ios - 在使用下一个事件之前，如何在 flattenMap block 中等待信号完成？

Question

这是我的伪代码:

[[[@[ctx1,ctx2]rac_sequence]signalWithScheduler:RACScheduler.immediateScheduler]

flattenMap:^RACStream *(id ctx) 
{
    // first flatten map
    return [RACSignal createSignal:^(id <RACSubscriber> subscriber) 
    {
        [executeRequestAsynch 
             onDone:^{
             [subscriber sendNext:ctx];  
             [subscriber sendCompleted]; 
    }
    ]

    }]
    flattenMap:^RACStream *(id ctx) {

    // second flattenMap  
    }];

}

现在这是我希望在订阅时发生的事情:

1)  ctx1 should get fed to the first flattenMap block
2)  the Server Request "executeRequestAsynch" should be called
3)  on completion of the serverRequest the second flattenMap should be called with ctx1
4)  ctx2 should get fed to the first flattenMap block
5)  the Server Request "executeRequestAsynch" should be called
6)  on completion of the serverRequest the second flattenMap should be called with ctx1

但是这种情况发生了:

1)  ctx1 gets fed to the first flattenMap block
2)  the Server Request "executeRequestAsynch" is called
3)  ctx2 gets fed to the first flattenMap block
4)  the Server Request "executeRequestAsynch" is called
5)   on completion of the serverRequest the second flattenMap is called with ctx1
6)  on completion of the serverRequest the second flattenMap is called with ctx2

如何实现第一个场景？

---

回答这个似乎可以解决问题!谢谢:输出是:

 Step 1
   Step 1 child
     Step 1    child child
 Step 2
   Step 2 child
     Step 2    child child
 done all

我仍然想知道 defer 在下面的 senario 中会实现什么

-(RACSignal*) executeRequestAsynch:(NSString*) ctx {
 return [RACSignal createSignal:^RACDisposable * (id<RACSubscriber> subscriber) {
     NSLog(@"  %@ child",ctx);
     [subscriber sendCompleted];
                 return nil;
        }];

}

-(RACSignal*) executeRequestAsynch2:(NSString*) ctx {
return [RACSignal createSignal:^RACDisposable * (id<RACSubscriber> subscriber) {
    NSLog(@"  %@    child child",ctx);
    [subscriber sendCompleted];
    return nil;
}];

 }

  - (void)viewDidLoad
{
    [super viewDidLoad];
    RACSignal *contexts = [[@[ @"Step 1", @"Step 2" ] 
    rac_sequence]  signalWithScheduler:RACScheduler.immediateScheduler];
    RACSignal *ne = [[contexts map:^(id ctx) {
    NSLog(@"%@",ctx);
    return [[self executeRequestAsynch:ctx] concat: [self executeRequestAsynch2:ctx ]];}]concat] ;
   [ne subscribeCompleted:^{
    NSLog(@"done all");
}];
}

here is my final real solution

Answer 1

我不能 100% 确定我理解您想要实现的目标。我的解释是这样的:

RACSignal *contexts = [@[ ctx1, ctx2 ] things.rac_sequence signalWithScheduler:RACScheduler.immediateScheduler];
[[contexts map:^(id ctx) {
    return [executeRequestAsynch concat:[RACSignal defer:^{
        // Returns another signal involving ctx?
    }]];
}] concat];

这会将每个上下文映射到 executeRequestAsynch，然后将它们连接起来，以便串行完成。

是这样还是我误解了你的意思？