- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
鉴于有 3 张 table
@Entity
data class Pet(
@PrimaryKey
val id: String,
val name: String,
val colour: String,
height: Int
ownerId: String,
householdId: String
)
@Entity
data class Owner(
@PrimaryKey
val id: String,
val name: String,
val address: String,
val telephoneNumer: Int
)
@Entity
data class Household(
@PrimaryKey
val id: String,
val name: String,
val address: String,
val region: String
)
执行返回
Pet
中的所有条目的查询的最佳方式是什么? ,
Owner
和
Household
在哪里
name
包含“Dav”
Pet("Dave"...) // Pet
Owner("David"...) //Owner
Owner("Davina"...) //Owner
Household("Davenport Close"...) //Address
最佳答案
如果您想要的只是一个 LIST,那么您可以使用自定义 POJO 和使用 UNION 的查询。
例如,您可以有一个 POJO,例如:-
data class searchPOJO(
val id: String,
val type: String,
val name: String
)
和一个查询,例如:-
@Query("SELECT id,'Pet' AS type,name FROM pet WHERE name LIKE :search UNION SELECT id, 'owner' AS type, name FROM owner WHERE name LIKE :search UNION SELECT id, 'household' AS type, name FROM household WHERE name LIKE :search")
abstract fun search(search: String): List<searchPOJO>
你会像这样调用它:-
dao.search("%dav%")
db = TheDatabase.getInstance(this)
dao = db.getAllDao()
dao.insert(Pet("pet1","Dave","brown",10,"Fred","1 Maria Cresent"))
dao.insert(Pet("pet2","George","red",11,"Bert","2 Somewhere Place"))
dao.insert(Owner("owner1","David","10 Somewhere Place",1111111111))
dao.insert(Owner("owner2","Laura","14 Overthere Way", 222222222))
dao.insert(Owner("owner3","Davina","3 Wayland Way",333333333))
dao.insert(Household("h1","Davenport Close","1 Davenport Close","region 1"))
dao.insert(Household("h2","Daintree House","100 Smith Street","region 2"))
for(s: searchPOJO in dao.search("%dav%")) {
Log.d("DBINFO","ID is ${s.id} Name is ${s.name} Type is ${s.type}")
}
结果是 :-
D/DBINFO: ID is h1 Name is Davenport Close Type is household
D/DBINFO: ID is owner1 Name is David Type is owner
D/DBINFO: ID is owner3 Name is Davina Type is owner
D/DBINFO: ID is pet1 Name is Dave Type is Pet
如果您想要在每次查找时返回每个对象中的一个(1 个有效对象和其他 2 个无效对象(例如空白))并指示返回的类型,那么您可以使用类似以下的 POJO:-
data class searchPOJOAll (
val type: String,
@Embedded(prefix = Companion.pet_prefix)
val pet: Pet,
@Embedded(prefix = owner_prefix)
val owner: Owner,
@Embedded(prefix = household_prefix)
val household: Household
) {
companion object {
const val pet_prefix: String = "pet_"
const val owner_prefix: String = "owner_"
const val household_prefix: String = "household_"
}
}
@Query("SELECT 'PET' AS type, " +
" id AS " + searchPOJOAll.pet_prefix + "id," +
" name AS " + searchPOJOAll.pet_prefix + "name," +
" colour AS " + searchPOJOAll.pet_prefix + "colour," +
" height AS " + searchPOJOAll.pet_prefix + "height," +
" ownerId AS " + searchPOJOAll.pet_prefix + "ownerId," +
" householdId AS " + searchPOJOAll.pet_prefix + "householdId," +
" '' AS " + searchPOJOAll.owner_prefix + "id," +
" '' AS " + searchPOJOAll.owner_prefix + "name," +
" '' AS " + searchPOJOAll.owner_prefix + "address," +
" '' AS " + searchPOJOAll.owner_prefix + "telephoneNumer," +
" '' AS " + searchPOJOAll.household_prefix + "id," +
" '' AS " + searchPOJOAll.household_prefix + "name," +
" '' AS " + searchPOJOAll.household_prefix + "address," +
" '' AS " + searchPOJOAll.household_prefix + "region" +
" FROM pet WHERE name LIKE :search " +
