Swift 同步进行多个地图局部搜索
我尝试同步进行多个搜索(我的意思是一个接一个,在运行下一个请求之前等待上一个请求完成),并阻塞,直到所有操作完成后再继续 但本地搜索的完成句柄看起来像被阻止了,一旦信号量放弃,它就会运行。我多次尝试都没有成功 我的代码和日志如下(您可以复制/粘贴到游乐场): [更新] 预期输出应在最后一行显示无Swift 同步进行多个地图局部搜索,swift,asynchronous,mklocalsearchrequest,Swift,Asynchronous,Mklocalsearchrequest,我尝试同步进行多个搜索(我的意思是一个接一个,在运行下一个请求之前等待上一个请求完成),并阻塞,直到所有操作完成后再继续 但本地搜索的完成句柄看起来像被阻止了,一旦信号量放弃,它就会运行。我多次尝试都没有成功 我的代码和日志如下(您可以复制/粘贴到游乐场): [更新] 预期输出应在最后一行显示无***警告和全部完成,如下所示(编号的确切顺序取决于网络条件): [更新2]取消注释行时输出的值//queue.maxConcurrentOperationCount=1 搜索(:in:centered:
***警告
和全部完成
,如下所示(编号的确切顺序取决于网络条件):
[更新2]取消注释行时输出的值//queue.maxConcurrentOperationCount=1
搜索(:in:centered:id:):第0部分的本地搜索2020-03-28 23:49:41+0000
搜索(:in:centered:id:):等待第0部分完成2020-03-28 23:49:41+0000
搜索(:in:centered:id:):***警告:第0次超时,作业未完成2020-03-28 23:49:46+0000
搜索(:in:centered:id:):第1部分的本地搜索2020-03-28 23:49:46+0000
搜索(:in:centered:id:):等待第1部分完成2020-03-28 23:49:46+0000
搜索(:in:centered:id:):***警告:第1次超时,作业未完成2020-03-28 23:49:51+0000
搜索(:in:centered:id:):第二部分的本地搜索2020-03-28 23:49:51+0000
搜索(:in:centered:id:):等待第2部分完成2020-03-28 23:49:51+0000
搜索(:in:centered:id:):***警告:第2次超时,作业未完成2020-03-28 23:49:56+0000
搜索(:in:centered:id:):第三部分的本地搜索2020-03-28 23:49:56+0000
搜索(:in:centered:id:):等待第三部分完成2020-03-28 23:49:56+0000
搜索(:in:centered:id:):***警告:第三次超时,作业未完成2020-03-28 23:50:01+0000
全部完成2020-03-2823:50:01+0000
0获得可选(10)项2020-03-28 23:50:02+0000
3个可选(10)项2020-03-28 23:50:02+0000
2个可选(10)项2020-03-28 23:50:02+0000
1个可选(10)项2020-03-28 23:50:02+0000
注意:顺便说一句,我还在每次打印的末尾添加了
\(Date())
如果您希望这些操作以串行方式运行,则必须指定队列一次只能运行一个,例如
queue.maxConcurrentOperationCount = 1
而且,正如您所发现的,您希望避免使用addOperations
的waitUntilFinished
选项,因为这会在操作完成之前阻塞当前线程。相反,使用完成处理程序模式
以下是我使用的代码:
func performMultipleSearches(completion: @escaping () -> Void) {
let searches = ["restaurant", "coffee", "hospital", "natural history museum"]
let queue = OperationQueue()
queue.maxConcurrentOperationCount = 1
for (i, searchText) in searches.enumerated() {
queue.addOperation {
self.search(searchText, in: self.mapView.region, id: i)
}
}
queue.addOperation {
completion()
}
}
func search(_ query: String, in region: MKCoordinateRegion, id: Int) {
let semaphore = DispatchSemaphore(value: 0)
os_log("%d starting", id)
let request = MKLocalSearch.Request()
request.naturalLanguageQuery = query
request.region = region
if #available(iOS 13, *) {
request.resultTypes = .pointOfInterest
}
let search = MKLocalSearch(request: request)
search.start { response, error in
defer { semaphore.signal() }
guard let mapItems = response?.mapItems else {
os_log(" %d failed", id)
return
}
os_log(" %d succeeded, found %d:", id, mapItems.count)
}
os_log(" %d waiting", id)
guard semaphore.wait(timeout: .now() + 5) == .success else {
os_log(" %d timedout", id)
return
}
os_log(" %d done", id)
}
这产生了:
2020-03-28 16:16:25.219565-0700 MyApp[46601:2107182] 0 starting
2020-03-28 16:16:25.220018-0700 MyApp[46601:2107182] 0 waiting
2020-03-28 16:16:25.438121-0700 MyApp[46601:2107033] 0 succeeded, found 10:
2020-03-28 16:16:25.438269-0700 MyApp[46601:2107182] 0 done
2020-03-28 16:16:25.438436-0700 MyApp[46601:2107182] 1 starting
2020-03-28 16:16:25.438566-0700 MyApp[46601:2107182] 1 waiting
