ios - 在 iOS 中显示相同文本的多个信息窗口

Question

我已经实现了 infoWindow 来显示我的多个标记的数据。但是我只能通过信息窗口显示所有相同的数据。

如何显示与该标记相关的数据，以便不重复？

这是我的代码:

     for (int i = 0 ; i < jsonDataDict.count; i ++) {
        NSDictionary *newDict = [jsonDataDict objectAtIndex:i ];

        double latitude = [[newDict valueForKey:@"lat"]doubleValue];
        double longitute = [[newDict valueForKey:@"lng"]doubleValue];


      nameStr = [newDict valueForKey:@"name"];


      countJson = i ;

        CLLocationCoordinate2D position = CLLocationCoordinate2DMake(latitude, longitute);
        GMSMarker *marker = [GMSMarker markerWithPosition:position];
        marker.title = [newDict valueForKey:@"name"];
      //  marker.icon = [UIImage imageNamed:@"pin11.png"];
        marker.icon = [self image:[UIImage imageNamed:@"pinPopoye.png"] scaledToSize:CGSizeMake(75.0f, 60.0f)];
        marker.appearAnimation = kGMSMarkerAnimationPop;
        marker.infoWindowAnchor = CGPointMake(1.1, 0.70);
        marker.map = self.mapView;
        [self mapView:self.mapView markerInfoWindow:marker];


    }
}

- (UIView *)mapView:(GMSMapView *)mapView markerInfoWindow:(GMSMarker *)marker
{    UIView *customView = [[UIView alloc] initWithFrame:CGRectMake(0, 0, 60, 25)];
   customView.backgroundColor = [UIColor colorWithRed:71.0/255.0 green:65.0/255.0 blue:65.0/255.0 alpha:0.8];
    customView.layer.cornerRadius = 5;
    customView.layer.masksToBounds = YES;

    //  Orange Color ==== [UIColor colorWithRed:254.0/255.0 green:165.0/255.0 blue:4.0/255.0 alpha:0.5];
       UILabel *nameLabel = [[UILabel alloc]initWithFrame:CGRectMake(0, 0, 60, 10)];
   nameLabel.text = nameStr;
   nameLabel.textColor = [UIColor whiteColor];
   nameLabel.textAlignment = NSTextAlignmentCenter;
   nameLabel.font = [UIFont systemFontOfSize:8.0];
    [customView addSubview:nameLabel];

      return customView;
}

Answer 1

替换这条语句:

nameLabel.text = nameStr;

与:

nameLabel.text = marker.title;

问题是您使用了一个共享的 NSString，nameStr，它在您的 for 循环的每次迭代中都会被覆盖。因此，所有标签在最终显示时共享相同的字符串值。你也可以这样做:

nameLabel.text = [nameStr copy];

它应该可以工作——但我认为在您的代码中使用 nameStr 只是之前一些“hack”的残余。