: 'POST',
    headers: {'Content-Type':'application/x-www-form-urlencoded'}, // this line is important, if this content-type is not set it wont work
    body: 'foo=bar&blah=1'
});

构建body字符串的另一种方法是使用库，而不是像上面那样将其输入。例如query-string或qs包中的stringify函数。所以使用它看起来像这样:

import queryString from 'query-string'; // import the queryString class

fetch('url here', {
    method: 'POST',
    headers: {'Content-Type':'application/x-www-form-urlencoded'}, // this line is important, if this content-type is not set it wont work
    body: queryString.stringify({for:'bar', blah:1}) //use the stringify object of the queryString class
});

**//POST a request**


const createTodo = async (todo) =>  {
    let options = {
        method: "POST",
        headers: {
            "Content-Type":"application/json",
        },
        body: JSON.stringify(todo)      
    }
    let p = await fetch("https://jsonplaceholder.typicode.com/posts", options);
    let response = await p.json();
    return response;
}

**//GET request**

const getTodo = async (id) => {
    let response = await fetch('https://jsonplaceholder.typicode.com/posts/' + id);
  let r = await response.json();
  return r;
}
const mainFunc = async () => {
    let todo = {
            title: "milan7",
            body: "dai7",
            userID: 101
        }
    let todor = await createTodo(todo);
    console.log(todor);
    console.log(await getTodo(5));
}
mainFunc()

在花了一些时间后，反向工程jsFiddle，试图生成有效负载-有一个效果。

请采取眼睛(照顾)在线返回response.json();这里的回应不是回应，而是承诺。

var json = {
    json: JSON.stringify({
        a: 1,
        b: 2
    }),
    delay: 3
};

fetch('/echo/json/', {
    method: 'post',
    headers: {
        'Accept': 'application/json, text/plain, */*',
        'Content-Type': 'application/json'
    },
    body: 'json=' + encodeURIComponent(JSON.stringify(json.json)) + '&delay=' + json.delay
})
.then(function (response) {
    return response.json();
})
.then(function (result) {
    alert(result);
})
.catch (function (error) {
    console.log('Request failed', error);
});

jsFiddle: http://jsfiddle.net/egxt6cpz/46/ && Firefox > 39 && Chrome > 42

// extend FormData for direct use of js objects
Object.defineProperties(FormData.prototype, { 
    load: {
       value: function (d) {
                   for (var v in d) {
                      this.append(v, typeof d[v] === 'string' ? d[v] : JSON.stringify(d[v]));
                   }
               }
           }
   })

var F = new FormData;
F.load({A:1,B:2});

fetch('url_target?C=3&D=blabla', {
        method: "POST", 
          body: F
     }).then( response_handler )

你可以用await/async做得更好。

http请求参数:

const _url = 'https://jsonplaceholder.typicode.com/posts';
let _body = JSON.stringify({
  title: 'foo',
  body: 'bar',
  userId: 1,
});
  const _headers = {
  'Content-type': 'application/json; charset=UTF-8',
};
const _options = { method: 'POST', headers: _headers, body: _body };

使用干净的async/await语法:

const response = await fetch(_url, _options);
if (response.status >= 200 && response.status <= 204) {
  let data = await response.json();
  console.log(data);
} else {
  console.log(`something wrong, the server code: ${response.status}`);
}

使用老式fetch().then().then():

fetch(_url, _options)
  .then((res) => res.json())
  .then((json) => console.log(json));

你只需要检查响应是否正常，因为调用没有返回任何东西。

var json = {
    json: JSON.stringify({
        a: 1,
        b: 2
    }),
    delay: 3
};

fetch('/echo/json/', {
    method: 'post',
    headers: {
        'Accept': 'application/json, text/plain, */*',
        'Content-Type': 'application/json'
    },
    body: 'json=' + encodeURIComponent(JSON.stringify(json.json)) + '&delay=' + json.delay
})
.then((response) => {if(response.ok){alert("the call works ok")}})
.catch (function (error) {
    console.log('Request failed', error);
});