2020-03-28 16:16:25.639198-0700 MyApp[46601:2107033] 1 succeeded, found 10:
2020-03-28 16:16:25.639357-0700 MyApp[46601:2107182] 1 done
2020-03-28 16:16:25.639490-0700 MyApp[46601:2107182] 2 starting
2020-03-28 16:16:25.639598-0700 MyApp[46601:2107182] 2 waiting
2020-03-28 16:16:25.822085-0700 MyApp[46601:2107033] 2 succeeded, found 10:
2020-03-28 16:16:25.822274-0700 MyApp[46601:2107182] 2 done
2020-03-28 16:16:25.822422-0700 MyApp[46601:2107162] 3 starting
2020-03-28 16:16:25.822567-0700 MyApp[46601:2107162] 3 waiting
2020-03-28 16:16:26.015566-0700 MyApp[46601:2107033] 3 succeeded, found 1:
2020-03-28 16:16:26.015696-0700 MyApp[46601:2107162] 3 done
2020-03-28 16:16:26.015840-0700 MyApp[46601:2107162] all done
无论如何,我不会使用信号量,而是使用异步
操作
子类。例如,您可以使用异步操作
类,然后执行以下操作:
class SearchOperation: AsynchronousOperation {
let identifier: Int
let searchText: String
let region: MKCoordinateRegion
init(identifier: Int, searchText: String, region: MKCoordinateRegion) {
self.identifier = identifier
self.searchText = searchText
self.region = region
super.init()
}
override func main() {
os_log("%d started", identifier)
let request = MKLocalSearch.Request()
request.naturalLanguageQuery = searchText
request.region = region
if #available(iOS 13, *) {
request.resultTypes = .pointOfInterest
}
let search = MKLocalSearch(request: request)
search.start { response, error in
defer { self.finish() }
guard let mapItems = response?.mapItems else {
os_log(" %d failed", self.identifier)
return
}
os_log(" %d succeeded, found %d:", self.identifier, mapItems.count)
}
}
}
然后
let searches = ["restaurant", "coffee", "hospital", "natural history museum"]
let queue = OperationQueue()
queue.maxConcurrentOperationCount = 1
for (i, searchText) in searches.enumerated() {
queue.addOperation(SearchOperation(identifier: i, searchText: searchText, region: mapView.region))
}
queue.addOperation {
completion()
}
如果希望这些操作以串行方式运行,则必须指定队列一次只能运行一个,例如
queue.maxConcurrentOperationCount = 1
而且,正如您所发现的,您希望避免使用addOperations
的waitUntilFinished
选项,因为这会在操作完成之前阻塞当前线程。相反,使用完成处理程序模式
以下是我使用的代码:
func performMultipleSearches(completion: @escaping () -> Void) {
let searches = ["restaurant", "coffee", "hospital", "natural history museum"]
let queue = OperationQueue()
queue.maxConcurrentOperationCount = 1
for (i, searchText) in searches.enumerated() {
queue.addOperation {
self.search(searchText, in: self.mapView.region, id: i)
}
}
queue.addOperation {
completion()
}
}
func search(_ query: String, in region: MKCoordinateRegion, id: Int) {
let semaphore = DispatchSemaphore(value: 0)
os_log("%d starting", id)
let request = MKLocalSearch.Request()
request.naturalLanguageQuery = query
request.region = region
if #available(iOS 13, *) {
request.resultTypes = .pointOfInterest
}
let search = MKLocalSearch(request: request)
search.start { response, error in
defer { semaphore.signal() }
guard let mapItems = response?.mapItems else {
os_log(" %d failed", id)
return
}
os_log(" %d succeeded, found %d:", id, mapItems.count)
}
os_log(" %d waiting", id)
guard semaphore.wait(timeout: .now() + 5) == .success else {
os_log(" %d timedout", id)
return
}
os_log(" %d done", id)
}
这产生了:
2020-03-28 16:16:25.219565-0700 MyApp[46601:2107182] 0 starting
