异步函数必须返回一个布尔值 [英] Async function must return a boolean value
问题描述
我有一个方法可以在表单标签中调用 onsubmit 事件.
I have a method that I am calling on the onsubmit event in the form tag.
所以我需要从方法返回一个真或假.
So I need a true or false to be returned from the method.
我使用 API 来检索数据,并根据 API 的响应,返回 true 或 false.但是因为它是一个正在运行的异步函数,所以我无法正确等待 API 的响应,分析它然后返回我的决定.
I use an API to retrieve data, and according to the response from the API, I return true or false. But because it is an async function thats running, I cant get it right to wait for the response from the API, analyze it and then return my decision.
关于如何解决这个问题的任何想法
Any ideas on how I could solve this problem
function GetPolygonID()
{
document.getElementById("displayerror").innerHTML = "";
var retrievedpoly = document.getElementById('polygondetails').value;
var parts = retrievedpoly.split('coordinates');
var parttoadd = parts[1].substring(0, parts[1].length - 2) + "}";
console.log(parttoadd);
var myx = '{"name":"Polygon OneTwoThree","geo_json":{"type":"Feature","properties":{},"geometry":{"type":"Polygon","coordinates' + parttoadd;
var url = 'http://api.agromonitoring.com/agro/1.0/polygons?appid=apiid';
const request = async() => {
const response = await fetchPoly(url, myx);
const data = await response.json();
const errorCheck = await CheckInfo(data);
console.log("2: " + errorCheck);
return await errorCheck;
};
return request();
}
function CheckInfo(data)
{
let flag = false;
console.log(data);
if (JSON.stringify(data).includes("Geo json Area is invalid. Available range: 1 - 3000 ha"))
{
var myval = JSON.stringify(data);
//myval = myval.replace(/\\n/g,"<br/>");
parts = myval.split("\\n ").join(",").split("\\n");
console.log(parts);
var todisplay = parts[1].substring(10);
todisplay += ("<br/>" + parts[2].substring(10).replace(",", "<br/>").replace("c", "C"));
console.log(todisplay);
document.getElementById("displayerror").innerHTML = todisplay;
} else
{
flag = true;
}
console.log("1:" + flag);
return flag;
}
function fetchPoly(url, data)
{
return fetch(url, {
method: "POST", // *GET, POST, PUT, DELETE, etc.
mode: "cors", // no-cors, cors, *same-origin
cache: "no-cache", // *default, no-cache, reload, force-cache, only-if-cached
credentials: "same-origin", // include, *same-origin, omit
headers: {
"Content-Type": "application/json"
// "Content-Type": "application/x-www-form-urlencoded",
},
redirect: "follow", // manual, *follow, error
referrer: "no-referrer", // no-referrer, *client
body: data // body data type must match "Content-Type" header
});
}
最初我确实用 .then() 尝试过,然后我像这样分解了它,因为我认为在这里返回一个值会更容易.
I did try it with .then(), originally, then I broke it down like this, as I thought it would be easier to return a value here.
本质上我需要 GetPolygonID() 来返回它从 CheckInfo() 获取的布尔值.CheckInfo() 决定表单是否应该提交
Essentially I need GetPolygonID() to return a boolean which it gets from CheckInfo(). CheckInfo() determines if the form should submit or not
有没有想过如何解决这个问题?
Any thought on how I could fix this?
谢谢
推荐答案
GetPolygonID()
函数返回一个 Promise,因此必须使用 await
调用它,或者您可以调用 then
:
GetPolygonID()
function returns a Promise, so it must be either called with await
or you can call then
upon it:
var res = await GetPolygonID();
GetPolygonID().then(res => console.log(res));
你可以使整个函数async
:
async function GetPolygonID() {
document.getElementById("displayerror").innerHTML = "";
var retrievedpoly = document.getElementById('polygondetails').value;
var parts = retrievedpoly.split('coordinates');
var parttoadd = parts[1].substring(0, parts[1].length - 2) + "}";
console.log(parttoadd);
var myx = '{"name":"Polygon OneTwoThree","geo_json":{"type":"Feature","properties":{},"geometry":{"type":"Polygon","coordinates' + parttoadd;
var url = 'http://api.agromonitoring.com/agro/1.0/polygons?appid=apiid';
const response = await fetchPoly(url, myx);
const data = response.json();
const errorCheck = CheckInfo(data);
console.log("2: " + errorCheck);
return errorCheck;
}
使用 async
函数进行表单验证,您可以这样做:
Using an async
function for a form validation, you can do this:
function onSubmit(form) {
GetPolygonID().then(res => res ? form.submit() : null);
return false;
}
...
<form method="POST" onsubmit="return onSubmit(this);">
...
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