Android Room - 使用 LIKE 选择查询 [英] Android Room - Select query with LIKE

问题描述

我正在尝试进行查询以搜索名称包含文本的所有对象:

I'm trying to make a query to search all objects whose names contain text:

@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?

推荐答案

您应该将 % 字符括在输入查询中 - 而不是查询本身.

You should enclose the % characters in your input query - not in the query itself.

例如试试这个:

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

那么您的 String search 值应该如下所示:

Then your String search value should look like:

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

此外，绑定参数名称应该与变量名称相匹配，因此它应该看起来像:

Furthermore, the binding parameter name should match the variable name, so rather than arg0 it should look like:

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

这篇关于Android Room - 使用 LIKE 选择查询的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！