2020-03-28 16:16:25.220018-0700 MyApp[46601:2107182] 0 waiting
2020-03-28 16:16:25.438121-0700 MyApp[46601:2107033] 0 succeeded, found 10:
2020-03-28 16:16:25.438269-0700 MyApp[46601:2107182] 0 done
2020-03-28 16:16:25.438436-0700 MyApp[46601:2107182] 1 starting
2020-03-28 16:16:25.438566-0700 MyApp[46601:2107182] 1 waiting
2020-03-28 16:16:25.639198-0700 MyApp[46601:2107033] 1 succeeded, found 10:
2020-03-28 16:16:25.639357-0700 MyApp[46601:2107182] 1 done
2020-03-28 16:16:25.639490-0700 MyApp[46601:2107182] 2 starting
2020-03-28 16:16:25.639598-0700 MyApp[46601:2107182] 2 waiting
2020-03-28 16:16:25.822085-0700 MyApp[46601:2107033] 2 succeeded, found 10:
2020-03-28 16:16:25.822274-0700 MyApp[46601:2107182] 2 done
2020-03-28 16:16:25.822422-0700 MyApp[46601:2107162] 3 starting
2020-03-28 16:16:25.822567-0700 MyApp[46601:2107162] 3 waiting
2020-03-28 16:16:26.015566-0700 MyApp[46601:2107033] 3 succeeded, found 1:
2020-03-28 16:16:26.015696-0700 MyApp[46601:2107162] 3 done
2020-03-28 16:16:26.015840-0700 MyApp[46601:2107162] all done
无论如何,我不会使用信号量,而是使用异步
操作
子类。例如,您可以使用异步操作
类,然后执行以下操作:
class SearchOperation: AsynchronousOperation {
let identifier: Int
let searchText: String
let region: MKCoordinateRegion
init(identifier: Int, searchText: String, region: MKCoordinateRegion) {
self.identifier = identifier
self.searchText = searchText
self.region = region
super.init()
}
override func main() {
os_log("%d started", identifier)
let request = MKLocalSearch.Request()
request.naturalLanguageQuery = searchText
request.region = region
if #available(iOS 13, *) {
request.resultTypes = .pointOfInterest
}
let search = MKLocalSearch(request: request)
search.start { response, error in
defer { self.finish() }
guard let mapItems = response?.mapItems else {
os_log(" %d failed", self.identifier)
return
}
os_log(" %d succeeded, found %d:", self.identifier, mapItems.count)
}
}
}
然后
let searches = ["restaurant", "coffee", "hospital", "natural history museum"]
let queue = OperationQueue()
queue.maxConcurrentOperationCount = 1
for (i, searchText) in searches.enumerated() {
queue.addOperation(SearchOperation(identifier: i, searchText: searchText, region: mapView.region))
}
queue.addOperation {
completion()
}
我已经试过了(参见代码中的注释)。对我来说,这就像在search.start()中阻止了趋势一样,因为在信号量过期之前,永远不会触发完成闭包(应该在主线程上运行)。在操场上试一试。使用
maxConcurrentOperationCount
set对我来说效果很好。我更新了预期结果,以便更清楚地获得什么,并用我的输出更新了我的结果。但是,maxConcurrentOperationCount=1
是必需的。否则,它将同时运行它们。对我来说,这就像在search.start()中阻止了趋势一样,因为在信号量过期之前,永远不会触发完成闭包(应该在主线程上运行)。在操场上试一试。使用maxConcurrentOperationCount
set对我来说效果很好。我更新了预期结果,以便更清楚地获得什么,并用我的输出更新了我的结果。但是,maxConcurrentOperationCount=1
是必需的。否则,它将同时运行它们。但是,如果您使用maxConcurrentOperationCount=1
(这对于不同时运行搜索非常重要),那么预期的结果就不应该是这样的。在上一次搜索完成之前,您不应该看到开始下一次搜索。我已在取消注释max=1时使用输出进行了更新,以便查看。我正在查看你的代码:情况不同。在您的中,您可以进行不同的查询。在我的世界里,只有地区在变化。因此,我正在进行更改,以查看eta的行为是否相同。是的,既然您使用了max=1,那么您将看到您的输出是有序的,这是完全有意义的。唯一让人好奇的是他们为什么要超时。你是否在应用程序中而不是在游乐场中尝试过这一点?你试过把你的超时时间从5秒增加到更大的吗?是的,这是最初的点。为什么会超时。在真实设备上也是如此。如果我设置了长时间的超时,同样的行为。通过删除设置waitUntilFinished:false,超时消失了。但是,我当然失去了我的障碍。你列出了你期望的结果。但是,如果您使用maxConcurrentOperationCount=1
(这对于不同时运行搜索非常重要),那么预期的结果就不应该是这样的。在上一次搜索完成之前,您不应该看到启动下一次搜索