在我的下面的函数中,addToFlatList
只调用一次,即使我知道我的数据库中有几个项目要添加。似乎拳头addToFlatList
永远不会解决?我究竟做错了什么?
photosSnapshot.forEach(async function(childSnapshot) {
await addToFlatList(childSnapshot.key, childSnapshot.val())(dispatch);
});
addToFlatList函数:
const addToFlatList = (photoId, photoObj) => async(dispatch) => {
database.ref('users').child(photoObj.author).once('value').then((userSnapshot) => {
var userInfo = userSnapshot.val();
dispatch({type: "GOT_USER", payload: userInfo});
}).catch(error => {
dispatch({type: "GOT_ERROR"});
});
}
更新:
试图像这样返回派遣。 addToFlatList
仍然只被召唤一次。
const addToFlatList = async(photoId, photoObj) => {
return (dispatch) => {
database.ref('users').child(photoObj.author).once('value').then((userSnapshot) => {
var userInfo = userSnapshot.val();
dispatch({type: "GOT_USER", payload: userInfo});
}).catch(error => {
dispatch({type: "GOT_ERROR"});
});
}
}
还试过这个:
const addToFlatList = async(photoId, photoObj) => {
database.ref('users').child(photoObj.author).once('value').then((userSnapshot) => {
return (dispatch) => {
// never hit this point
var userInfo = userSnapshot.val();
dispatch({type: "GOT_USER", payload: userInfo});
}
}).catch(error => {
dispatch({type: "GOT_ERROR"});
});
}
你必须回复承诺:
const addToFlatList = (photoId, photoObj) => (dispatch) => {
return database.ref('users').child(photoObj.author).once('value').then((userSnapshot) => {
// ^^^^^^
var userInfo = userSnapshot.val();
return dispatch({type: "GOT_USER", payload: userInfo});
}).catch(error => {
return dispatch({type: "GOT_ERROR"});
});
};
或者,您必须等待承诺,以便您的async
函数不会过早结束:
const addToFlatList = (photoId, photoObj) => async (dispatch) => {
try {
const userSnapshot = await database.ref('users').child(photoObj.author).once('value');
// ^^^^^
var userInfo = userSnapshot.val();
return dispatch({type: "GOT_USER", payload: userInfo});
} catch(error) {
return dispatch({type: "GOT_ERROR"});
}
}