我使用fetch api来获取可能返回的URL:
响应:status = 200,json body = {'user':'abc','id':1}
要么
回复:状态= 400,json body = {'reason':'某种原因'}
要么
回复:状态= 400,json body = {'reason':'其他原因'}
我想创建一个单独的函数 request()
,我在代码的各个部分使用如下:
request('http://api.example.com/').then(
// status 200 comes here
data => // do something with data.id, data.user
).catch(
// status 400, 500 comes here
error => // here error.reason will give me further info, i also want to know whether status was 400 or 500 etc
)
我无法进行200到400,500之间的分割(我试过抛出一个错误) . 当我抛出错误时,我发现很难仍然提取JSON主体(用于error.reason) .
我目前的代码如下:
import 'whatwg-fetch';
/**
* Requests a URL, returning a promise
*/
export default function request(url, options={}) {
console.log('sending api request, url = ' + url)
return fetch(url, options)
.then(checkStatus)
.then(parseJSON)
.then((data) => ({data}))
.catch((err) => ({err}));
}
function checkStatus(response) {
if (response.status >= 200 && response.status < 300) {
return response;
}
const error = new Error(response.statusText);
error.response = response;
throw error;
}
function parseJSON(response) {
return response.json(); // json() is a promise itself
}
我试图通过执行如下操作来解决此问题,通过反转 .then()
调用的顺序,但不起作用
export default function request(url, options) {
return fetch(url, options)
.then(parseJSON) // note that now first calling parseJSON to get not just JSON but also status.
.then(checkStatus) // i.e. Inverted order of the two functions from before
.then((data) => ({data}))
.catch((err) => ({err}));
}
function checkStatus({data, status}) {
if (status >= 200 && status < 300) {
return data;
}
else {
// const error = new Error(response.statusText);
const error = new Error("Something went wrong");
// error.response = response;
error.data = data;
throw error;
}
}
function parseJSON(response) {
let jsonBody
response.json().then(json => {
jsonBody = json // this does not help, i thought it will make jsonBody fill up, but seems its in a diff thread
})
return {
data: jsonBody,
status: response.status // my aim is to send a whole dict with status and data to send it to checkStatus, but this does not work
}
}
2 回答
response.json()
返回异步结果 . 您没有将parseJSON
处的对象从.then()
链接到response.json()
. 要纠正该问题,您可以在parseJSON
调用时返回response.json()
promise并返回包含data
和status
的对象,从.then()
链接到response.json()
这里有一个稍微不同的方法:使用单行程我用ok,status和json-as-object(不是promise)创建一个类似响应的promise,然后我决定如何处理这个对象 . 一般来说,如果response.ok为false,我会拒绝响应,否则我只使用json-data解析 . 网络错误/ json-parse-errors像往常一样被拒绝 .