Android Room-使用LIKE选择查询 [英] Android Room - Select query with LIKE
本文介绍了Android Room-使用LIKE选择查询的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试查询所有名称包含文本的对象:
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
该如何解决?
推荐答案
您应该在输入查询中添加%
字符,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>>
然后您的String search
值应类似于:
search = "%fido%";
loadHamsters(search);
此外,绑定参数名称应与变量名称匹配,因此,与arg0
相比,它应类似于:
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!
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