iOS多个网络请求完成后执行下一步

在开发中,我们很容易遇到这样的需求,需要我们同时做多个网络请求,所有网络请求都完成后才能进行下一步的操作。如下载多个图片,下载完了才能展示。今天我们就来研究一下这个问题的解决方案。

1.首先,我们创建一个项目,然后做一般性的做法,不做任何处理去连续请求一个接口10次:

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//1.无处理
UIButton *Btn1 = [UIButton buttonWithType:UIButtonTypeCustom];
Btn1.frame = CGRectMake(100, 100, 100, 40);
Btn1.backgroundColor = [UIColor grayColor];
[Btn1 setTitle:@"noConduct" forState:UIControlStateNormal];
[Btn1 addTarget:self action:@selector(Btn1) forControlEvents:UIControlEventTouchUpInside];
[self.view addSubview:Btn1];
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//1.noConduct
-(void)Btn1{
NSString *str = @"http://www.jianshu.com/p/6930f335adba";
NSURL *url = [NSURL URLWithString:str];
NSURLRequest *request = [NSURLRequest requestWithURL:url];
NSURLSession *session = [NSURLSession sharedSession];
for (int i=0; i<10; i++) {
NSURLSessionDataTask *task = [session dataTaskWithRequest:request completionHandler:^(NSData * _Nullable data, NSURLResponse * _Nullable response, NSError * _Nullable error) {
NSLog(@"%d---%d",i,i);
}];
[task resume];
}
NSLog(@"end");
}
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2017-12-04 17:10:10.503 DownImage[3289:261033] end
2017-12-04 17:10:10.676 DownImage[3289:261080] 0---0
2017-12-04 17:10:10.704 DownImage[3289:261080] 1---1
2017-12-04 17:10:10.754 DownImage[3289:261096] 4---4
2017-12-04 17:10:10.760 DownImage[3289:261080] 2---2
2017-12-04 17:10:10.800 DownImage[3289:261096] 5---5
2017-12-04 17:10:10.840 DownImage[3289:261080] 7---7
2017-12-04 17:10:10.844 DownImage[3289:261082] 6---6
2017-12-04 17:10:10.846 DownImage[3289:261096] 3---3
2017-12-04 17:10:10.888 DownImage[3289:261096] 8---8
2017-12-04 17:10:10.945 DownImage[3289:261080] 9---9
作者:_清墨
链接:https://www.jianshu.com/p/fb4fb80aefb8
來源:简书
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

很明显,无任何处理情况下,end最先被打印出来,由于网络请求的异步回调,然后各个网络请求的回调顺序是无序的。

2.使用GCD的dispatch_group_t

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//2.group--
-(void)Btn2{
NSString *str = @"http://www.jianshu.com/p/6930f335adba";
NSURL *url = [NSURL URLWithString:str];
NSURLRequest *request = [NSURLRequest requestWithURL:url];
NSURLSession *session = [NSURLSession sharedSession];
dispatch_group_t downloadGroup = dispatch_group_create();
for (int i=0; i<10; i++) {
dispatch_group_enter(downloadGroup);
NSURLSessionDataTask *task = [session dataTaskWithRequest:request completionHandler:^(NSData * _Nullable data, NSURLResponse * _Nullable response, NSError * _Nullable error) {
NSLog(@"%d---%d",i,i);
dispatch_group_leave(downloadGroup);
}];
[task resume];
}
dispatch_group_notify(downloadGroup, dispatch_get_main_queue(), ^{
NSLog(@"end");
});
}
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2017-12-04 17:14:46.984 DownImage[3289:265374] 2---2
2017-12-04 17:14:46.987 DownImage[3289:265370] 1---1
2017-12-04 17:14:47.052 DownImage[3289:265383] 5---5
2017-12-04 17:14:47.065 DownImage[3289:265370] 4---4
2017-12-04 17:14:47.111 DownImage[3289:265379] 3---3
2017-12-04 17:14:47.121 DownImage[3289:265383] 6---6
2017-12-04 17:14:47.169 DownImage[3289:265383] 7---7
2017-12-04 17:14:47.192 DownImage[3289:265370] 9---9
2017-12-04 17:14:47.321 DownImage[3289:265383] 8---8
2017-12-04 17:14:47.747 DownImage[3289:265374] 0---0
2017-12-04 17:14:47.747 DownImage[3289:261033] end

