如何在Firebase表上进行连接-howtodojoinsonFirebasetables

7  

Your code snippet has a nasty side-effect:

你的代码片段有一个讨厌的副作用:

var userId;
tableOne.on('value', function (snapshot) {
    userId = snapshot.val().userId; // line 1 (results like 1,2,3,4,5,6)
    anotherTable.child('userdetails').child(userId).once('value', function(mediaSnap) {
        console.log(userId + ":" + mediaSnap.val().name);
    });
});

You're not declaring userId as a variable, which means that it becomes a global variable in Javascript. And since the callback function executes asynchronously, there is a good chance that the global values will have changed by the time you need it.

您没有将userId声明为变量,这意味着它将成为Javascript中的全局变量。由于回调函数是异步执行的,因此很有可能在需要时全局值会发生变化。

The solution is simply to make userId a local variable of the callback function:

解决方案只是使userId成为回调函数的局部变量:

tableOne.on('value', function (snapshot) {
    var userId = snapshot.val().userId; // line 1 (results like 1,2,3,4,5,6)
    anotherTable.child('userdetails').child(userId).once('value', function(mediaSnap) {
        console.log(userId + ":" + mediaSnap.val().name);
    });
});

This will ensure that each value of userId is captured inside the function.

这将确保在函数内捕获userId的每个值。

0  

Solution for list join:

列表连接的解决方案:

tableOne.orderByKey().on("value", function (snapshot) {
    //console.log(snapshot.val());
    snapshot.forEach(function (data) {
        tableTwo.once('value').then(function (info) {
            info = info.val();
        });
    });
});