" UNION SELECT 'OWNER' AS type, " +
" '' AS " + searchPOJOAll.pet_prefix + "id," +
" '' AS " + searchPOJOAll.pet_prefix + "name," +
" '' AS " + searchPOJOAll.pet_prefix + "colour," +
" '' AS " + searchPOJOAll.pet_prefix + "height," +
" '' AS " + searchPOJOAll.pet_prefix + "ownerId," +
" '' AS " + searchPOJOAll.pet_prefix + "householdId," +
" id AS " + searchPOJOAll.owner_prefix + "id," +
" name AS " + searchPOJOAll.owner_prefix + "name," +
" address AS " + searchPOJOAll.owner_prefix + "address," +
" telephoneNumer AS " + searchPOJOAll.owner_prefix + "telephoneNumer," +
" '' AS " + searchPOJOAll.household_prefix + "id," +
" '' AS " + searchPOJOAll.household_prefix + "name," +
" '' AS " + searchPOJOAll.household_prefix + "address," +
" '' AS " + searchPOJOAll.household_prefix + "region" +
" FROM owner WHERE name LIKE :search " +
" UNION SELECT 'HOUSEHOLD' AS type," +
" '' AS " + searchPOJOAll.pet_prefix + "id," +
" '' AS " + searchPOJOAll.pet_prefix + "name," +
" '' AS " + searchPOJOAll.pet_prefix + "colour," +
" '' AS " + searchPOJOAll.pet_prefix + "height," +
" '' AS " + searchPOJOAll.pet_prefix + "ownerId," +
" '' AS " + searchPOJOAll.pet_prefix + "householdId," +
" '' AS " + searchPOJOAll.owner_prefix + "id," +
" '' AS " + searchPOJOAll.owner_prefix + "name," +
" '' AS " + searchPOJOAll.owner_prefix + "address," +
" '' AS " + searchPOJOAll.owner_prefix + "telephoneNumer," +
" id AS " + searchPOJOAll.household_prefix + "id," +
" name AS " + searchPOJOAll.household_prefix + "name," +
" address AS " + searchPOJOAll.household_prefix + "address," +
" region AS " + searchPOJOAll.household_prefix + "region" +
" FROM household WHERE name LIKE :search"
)
abstract fun searchAll(search: String): List<searchPOJOAll>
扩展之前的演示代码:-
/* List of all 3 object types (only one of which is of use) */
for(s: searchPOJOAll in dao.searchAll("%dav%")) {
if(s.type == "PET") {
Log.d("DBINFO","${s.type}>>>" + getPetString(s.pet))
}
if (s.type == "OWNER") {
Log.d("DBINFO","${s.type}>>>" + getOwnerString(s.owner))
}
if (s.type == "HOUSEHOLD") {
Log.d("DBINFO","${s.type}>>>" +getHouseholdString(s.household))
}
}
并具有以下功能:-
fun getPetString(pet: Pet):String {
return "ID is ${pet.id} Petname = ${pet.name} colour is ${pet.colour} height is ${pet.height} ownerID is ${pet.ownerId} householId is ${pet.householdId}"
}
fun getOwnerString(owner: Owner): String {
return "ID is ${owner.id} Ownername is ${owner.name} address is ${owner.address} telno is ${owner.telephoneNumer}"
}
fun getHouseholdString(household: Household): String {
return "ID is ${household.id} name is ${household.name} address is ${household.address} region is ${household.region}"
}
结果(对于相同的数据)将是:-
D/DBINFO: HOUSEHOLD>>>ID is h1 name is Davenport Close address is 1 Davenport Close region is region 1
D/DBINFO: OWNER>>>ID is owner1 Ownername is David address is 10 Somewhere Place telno is 1111111111
D/DBINFO: OWNER>>>ID is owner3 Ownername is Davina address is 3 Wayland Way telno is 333333333
D/DBINFO: PET>>>ID is pet1 Petname = Dave colour is brown height is 10 ownerID is Fred householId is 1 Maria Cresent
尽管根据评论Android文档中提供的示例似乎创建了一个新的组合模型,而不是使用POJO从上述3个表中返回单独的实体,例如:-
data class PetWithOwnerAndHousehold (
@Embedded
val pet: Pet,