从上两次输出可以看出,end确实是在所有网络请求之后才输出出来,这也是符合了我们的需求。

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dispatch_group_t downloadGroup = dispatch_group_create();
dispatch_group_enter(downloadGroup);
dispatch_group_leave(downloadGroup);
dispatch_group_notify(downloadGroup, dispatch_get_main_queue(), ^{
});

对以上4行代码可理解为:创建一个dispatch_group_t, 每次网络请求前先dispatch_group_enter,请求回调后再dispatch_group_leave,对于enter和leave必须配合使用,有几次enter就要有几次leave,否则group会一直存在。当所有enter的block都leave后,会执行dispatch_group_notify的block。

3.使用GCD的信号量dispatch_semaphore_t

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//3.semaphore--
-(void)Btn3{
NSString *str = @"http://www.jianshu.com/p/6930f335adba";
NSURL *url = [NSURL URLWithString:str];
NSURLRequest *request = [NSURLRequest requestWithURL:url];
NSURLSession *session = [NSURLSession sharedSession];
dispatch_semaphore_t sem = dispatch_semaphore_create(0);
for (int i=0; i<10; i++) {
NSURLSessionDataTask *task = [session dataTaskWithRequest:request completionHandler:^(NSData * _Nullable data, NSURLResponse * _Nullable response, NSError * _Nullable error) {
NSLog(@"%d---%d",i,i);
count++;
if (count==10) {
dispatch_semaphore_signal(sem);
count = 0;
}
}];
[task resume];
}
dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER);
dispatch_async(dispatch_get_main_queue(), ^{
NSLog(@"end");
});
}
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dispatch_semaphore_t sem = dispatch_semaphore_create(0);
dispatch_semaphore_signal(sem);
dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER);
对这三句代码可以这样理解:dispatch_semaphore信号量为基于计数器的一种多线程同步机制。如果semaphore计数大于等于1,计数-1,返回,程序继续运行。如果计数为0,则等待。dispatch_semaphore_signal(semaphore)为计数+1操作,dispatch_semaphore_wait(sema, DISPATCH_TIME_FOREVER)为设置等待时间,这里设置的等待时间是一直等待。
对于以上代码通俗一点就是,开始为0,等待,等10个网络请求都完成了,dispatch_semaphore_signal(semaphore)为计数+1,然后计数-1返回,程序继续执行。 (这里也就是为什么有个count变量的原因,记录网络回调的次数,回调10次之后再发信号量,使后面程序继续运行)。
4.考虑新需求,10个网络请求顺序回调。

需求需要顺序回调,即执行完第一个网络请求后,第二个网络请求回调才可被执行,简单来讲就是输出得是0,1,2,3…9这种方式的。

对于这个需求我也是根据自己最近做的项目来提的,因为网络请求回调的异步性,我们虽可以控制网络请求的顺序执行,却不能控制它的完成回调顺序。这就有点伤了,目前我项目是找到了解决方案,但这个问题还没有找到解决办法,提出来跟大家讨论一下。(请忽略网络请求执行,回调,在回调里请求下一个接口的办法,讨论还有没有别的方法,最好show the code).

最后,贴点NSOperation的代码,为了解决新需求所写,由于网络请求回调异步性不能满足需求,但若不是网络请求等异步回调的方式,这样的做法是可以的,大家可以试试.

