带有参数的Android Room @Delete [英] Android Room @Delete with parameters

问题描述

我知道我不能在查询中使用DELETE(顺便说一句是很遗憾的)，我会收到以下错误消息:

I know I can't use DELETE in a query (that is a shame by the way), I will get the following error:

<i>Error:error: Observable query return type (LiveData, Flowable etc) can only be used with SELECT queries that directly or indirectly (via @Relation, for example) access at least one table.</i>

但是我不能使用@Delete(WHERE... xxx) 那么如何通过参数删除特定行呢?

But I can't use @Delete(WHERE... xxx) So how do I delete a specific row by a parameter?

推荐答案

房间的美在于我们玩的是物体.根据要求，您可以使用 对于Kotlin:

The beauty of room is, we play with the objects. As per requirement you can use for kotlin:

@Delete
fun delete(model: LanguageModel)

对于Java:

@Delete
void delete(LanguageModel model)

它将删除存储在数据库中的具有相同值的确切对象. LanguageModel是我的模型类，它可以完美运行.

it will delete the exact object which is stored in the db with the same values. LanguageModel is my model class and it works perfectly.

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