我有用户表和这段代码。
getOnline代码和连接按钮
var onlineStatus = firebase.database().ref("users/" + firebase.auth().currentUser.uid + "/online");
onlineStatus.set(1);
和
var dbUser = firebase.database();
var refUser = dbUser.ref("users");
refUser.orderByChild("online").equalTo(1).on("value", function(Data){
console.log(Data.val(), Data.key);
});
我已经可以看到在线= 1个用户,但我想随机获得1个在线= 1的用户。
我该怎么做?
更新
我想随意聊天。人们会根据标准进行匹配。喜欢Tinder App。想想看,有一个页面和按钮。当用户按下按钮时,它将匹配任何在线用户。和聊天。
firebase用户;
users
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email: "mail@hotmail.com"
online: 1
profile_picture:"https://scontent.xx.fbcdn.net"
username: "John Snow"
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email: "mail2@hotmail.com"
online: 1
profile_picture:"https://scontent.xx.fbcdn.net"
username: "Jane Snow"
答案 0 :(得分:6)
您可以使用Math.random()功能获取随机数(让我们称之为 n ),然后在线获取 n 用户:
refUser.orderByChild("online").equalTo(1).on("value", function(Data){
console.log(Data.val(), Data.key);
var totalOnline = Data.numChildren(); //get number of online users
var randomNr = Math.random() * totalOnline;
var userIndex = parseInt(randomNr, 10); //parse the random number from double to integer
var currentIndex = 0;
Data.forEach(function(snap){
if(currentIndex==userIndex){
var randomUser = snap.val();
//Do something with your random user
break;
}
currentIndex++;
});
});
另请注意,如果您拥有包含数千名用户的庞大应用,则可能需要将查询限制为50位用户以提高应用的效果:
refUser.orderByChild("online").equalTo(1).limitToFirst(50)
更新:要排除您自己的用户,您可以检查随机用户是否是您。如果是,请转到下一个:
if(currentIndex>=userIndex && snap.key != firebase.auth().currentUser.uid){
var randomUser = snap.val();
//Do something with your random user
break;
}
更新2 :我发现您无法在break
循环中使用forEach
。所以你可以使用提供的解决方案来解决这个问题here(使用BreakException):
refUser.orderByChild("online").equalTo(1).on("value", function(Data){
console.log(Data.val(), Data.key);
var totalOnline = Data.numChildren(); //get number of online users
var randomNr = Math.random() * totalOnline;
var userIndex = parseInt(randomNr, 10); //parse the random number from double to integer
var currentIndex = 0;
var BreakException = {};
try
{
Data.forEach(function(snap){
if(currentIndex==userIndex){
var randomUser = snap.val();
//Do something with your random user
throw BreakException;
}
currentIndex++;
});
}
catch(e){
if(e!== BreakException) throw e;
}
});