美好的一天,我一直想弄清楚如何从列表中获取单个对象,我做了google,但所有主题都展示了如何返回带有已排序对象或类似对象的List
。
我有一个User Class
class User() {
var email: String = ""
var firstname: String = ""
var lastname: String = ""
var password: String = ""
var image: String = ""
var userId: String = ""
constructor(email:String,
firstname: String,
lastname: String,
password: String,
image: String, userId : String) : this() {
this.email = email
this.firstname = firstname
this.lastname = lastname
this.password = password
this.image = image
this.userId = userId
}
}
在Java中,我会写类似
User getUserById(String id) {
User user = null;
for(int i = 0; i < myList.size;i++;) {
if(id == myList.get(i).getUserId())
user = myList.get(i)
}
return user;
}
我如何在Kotlin中获得相同的结果?
答案 0 :(得分:9)
您可以使用find
进行此操作,这将为您提供与给定谓词(或null
,如果没有匹配)相匹配的列表的第一个元素:
val user: User? = myList.find { it.userId == id }
或者,如果您确实确实需要与谓词匹配的最后一个元素(如Java示例代码一样),则可以使用last
:
val user: User? = myList.last { it.userId == id }
答案 1 :(得分:2)
如果您不想处理空对象,请尝试以下操作:
val index = myList.indexOfFirst { it.userId == id } // -1 if not found
if (index >= 0) {
val user = myList[index]
// do something with user
}
答案 2 :(得分:2)
简化
val user: User = myList.single { it.userId == id }
或者如果列表中没有您的过滤器
val user: User? = myList.singleOrNull{ it.userId == id }
答案 3 :(得分:0)
您还可以使用此扩展功能来返回对Pair(Boolean,T)
例如
blue_thermal_printer