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本片文章主要介绍Flow上下游处理，上游一个Flow使用map，上游两个Flow使用zip，上游三个Flow及以上使用combine 

1、下面代码展示了upStreamFlow作为上游，downStreamFlow作为下游，通过对upStreamFlow使用map操作符函数将upStreamFlow转换为新的Flow对象，每个元素都通过lambda表达式进行处理，并生成以“Number：”为开头的字符串。

package com.cqzimport kotlinx.coroutines.flow.*suspend fun main() {val upStreamFlow: Flow<Int> = flow {for (i in 1..5) {emit(i)}}val downStreamFlow: Flow<String> = upStreamFlow.map {"Number:$it"}upStreamFlow.collect {println("upStreamFlow:$it")}downStreamFlow.collect {println("downStreamFlow:$it")}}

运行以上代码将会得到以下结果

upStreamFlow:1
upStreamFlow:2
upStreamFlow:3
upStreamFlow:4
upStreamFlow:5
downStreamFlow:Number:1
downStreamFlow:Number:2
downStreamFlow:Number:3
downStreamFlow:Number:4
downStreamFlow:Number:5

2、上游两个Flow可以使用zip操作符函数将两个Flow合并为新的Flow对象。这个函数接受两个asdFlow作为参数，并通过lambda表达式将它们的元素进行组合。

package com.cqzimport kotlinx.coroutines.flow.*suspend fun main() {val upStreamFlow: Flow<Int> = flow {for (i in 1..5) {emit(i)}}val upStream2Flow: Flow<Int> = flow {for (i in 6..10) {emit(i)}}upStreamFlow.zip(upStream2Flow) { num1,num2->num1 + num2}.collect {println("zip:$it")}}

运行以上代码将会得到以下结果

zip:7
zip:9
zip:11
zip:13
zip:15

3、上游有三个Flow或以上的时候可以使用combine操作符函数，将三个Flow合并为一个Flow

package com.cqzimport kotlinx.coroutines.flow.*suspend fun main() {val flow1 = flowOf(1,2,3)val flow2 = flowOf(false,true,false)val flow3 = flowOf("A","B","C")val combinedFlow = combine(flow1,flow2,flow3) {i, b, s ->CombinedResult(i,b,s)}combinedFlow.collect {println("combinedFlow:$it")}
}
data class CombinedResult(val i: Int,val b: Boolean,val str: String)

运行以上代码将会得到以下结果

combinedFlow:CombinedResult(i=1, b=false, str=A)
combinedFlow:CombinedResult(i=2, b=false, str=A)
combinedFlow:CombinedResult(i=2, b=true, str=A)
combinedFlow:CombinedResult(i=2, b=true, str=B)
combinedFlow:CombinedResult(i=3, b=true, str=B)
combinedFlow:CombinedResult(i=3, b=false, str=B)
combinedFlow:CombinedResult(i=3, b=false, str=C)