在没有服务器端代码的情况下在 Firebase 中搜索 [英] Searching in Firebase without server side code

问题描述

我正在尝试从 Firebase 获取名称包含给定字符串的所有用户.例如，如果我有这些用户:

I am trying to get all the users having the name that contains a given string from Firebase. For example, if I have these users:

Devid, Andy, Bob

我希望所有用户的名称都包含D"，因此我希望结果如下:

I would like to get all the users having the name that contains a 'D' so I expect this as result:

Devid, Andy

这是我目前的 Firebase 结构:

This is my Firebase's structure at the moment:

由于 Firebase 区分大小写，我创建了一个包含小写名称的属性 name_.

Since Firebase is case sensitive I've created an attribute name_ that contains the lowercase name.

使用 startAt 和 endAt 我可以获得名称以定义的字符串开头的所有用户

Using startAt and endAt I can get all the users with the name starting with a defined string

ref.orderByChild("name_").startAt(text).endAt(text+"uf8ff").on('value', ...);

但这只会给我以给定字符串开头的名称的用户，例如，如果文本是D"，我会得到:

But this gives me only the users having the name that starts with a given string, for example if text is 'D' I'll get:

Devid

1) 目前我的查询意味着给我所有具有以给定字符串开头的 name_ 的用户"；有没有办法让它的意思是给我所有名称包含给定字符串的用户"?否

1) At the moment my query means, "give me all the users having name_ that starts with a given string" is there a way to make it mean "give me all the users which name contains a given string"? NO

Firebase 查询与全文搜索没有任何相似之处运营商.要完成这些，您要么必须集成一个外部全文搜索引擎，还是搞出一个很精心的自定义索引方案.Firebase 和索引/搜索

Firebase Queries don't have anything similar to full-text search operators. To accomplish those, you'll either have to integrate an external full-text search engine, or come up with a very elaborate custom indexing scheme. Firebase and indexing/search

2) 目前我不想拥有服务器端代码，有什么好的和有效的方式来实现自定义索引?

2) At the moment I don't want to have server side code, what can be a good and efficient way to implement custom indexes?

谢谢

推荐答案

好的 - 没有办法完全按照您当前的结构做您想做的事情.

Ok - there's no way to do exactly what you want with your current structure.

然而这只是我的脑海中浮现:

However this just popped into my head:

users:
  user_1234
    first_name: "Devid"
    components:
       "D": true
       "e": true
       "v": true
       "i": true
       "d": true
  user_5678
    first_name: "Andy"
    components:
       "A": true
       "n": true
       "d": true
       "y": true
  user_1010
    first_name: "Bob"
    components:
       "B": true
       "o": true
       "b": true

这里有一些 ObjC 代码来实现它(并且已经过测试！)

and here's some ObjC Code to make it happen (and it's tested!)

Firebase *ref = [myRootRef childByAppendingPath:@"users"];

FQuery *q1 = [ref queryOrderedByChild:@"components/b"];
FQuery *q2 = [q1 queryEqualToValue:@1];

[q2 observeEventType:FEventTypeChildAdded withBlock:^(FDataSnapshot *snapshot) {

    NSLog(@"%@", snapshot.value);

}];

此代码返回 Bob.

要获取所有 'd' 人，请将components/b"更改为components/d"

To get all of the 'd' people, change the "components/b" to "components/d"

您可以变得非常疯狂并添加更​​多组合以扩展您的搜索能力

You can get really crazy and add more combinations to expand your search capability

users:
  user_1234
    first_name: "Devid"
    components:
       "D": true
       "e": true
       "v": true
       "i": true
       "d": true
       "De": true
       "Dev": true
       "Devi": true
       "Devid": true
       "ev": true
       "evi": true
       "evid": true
       ... etc

编写几行代码来迭代名称并写出组合非常简单.

It would pretty simple to code up a few lines of code to iterate over the name and write out the combinations.

显然，将所有名字读入快照，将它们转储到数组中并(在 ObjC 中)使用 NSPredicate 来提取您需要的内容会更有效(如果您的数据集有限).

Obviously it would be way more efficient (if you have a limited data set) to just read all of the first names into snapshot, dump them into an array and (in ObjC) use an NSPredicate to pull out what you need.

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