我有一个聊天应用,当用户点击其他用户的个人资料时,我会使用以下代码检查此用户是否已经存在聊天:
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"<span class=sf-dump-key>abcd</span>" => "<span class=sf-dump-str title="4 characters">efgh</span>"
"<span class=sf-dump-key>wddf</span>" => "<span class=sf-dump-str title="8 characters">asdfsadf</span>"
"<span class=sf-dump-key>asfdf</span>" => "<span class=sf-dump-str title="8 characters">asdfsadf</span>"
"<span class=sf-dump-key>asedf</span>" => "<span class=sf-dump-str title="8 characters">asdfsadf</span>"
"<span class=sf-dump-key>asdif</span>" => "<span class=sf-dump-str title="8 characters">asdfsadf</span>"
"<span class=sf-dump-key>asodf</span>" => "<span class=sf-dump-str title="8 characters">asdfsadf</span>"
"<span class=sf-dump-key>aspdf</span>" => "<span class=sf-dump-str title="8 characters">asdfsadf</span>"
</samp>]
</pre><script>Sfdump("sf-dump-1820305802")</script>
这是我的数据库快照:
由于有一个名为“Tim Cook”的值,我点击了Tim Cook的个人资料,它应该打开现有的聊天活动,但每次打开新的聊天活动并且我的查询无法正常工作。我想知道我在做什么错误,这也是很好的技巧,还是我应该以另一种方式处理它? 提前谢谢。
答案 0 :(得分:0)
因为你想念孩子而无法工作。正如我在屏幕截图中看到的那样,chatId
丢失了。要解决此问题,请更改以下代码行:
FirebaseUser firebaseUser = FirebaseAuth.getInstance().getCurrentUser();
String uid = firebaseUser.getUid();
DatabaseReference rootRef = FirebaseDatabase.getInstance().getReference();
DatabaseReference userIdRef = rootRef.child("chats").child(uid);
Query query = userIdRef.orderByChild("otherUsersName").equalTo("Tim Cook");
ValueEventListener eventListener = new ValueEventListener() {
@Override
public void onDataChange(DataSnapshot dataSnapshot) {
for(DataSnapshot ds : dataSnapshot.getChildren()) {
if (ds.exists()) {
Log.d("TAG", ds.child("otherUsersName").getValue(String.class));
//open the exsiting chat
} else {
//start a new chat
}
}
}
@Override
public void onCancelled(DatabaseError databaseError) {}
};
query.addListenerForSingleValueEvent(eventListener);