Android Room - 使用LIKE选择查询

Question

我正在尝试进行查询以搜索名称中包含文本的所有对象:

@Query("SELECT * FROM hamster WHERE name LIKE %:arg0%")
fun loadHamsters(search: String?): Flowable<List<Hamster>>

消息:

Error:no viable alternative at input 'SELECT * FROM hamster WHERE name LIKE %'
Error:There is a problem with the query: [SQLITE_ERROR] SQL error or missing database (near "%": syntax error)
Error:Unused parameter: arg0

我也在尝试:

@Query("SELECT * FROM hamster WHERE name LIKE '%:arg0%'")
fun loadHamsters(search: String?): Flowable<List<Hamster>>

消息:

Error:Unused parameter: arg0

如何解决这个问题？

Answer 1

您可以使用SQLite字符串连接进行连接.

@Query("SELECT * FROM hamster WHERE name LIKE '%' || :search || '%'")
fun loadHamsters(search: String?): Flowable<List<Hamster>>

Answer 2

您应该%在输入查询中包含字符 - 而不是在查询本身中.

试试这个:

@Query("SELECT * FROM hamster WHERE name LIKE :arg0")
fun loadHamsters(search: String?): Flowable<List<Hamster>>

那你的String search价值应该是这样的:

search = "%fido%";
loadHamsters(search);

此外,绑定参数名称应该与变量名称匹配,因此arg0它应该看起来像:

@Query("SELECT * FROM hamster WHERE name LIKE :search")
fun loadHamsters(search: String?): Flowable<List<Hamster>>

Answer 3

dao 与 Room 数据库中使用的LIKE关键字的一些示例。

1. 从Search_Query开始

@Query("SELECT * FROM hamster WHERE name LIKE :search_query || '%'")
    fun loadHamsters(search_query: String?): Flowable<List<Hamster>>

2. 以Search_Query结尾

@Query("SELECT * FROM hamster WHERE name LIKE '%' || :search_query ")
   fun loadHamsters(search_query: String?): Flowable<List<Hamster>>

3. 获取所有具有Search_Query 的内容

@Query("SELECT * FROM hamster WHERE name LIKE '%' || :search_query || '%'")
   fun loadHamsters(search_query: String?): Flowable<List<Hamster>>

4. 使用 NOT LIKE 对于具有Search_Query 的数据除外

@Query("SELECT * FROM hamster WHERE name NOT LIKE '%' || :search_query || '%'")
   fun loadHamsters(search_query: String?): Flowable<List<Hamster>>