所以我尝试将大多数喜欢的数据排序到最少。如果没有对象名称,我该怎么做?因为目前我只有对象的ID?
这是我的Firebase json:
["-KfSODzRPjy46ayCFRBH": {
"_createdAt" = 1489773065234;
"_updatedAt" = 1489773065234;
content = (
"<null>",
Tittle,
"why not there is the nnnd ksksks lalalalalkdkdkdkdk kadsjfakl;",
whhhhhhalalallala
);
imageTitle = 1489773064177;
like = 2;
title = "Third Post";
},
["-KfSODzRPjy46ayCFRBH": {
"_createdAt" = 1489773065234;
"_updatedAt" = 1489773065234;
content = (
"<null>",
Tittle,
"why not there is the nnnd ksksks lalalalalkdkdkdkdk kadsjfakl;",
whhhhhhalalallala
);
imageTitle = 1489773064177;
like = 3;
title = "Third Post";
},
这就是我的阅读方式:
let handleref = databaseRef.child("v1/postmodel").observe(.value, with: { (snapshot) in
let postDict = snapshot.value as! [String : AnyObject]
print(postDict)
})