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//4.NSOperation
-(void)Btn4{
NSString *str = @"http://www.jianshu.com/p/6930f335adba";
NSURL *url = [NSURL URLWithString:str];
NSURLRequest *request = [NSURLRequest requestWithURL:url];
NSURLSession *session = [NSURLSession sharedSession];
NSMutableArray *operationArr = [[NSMutableArray alloc]init];
for (int i=0; i<10; i++) {
NSBlockOperation *operation = [NSBlockOperation blockOperationWithBlock:^{
NSURLSessionDataTask *task = [session dataTaskWithRequest:request completionHandler:^(NSData * _Nullable data, NSURLResponse * _Nullable response, NSError * _Nullable error) {
NSLog(@"%d---%d",i,i);
}];
[task resume];
//非网络请求
NSLog(@"noRequest-%d",i);
}];
[operationArr addObject:operation];
if (i0) {
NSBlockOperation *operation1 = operationArr[i-1];
NSBlockOperation *operation2 = operationArr[i];
[operation2 addDependency:operation1];
}
}
NSOperationQueue *queue = [[NSOperationQueue alloc]init];
[queue addOperations:operationArr waitUntilFinished:NO]; //YES会阻塞当前线程
#warning - 绝对不要在应用主线程中等待一个Operation,只能在第二或次要线程中等待。阻塞主线程将导致应用无法响应用户事件,应用也将表现为无响应。
}
2017-12-04 18:03:10.224 DownImage[3584:304363] noRequest-0
2017-12-04 18:03:10.226 DownImage[3584:304362] noRequest-1
2017-12-04 18:03:10.226 DownImage[3584:304363] noRequest-2
2017-12-04 18:03:10.231 DownImage[3584:304363] noRequest-3
2017-12-04 18:03:10.232 DownImage[3584:304362] noRequest-4
2017-12-04 18:03:10.233 DownImage[3584:304362] noRequest-5
2017-12-04 18:03:10.233 DownImage[3584:304363] noRequest-6
2017-12-04 18:03:10.234 DownImage[3584:304363] noRequest-7
2017-12-04 18:03:10.235 DownImage[3584:304363] noRequest-8
2017-12-04 18:03:10.236 DownImage[3584:304363] noRequest-9
2017-12-04 18:03:10.408 DownImage[3584:304597] 2---2
2017-12-04 18:03:10.408 DownImage[3584:304597] 0---0
2017-12-04 18:03:10.409 DownImage[3584:304597] 1---1
2017-12-04 18:03:10.461 DownImage[3584:304597] 5---5
2017-12-04 18:03:10.476 DownImage[3584:304363] 4---4
2017-12-04 18:03:10.477 DownImage[3584:304365] 6---6
2017-12-04 18:03:10.518 DownImage[3584:304365] 7---7
2017-12-04 18:03:10.537 DownImage[3584:304596] 8---8
2017-12-04 18:03:10.547 DownImage[3584:304362] 9---9
2017-12-04 18:03:11.837 DownImage[3584:304362] 3---3
5.还是使用信号量semaphore完成4的需求
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//5.semaphore--order
-(void)Btn5{
NSString *str = @"http://www.jianshu.com/p/6930f335adba";
NSURL *url = [NSURL URLWithString:str];
NSURLRequest *request = [NSURLRequest requestWithURL:url];
NSURLSession *session = [NSURLSession sharedSession];
dispatch_semaphore_t sem = dispatch_semaphore_create(0);
for (int i=0; i<10; i++) {
NSURLSessionDataTask *task = [session dataTaskWithRequest:request completionHandler:^(NSData * _Nullable data, NSURLResponse * _Nullable response, NSError * _Nullable error) {
NSLog(@"%d---%d",i,i);
dispatch_semaphore_signal(sem);
}];
[task resume];
dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER);
}
dispatch_async(dispatch_get_main_queue(), ^{
NSLog(@"end");
});
}
2017-12-05 10:17:28.175 DownImage[938:51296] 0---0
2017-12-05 10:17:28.331 DownImage[938:51289] 1---1
2017-12-05 10:17:28.506 DownImage[938:51289] 2---2
2017-12-05 10:17:28.563 DownImage[938:51289] 3---3
2017-12-05 10:17:28.662 DownImage[938:51289] 4---4
2017-12-05 10:17:28.733 DownImage[938:51296] 5---5
2017-12-05 10:17:28.792 DownImage[938:51296] 6---6
2017-12-05 10:17:28.856 DownImage[938:51286] 7---7
2017-12-05 10:17:29.574 DownImage[938:51289] 8---8
2017-12-05 10:17:29.652 DownImage[938:51286] 9---9
2017-12-05 10:17:29.653 DownImage[938:45252] end

