在消耗下一个事件之前，如何在flattenMap块中等待信号完成？

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  
    }];

}

现在这是我想要订阅时发生的事情：

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，发生这种情况：

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

如何使第一种情况发生？

---

回答这个似乎做好的工作！ 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

我仍然想知道延缓在senario底下会实现什么

-(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");
}];
}

这是我最终的解决方案

Answer 1

我不是100％肯定我了解您要完成的工作。 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，以便按顺序完成它们。

是这样吗，还是我误会了你？