javascript - Firebase JS API : How to avoid `off()` getting executed before `on()` is called/finished executing?

Question

在下面的代码中，off() 在 on() 之前执行。发生这种情况是因为

The DataSnapshot passed to the callback will be for the location at which on() was called. It won't trigger until the entire contents has been synchronized.

如文档中所述:https://firebase.google.com/docs/reference/js/firebase.database.Reference#on

quotesRef.orderByChild('index').on('value', function(snapshot) {
    snapshot.forEach(function(childSnapShot) {
        vm.allQuotes.push({
            key: childSnapShot.key,
            quoteTxt: childSnapShot.val().quote
        })
    })
})

quotesRef.off('value')

如何构造上述代码，以便仅在内容完全同步或实际调用 on 时调用 off()。

谢谢

Answer 1

要在从数据库读取数据后调用 off，请将其移至回调中:

quotesRef.orderByChild('index').on('value', function(snapshot) {
    snapshot.forEach(function(childSnapShot) {
        vm.allQuotes.push({
            key: childSnapShot.key,
            quoteTxt: childSnapShot.val().quote
        })
    })
    quotesRef.off('value')
})

但正如 André 评论的那样，这与使用 once() 完全相同:

quotesRef.orderByChild('index').once('value', function(snapshot) {
    snapshot.forEach(function(childSnapShot) {
        vm.allQuotes.push({
            key: childSnapShot.key,
            quoteTxt: childSnapShot.val().quote
        })
    })
})