@Relation( entity = Owner::class, parentColumn = "ownerId",entityColumn = "id")
val owner: Owner,
@Relation(entity = Household::class, parentColumn = "householdId", entityColumn = "id")
val household: Household
)
和一个查询/道,例如:-
@Query("SELECT pet.* FROM pet JOIN owner ON owner.id = pet.ownerId JOIN household ON household.id = pet.householdId WHERE pet.name LIKE :search OR owner.name LIKE :search OR household.name LIKE :search")
abstract fun getPetWithOwnerAndHousehold(search: String): List<PetWithOwnerAndHousehold>
将返回宠物、它的所有者和它的 houdehold(假设关系(pet->owener 和 pet->houdhold)是有效的(在上面的数据中它们不是))如果是 pat 名称、所有者的姓名或家庭名称匹配搜索项。
dao.insert(Pet("p1","adava","grey",12,"owner1","h2")) /* multiple hits pet name and owner name*/
dao.insert(Pet("p2","lady","blue",13,"owner2","h2")) /* no hits */
dao.insert(Pet("p3","X","pink",14,"owner2","h1")) /* one hit household name */
然后使用
查询的结果%dav% 因为搜索字符串是:-
D/DBINFO: Pet ID is p1 pet's name is adava OwnerID is owner1 owner's name is David etc. HouseholdID is h2 household's name is Daintree House
D/DBINFO: Pet ID is p3 pet's name is X OwnerID is owner2 owner's name is Laura etc. HouseholdID is h1 household's name is Davenport Close
关于android - 使用单个查询在 Android Room 中搜索多个表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/69245197/
我在我的应用程序中使用 Hibernate Search。其中一个子集合被映射为 IndexedEmbedded。子对象有两个字段,一个是 id,另一个是日期(使用日期分辨率到毫秒)。当我搜索 id=
The App Engine Search API有一个 GeoPoint 字段。可以用它来进行半径搜索吗?例如,给定一个 GeoPoint,查找位于特定半径内的所有文档。 截至目前,它看起来像 Ge
客户对我正在做的员工管理项目提出了这个新要求,以允许他们的用户进行自定义 bool 搜索。 基本上允许他们使用:AND、OR、NOT、括号和引号。 实现它的最佳方法是什么?我检查了 mysql,它们使
很想知道哪个更快 - 如果我有一个包含 25000 个键值对的数组和一个包含相同信息的 MySQL 数据库,搜索哪个会更快? 非常感谢大家! 最佳答案 回答这个问题的最好方法是执行基准测试。 关于ph
我喜欢 smartcase,也喜欢 * 和 # 搜索命令。但我更希望 * 和 # 搜索命令区分大小写,而/和 ?搜索命令遵循 smartcase 启发式。 是否有隐藏在某个地方我还没有找到的设置?我宁
我有以下 Marklogic 查询,当在查询控制台中运行时,它允许我检索具有管理员权限的系统用户: xquery version "1.0-ml"; import schema namespace b
我希望当您搜索例如“A”时,所有以“A”开头的全名都会出现。因此,如果名为“Andreas blabla”的用户将显示 我现在有这个: $query = "SELECT full_name, id,
我想在我的网站上添加对人名的搜索。好友列表已经显示在页面上。 我喜欢 Facebook 这样做的方式,您开始输入姓名,Facebook 只会显示与查询匹配的好友。 http://cl.ly/2t2V0
您好,我在我的网站上进行搜索时遇到此错误。 Fatal error: Uncaught Error: Call to undefined function mysql_connect() in /ho
声明( 叠甲 ):鄙人水平有限,本文为作者的学习总结,仅供参考。 1. 搜索介绍 搜索算法包括深度优先搜索(DFS)和广度优先搜索(BFS)这两种,从起点开始,逐渐扩大
我正在为用户列表使用 FuturBuilder。我通过 futur: fetchpost() 通过 API 获取用户。在专栏的开头,我实现了一个搜索栏。那么我该如何实现我的搜索栏正在搜索呢? Cont
我正在使用 MVC5,我想搜索结果并停留在同一页面,这是我在 Controller (LiaisonsProjetsPPController) 中执行搜索操作的方法: public ActionRes
Azure 搜索中的两种方法 Upload 与 MergeOrUpload 之间有什么区别。 他们都做完全相同的事情。即,如果文档不存在,它们都会上传文档;如果文档已经存在,则替换该文档。 由于这两种
实际上,声音匹配/搜索的当前状态是什么?我目前正在远程参与规划一个 Web 应用程序,该应用程序将包含和公开记录的短音频剪辑(最多 3-5 秒,人名)的数据库。已经提出了一个问题,是否可以实现基于用户
在商业应用程序中,具有数百个面并不罕见。当然,并非所有产品都带有所有这些标记。 但是在搜索时,我需要添加一个方面查询字符串参数,其中列出了我想要返回的所有方面。由于我事先不知道相关列表,因此我必须在查
当我使用nvcc 5.0编译.cu文件时,编译器会为我提供以下信息。 /usr/bin/ld: skipping incompatible /usr/local/cuda-5.0/lib/libcud
我正在使用基于丰富的 Lucene 查询解析器语法的 Azure 搜索。我将“~1”定义为距离符号的附加参数)。但我面临的问题是,即使存在完全匹配,实体也没有排序。 (例如,“blue~1”将返回“b
我目前有 3 个类,一个包含 GUI 的主类,我在其中调用此方法,一个包含数据的客户类,以及一个从客户类收集数据并将其放入数组列表的 customerList 类,以及还包含搜索数组列表方法。 我正在
假设我有多个 6 字符的字母数字字符串。 abc123、abc231、abc456、cba123、bac231 和 bac123 。 基本上我想要一个可以搜索和列出所有 abc 实例的选择语句。 我只
我有这个表 "Table"内容: +--------+ | Serial | +--------+ | d100m | <- expected result | D100M | <- expect
我是一名优秀的程序员,十分优秀!