[英]Returning results as a promise in Typescript
如何使下面的函数返回一个Promise,以便可以在调用此函数的Page中正确处理它?
getUploads() {
const rootDef = this.db.database.ref();
const uploadsRef = rootDef.child('userUploads').orderByChild('time');
const userRef = rootDef.child("userProfile");
var uploads = [];
uploadsRef.once("value").then((uploadSnaps) => {
uploadSnaps.forEach((uploadSnap) => {
var upload = uploadSnap.val();
userRef.child(uploadSnap.val().user).once("value").then((userSnap) => {
upload.displayName = userSnap.val().displayName;
upload.avatar = userSnap.val().avatar;
uploads.push(upload);
});
});
});
return uploads;
}
我尝试了以下操作,但显示错误。 我应该如何修改?
return new Promise((resolve, reject) => {
resolve(uploads);
});
我将按如下所示调用此方法。
this.db.getUploads().then((uploads) => {
this.allUploads = uploads;
console.log(this.allUploads);
});
我想你可以围绕你的方法的内容与
getUploads() {
return new Promise((resolve, reject) => {
// content of method
resolve(uploads); // instead of "return uploads"
});
}
您可以使用Promise.resolve :
Promise.resolve(value)方法返回使用给定值解析的Promise对象。 如果该值是一个thenable(即具有“ then”方法),则返回的Promise将“跟随”该thenable,并采用其最终状态; 否则,返回的承诺将被履行。
如此简单:
return Promise.resolve(uploads);
但是代码的问题是,在uploadsRef.once("value").then(...)
之前返回该值。
您应该简单地then
返回结果:
return uploadsRef.once("value").then((uploadSnaps) => {
...
return uploads
};
这应该正确处理异步调用并返回所有上载:
getUploads() {
const rootDef = this.db.database.ref();
const uploadsRef = rootDef.child('userUploads').orderByChild('time');
const userRef = rootDef.child("userProfile");
return uploadsRef.once("value").then((uploadSnaps) => {
return Promise.all(uploadSnaps.map(uploadSnap => {
var upload = uploadSnap.val();
return userRef.child(uploadSnap.val().user).once("value").then((userSnap) => {
upload.displayName = userSnap.val().displayName;
upload.avatar = userSnap.val().avatar;
return upload;
});
}));
});
}
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