Firebase检索除一名儿童外的所有儿童 [英] Firebase retrieving all children except one

问题描述

如何检索除Firebase数据库中与id值相等的子项之外的所有子项?
例如，在SQL中，我们可以这样:

How to retrieve all children except one which is equal with id value in Firebase Database?
For instance, in SQL we can do like this:

SELECT * FROM users WHERE id!=user_id

• 用户
  • K4i0dvMU8sMd1AA1lSLwQraHujC3
  • lIWobPViMmZcFiSsOIpvdflcZe53
  • rubSXsacSSYGoh1VwbXtWiFEsWt2
  • uMFNR972cTdr8UfEXoiyIe3EMYb2
  • users
    • K4i0dvMU8sMd1AA1lSLwQraHujC3
    • lIWobPViMmZcFiSsOIpvdflcZe53
    • rubSXsacSSYGoh1VwbXtWiFEsWt2
    • uMFNR972cTdr8UfEXoiyIe3EMYb2

    这是数据库的外观

    推荐答案

    您可以在客户端简单地过滤一个值

    You can simply do this client-side to filter one value

    (伪代码)

    for (DataSnapshot child : ref.getChildren()) {
        User user = child.getValue(User.class);
        if (!user.getId().equals(user_id)) { 
            list.add(value); 
    }
    

    不过，您正在显示firebase生成的UUID，所以我不知道您希望如何提前知道该内容以过滤任何内容.

    You're showing the UUID that firebase generated, though, so I don't know how you expect to know that ahead of time to filter anything.

    如果这些对象中有一些嵌套的user_id键，或者您可以重做数据库以存储已知的用户ID，则可以使用它.

    If there is some nested user_id key within those objects, or you can redo your database to store a known user id, then use that.

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