Mavzu: Android ilovalarini ishlab chiqishda Kotlin dasturlash tilidan foydalanish

Oqim ishga tushirilmoqda

Shuni ta'kidlash kerakki, qabul qilingan ma'lumotlarga terminal operatsiyasi qo'llanilmaguncha, asenkron ip ishlamaydi, masalan collect():

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import kotlinx.coroutines.flow.* suspend fun main(){ val numberFlow = getNumbers() // поток создан, но не запущен println("numberFlow has created") println("launch collect function") numberFlow.collect { number -> println(number) } // запуск потока } fun getNumbers() = flow{ println("numberFlow has started") for(item in 1..5){ emit(item * item) } }

Dasturning konsol chiqishi:
numberFlow yaratdi
yig'ish funktsiyasini ishga tushiring
numberFlow boshlandi
1
4
to'qqiz
16
25

Birinchi, oxirgi, yagona funktsiyalar

Usul first()ro'yxatning birinchi ob'ektini oladi:

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import kotlinx.coroutines.flow.* suspend fun main(){ val userFlow = listOf("Tom", "Bob", "Kate", "Sam", "Alice").asFlow() val firstUser = userFlow.first() println("First User: $firstUser") // First User: Tom }

Usul first(), shuningdek , mantiqiy ob'ektni qaytaradigan parametr sifatida shart funktsiyasini olishi mumkin. Keyin first() bu shartga mos keladigan oqimning birinchi elementini qaytaradi:

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import kotlinx.coroutines.flow.* suspend fun main(){ val userFlow = listOf("Tom", "Bob", "Kate", "Sam", "Alice").asFlow() val firstUser = userFlow.first{ name-> name.length > 3} println("First User: $firstUser") // First User: Kate }

Bu erda biz 3 belgidan ortiq uzunlikdagi birinchi elementni olamiz. Oqimda bu "Keyt" qatori.
Biroq, oqim elementlarining hech biri shartga javob bermasa, vaziyat yuzaga kelishi mumkin:

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val userFlow = listOf("Tom", "Bob", "Sam").asFlow() val firstUser = userFlow.first{ name-> name.length > 3}

Yoki oqim juda bo'sh:

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val userFlow = listOf().asFlow() val firstUser = userFlow.first()

Ikkala holatda ham dastur ishlayotganda istisno qilinadi. Bunday holda, biz istisnoni, albatta, ishlatishimiz mumkin try..catch. Biroq, muqobil sifatida, Kotlin oqim bo'sh bo'lsa yoki uning elementlaridan hech biri shartga mos kelmasa, firstOrNull()qaytaradigan usulni taqdim etadi null:

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val userFlow = listOf().asFlow() val firstUser1 = userFlow.firstOrNull() val firstUser2 = userFlow.firstOrNull{ name-> name.length > 3}

Bunday holda, biz natija qiymatini tekshirishimiz mumkin null:

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if(firstUser1 == null) println("User not found") else println("First User: $firstUser1")