我们对比 3 的代码,3 中我们只在最后也就是循环结束后调用dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER),循环中当网络请求回调10次(也就是都回调完)后,使传入的信号量加1:( dispatch_semaphore_signal(sem) ),这时等待结束,然后进行后续的操作。

在这个方法里,我们每一次遍历,都让其dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER),这个时候线程会等待,阻塞当前线程,直到dispatch_semaphore_signal(sem)调用之后,而我们dispatch_semaphore_signal(sem)是在网络请求的回调里调用的,所以这个方法的逻辑是:

遍历—发起任务—等待—任务完成信号量加1—等待结束,开始下一个任务
发起任务—等待—任务完成信号量加1—等待结束,开始下一个任务
发起任务—等待—任务完成信号量加1—等待结束,开始下一个任务

这样循环的模式,一个任务完成才能接着去做下面的任务,满足我们的需求。

但我们也要发现这样一个问题,我们使用这种方式,可以明显感觉出整个过程需要花费的时间大大增加了,不像我们 3 中同时(几乎)开启任务等待完成回调,这里是一个网络请求发出,等待,完成后发出第二个网络请求,等待,完成后再发出第三个,这样我们等待的时间是10个网络请求每一个回调时间的和,在时间上大大增加了消耗,而且对于dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER),它是会阻塞线程的,我们如果需要在网络请求完成后修改UI,那这种方式会影响我们的界面交互,接下来我们对比一下两者时间消耗:

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3-------------3----------3-------
2017-12-05 10:29:51.178 DownImage[938:56971] 2---2
2017-12-05 10:29:51.193 DownImage[938:57200] 0---0
2017-12-05 10:29:51.202 DownImage[938:56631] 3---3
2017-12-05 10:29:51.248 DownImage[938:56971] 1---1
2017-12-05 10:29:51.262 DownImage[938:56971] 5---5
2017-12-05 10:29:51.291 DownImage[938:56631] 6---6
2017-12-05 10:29:51.375 DownImage[938:56631] 7---7
2017-12-05 10:29:51.384 DownImage[938:56631] 4---4
2017-12-05 10:29:51.434 DownImage[938:56971] 8---8
2017-12-05 10:29:51.487 DownImage[938:57199] 9---9
2017-12-05 10:29:51.488 DownImage[938:45252] end
5-------------5----------5-------
2017-12-05 10:29:52.190 DownImage[938:56631] 0---0
2017-12-05 10:29:52.304 DownImage[938:57199] 1---1
2017-12-05 10:29:52.432 DownImage[938:56971] 2---2
2017-12-05 10:29:52.520 DownImage[938:56971] 3---3
2017-12-05 10:29:52.576 DownImage[938:56631] 4---4
2017-12-05 10:29:52.628 DownImage[938:56971] 5---5
2017-12-05 10:29:52.706 DownImage[938:56631] 6---6
2017-12-05 10:29:52.764 DownImage[938:56971] 7---7
2017-12-05 10:29:52.853 DownImage[938:56631] 8---8
2017-12-05 10:29:52.925 DownImage[938:56971] 9---9
2017-12-05 10:29:52.926 DownImage[938:45252] end

看得出3花费时间为51.488 - 51.178约300多ms
— —5花费时间为52.926 - 52.190约700多ms

所以大家还请谨慎使用。

https://www.jianshu.com/p/fb4fb80aefb8