0

我想使用 mapkit 框架获取附近位置的数组。因此，当用户在文本字段中键入时，我会调用以下函数。

- (void)searchForLocations:(NSString *)string
{
    [NSObject cancelPreviousPerformRequestsWithTarget:self selector:@selector(search:) object:nil];
    [self performSelectorInBackground:@selector(search:) withObject:string];
}

- (void)search :(NSString *)string
{
    MKLocalSearchRequest *request = [[MKLocalSearchRequest alloc] init];
    request.naturalLanguageQuery = string;
    MKCoordinateRegion region;
    MKCoordinateSpan span;
    span.latitudeDelta = 0.05;
    span.longitudeDelta = 0.05;

    region.span = span;
    region.center = newLocation.coordinate;

    request.region = region;

    MKLocalSearch *search = [[MKLocalSearch alloc]initWithRequest:request];

    [search startWithCompletionHandler:^(MKLocalSearchResponse
                                         *response, NSError *error) {
        if (response.mapItems.count == 0)
        {
            NSLog(@"No Matches");
        }
        else
        {

            NSLog(@"name = %@", item.name);
            NSLog(@"Phone = %@", item.phoneNumber);
        }
    }];
}

如您所见，如果有新的输入文本出现，我想取消先前的搜索。但是之前的搜索并没有取消。如何取消之前的搜索？

提前致谢。

4

3 回答 3

2

MKLocalSearch 上有一个取消方法。你试过那个吗？

编辑：啊，对不起，我是愚蠢的。您需要以某种方式保留对旧搜索的引用才能取消它。将它放在一个属性中，当搜索完成时您可以清除（即设置为 nil）。每当你调用搜索函数时，取消之前的搜索函数（不需要“如果”，nil 吞下所有），然后创建一个新的

@property (nonatiomic, strong) MKLocalSearch *previousSearch;

- (void)search :(NSString *)string
{
    [self.previousSearch cancel];
    MKLocalSearchRequest *request = [[MKLocalSearchRequest alloc] init];
    request.naturalLanguageQuery = string;
    MKCoordinateRegion region;
    MKCoordinateSpan span;
    span.latitudeDelta = 0.05;
    span.longitudeDelta = 0.05;

    region.span = span;
    region.center = newLocation.coordinate;

    request.region = region;

    MKLocalSearch *search = [[MKLocalSearch alloc]initWithRequest:request];

    [search startWithCompletionHandler:^(MKLocalSearchResponse
                                     *response, NSError *error) {
        self.previousSearch = nil;
        if (response.mapItems.count == 0)
        {
            NSLog(@"No Matches");
        }
        else
        {

            NSLog(@"name = %@", item.name);
            NSLog(@"Phone = %@", item.phoneNumber);
        }
    }];

    self.previousSearch = search;
}
于 2015-07-30T09:21:55.210 回答
0

该MKLocalSearch对象有一个cancel方法，您可以使用它来取消挂起的搜索。您不能像尝试那样简单地取消选择器，因为该选择器将很快完成，并MKLocalSearch在后台调度。

您将需要一个属性来存储您的搜索对象，以便您可以判断它是否仍在搜索并在需要时取消搜索。

@property (strong,nonatomic) MKLocalSearch *localSearch;

- (void)search :(NSString *)string
{
    MKLocalSearchRequest *request = [[MKLocalSearchRequest alloc] init];
    request.naturalLanguageQuery = string;
    MKCoordinateRegion region;
    MKCoordinateSpan span;
    span.latitudeDelta = 0.05;
    span.longitudeDelta = 0.05;

    region.span = span;
    region.center = newLocation.coordinate;

    request.region = region;

    if (self.localSearch!=nil) {
        if (self.localSearch.searching) {
            [self.localSearch cancel];
            self.localSearch=nil;
        }
    }

    self.localSearch = [[MKLocalSearch alloc]initWithRequest:request];

    [search startWithCompletionHandler:^(MKLocalSearchResponse
                                         *response, NSError *error) {
        if (response.mapItems.count == 0)
        {
            NSLog(@"No Matches");
        }
        else
        {

            NSLog(@"name = %@", item.name);
            NSLog(@"Phone = %@", item.phoneNumber);
        }
    }];
}
于 2015-07-30T09:50:20.503 回答
-1

那么你可以使用 NSBlockOperation。假设您用于 NSBlockOperation 的全局变量是 blockOperation。然后 ：

[search startWithCompletionHandler:^(MKLocalSearchResponse
                                     *response, NSError *error) {
    blockOperation = NSBlockOperation(block: { () -> Void in
         if (response.mapItems.count == 0) {
            NSLog(@"No Matches");
         } else {
            NSLog(@"name = %@", item.name);
            NSLog(@"Phone = %@", item.phoneNumber);
         }
    })
}];

然后你可以使用这个 blockOperation 在任何你想要的地方取消，如下所示：

blockOperation.cancel()
于 2015-07-30T09:25:50.363 回答