试图安慰group2
成员,所以在这种情况下结果应该是user2
的,如果它不存在,则不显示:
{
"users":{
"user1":{
"username":"john",
"groups":{
"group1":true,
"group3":true
}
},
"user2": ...,
}
"groups": {
"group1"{
...
},
"group2"{
"group_name":"Moderators",
"members":{
"user2":true
}
}
}
}
我试过:
const ref = firebase.database().ref('groups');
ref.orderByChild("user2").on("child_added", function (snapshot) {
console.log(snapshot.key)
});
但没有运气
如果您尝试列出组 2 的成员:
const ref = firebase.database().ref('groups');
ref.child("group2/members").on("child_added", function (snapshot) {
console.log(snapshot.key)
});
这将打印:
用户2