我试图使用fetch POST一个JSON对象。
根据我的理解，我需要将一个字符串化的对象附加到请求的主体，例如:
fetch("/echo/json/",
{
headers: {
'Accept': 'application/json',
'Content-Type': 'application/json'
},
method: "POST",
body: JSON.stringify({a: 1, b: 2})
})
.then(function(res){ console.log(res) })
.catch(function(res){ console.log(res) })
当使用jsfiddle的JSON回显时，我希望看到我发送的对象({a: 1, b: 2})回来，但这不会发生- chrome devtools甚至不显示JSON作为请求的一部分，这意味着它没有被发送。
当前回答
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
});
其他回答
我认为，我们不需要将JSON对象解析为字符串，如果远程服务器接受JSON到他们的请求，只需运行:
const request = await fetch ('/echo/json', {
headers: {
'Content-type': 'application/json'
},
method: 'POST',
body: { a: 1, b: 2 }
});
比如curl请求
curl -v -X POST -H 'Content-Type: application/json' -d '@data.json' '/echo/json'
如果远程服务不接受json文件作为主体，只需发送一个dataForm:
const data = new FormData ();
data.append ('a', 1);
data.append ('b', 2);
const request = await fetch ('/echo/form', {
headers: {
'Content-type': 'application/x-www-form-urlencoded'
},
method: 'POST',
body: data
});
比如curl请求
curl -v -X POST -H 'Content-type: application/x-www-form-urlencoded' -d '@data.txt' '/echo/form'
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
2021年的答案:以防你在这里寻找如何使用async/await或promises来进行GET和POST Fetch api请求，而不是axios。
我使用jsonplaceholder伪API来演示:
使用async/await获取api GET请求:
const asyncGetCall = async () => {
try {
const response = await fetch('https://jsonplaceholder.typicode.com/posts');
const data = await response.json();
// enter you logic when the fetch is successful
console.log(data);
} catch(error) {
// enter your logic for when there is an error (ex. error toast)
console.log(error)
}
}
asyncGetCall()
使用async/await获取api POST请求:
const asyncPostCall = async () => {
try {
const response = await fetch('https://jsonplaceholder.typicode.com/posts', {
method: 'POST',
headers: {
'Content-Type': 'application/json'
},
body: JSON.stringify({
// your expected POST request payload goes here
title: "My post title",
body: "My post content."
})
});
const data = await response.json();
// enter you logic when the fetch is successful
console.log(data);
} catch(error) {
// enter your logic for when there is an error (ex. error toast)
console.log(error)
}
}
asyncPostCall()
使用Promises获取请求:
fetch('https://jsonplaceholder.typicode.com/posts')
.then(res => res.json())
.then(data => {
// enter you logic when the fetch is successful
console.log(data)
})
.catch(error => {
// enter your logic for when there is an error (ex. error toast)
console.log(error)
})
使用承诺POST请求:
fetch('https://jsonplaceholder.typicode.com/posts', {
method: 'POST',
headers: {
'Content-Type': 'application/json',
},
body: JSON.stringify({
// your expected POST request payload goes here
title: "My post title",
body: "My post content."
})
})
.then(res => res.json())
.then(data => {
// enter you logic when the fetch is successful
console.log(data)
})
.catch(error => {
// enter your logic for when there is an error (ex. error toast)
console.log(error)
})
使用Axios获取请求:
const axiosGetCall = async () => {
try {
const { data } = await axios.get('https://jsonplaceholder.typicode.com/posts')
// enter you logic when the fetch is successful
console.log(`data: `, data)
} catch (error) {
// enter your logic for when there is an error (ex. error toast)
console.log(`error: `, error)
}
}
axiosGetCall()
使用Axios的POST请求:
const axiosPostCall = async () => {
try {
const { data } = await axios.post('https://jsonplaceholder.typicode.com/posts', {
// your expected POST request payload goes here
title: "My post title",
body: "My post content."
})
// enter you logic when the fetch is successful
console.log(`data: `, data)
} catch (error) {
// enter your logic for when there is an error (ex. error toast)
console.log(`error: `, error)
}
}
axiosPostCall()
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);
});
