在Xcode中的两个坐标之间绘制一条折线 [英] Draw a polyline between two coordinates in Xcode
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问题描述
我想在地图上的两个或多个给定坐标之间绘制一条多义线(多义线而不是路线).想象一下,我在地图上有2个掉线的引脚,我需要从第一个掉线的引脚到第二个掉线的引脚绘制一条直线.
I would like to draw a POLY-LINE between two or more given coordinates on the map(poly-line and not the route). Imagine I have 2 dropped pins on the map and I need to draw a straight line from the first dropped pin to the second dropped pin.
推荐答案
在界面文件中
MKPolyline* _routeLine;
MKPolylineView* _routeLineView;
在实施文件中
将所有坐标存储在
NSMutablrArray *routeLatitudes
然后
MKMapPoint* pointArr = malloc(sizeof(CLLocationCoordinate2D) * [routeLatitudes count]);
for(int idx = 0; idx < [routeLatitudes count]; idx++)
{
CLLocationCoordinate2D workingCoordinate;
workingCoordinate.latitude=[[routeLatitudes objectAtIndex:idx] doubleValue];
workingCoordinate.longitude=[[routeLongitudes objectAtIndex:idx] doubleValue];
MKMapPoint point = MKMapPointForCoordinate(workingCoordinate);
pointArr[idx] = point;
}
// create the polyline based on the array of points.
routeLine = [MKPolyline polylineWithPoints:pointArr count:[routeLatitudes count]];
[mapViewHome addOverlay:self.routeLine];
free(pointArr);
和覆盖代表
- (MKOverlayView *)mapView:(MKMapView *)mapView viewForOverlay:(id <MKOverlay>)overlay
{
MKOverlayView* overlayView = nil;
if(overlay == routeLine)
{
routeLineView = [[[MKPolylineView alloc] initWithPolyline:self.routeLine] autorelease];
routeLineView.fillColor = [UIColor colorWithRed:0.945 green:0.027 blue:0.957 alpha:1];
routeLineView.strokeColor = [UIColor colorWithRed:0.945 green:0.027 blue:0.957 alpha:1];
routeLineView.lineWidth = 4;
overlayView = routeLineView;
}
return overlayView;
}
希望这会有所帮助
修改后的代码
这是获取NsMutableArray坐标的代码.
here is the code to get NsMutableArray of coordinates.
调用此功能
NSString * saddr = [NSString stringWithFormat:@"%f,%f",StartCoordinate.latitude, StartCoordinate.longitude];
NSString* daddr = [NSString stringWithFormat:@"%f,%f",EndCoordinate.latitude, EndCoordinate.longitude];
routeLatitudes=[[[self getDirectionRoutesFrom:[saddr copy] to:[daddr mutableCopy]] mutableCopy] retain];
功能定义
-(NSMutableArray *)getDirectionRoutesFrom:(NSString *)saddr1 to:(NSString *)daddr
{
NSString* apiUrlStr = [NSString stringWithFormat:@"http://maps.google.com/maps?output=dragdir&saddr=%@&daddr=%@", saddr1, daddr];
NSURL* apiUrl = [NSURL URLWithString:apiUrlStr];
//NSString *apiResponse = [NSString stringWithContentsOfURL:apiUrl];
NSString *apiResponse = [NSString stringWithContentsOfURL:apiUrl encoding:NSUTF8StringEncoding error:nil];
NSString* encodedPoints = [apiResponse stringByMatching:@"points:\\\"([^\\\"]*)\\\"" capture:1L];
//NSMutableArray *temparr=[[MapViewController decodePolyLine:[encodedPoints mutableCopy]] retain];
return [[self decodePolyLine:[encodedPoints mutableCopy]] retain];
//return temparr;
}
和
-(NSMutableArray *)decodePolyLine: (NSMutableString *)encoded {
[encoded replaceOccurrencesOfString:@"\\\\" withString:@"\\"
options:NSLiteralSearch
range:NSMakeRange(0, [encoded length])];
NSInteger len = [encoded length];
NSInteger index = 0;
NSMutableArray *array = [[[NSMutableArray alloc] init] autorelease];
NSInteger lat=0;
NSInteger lng=0;
while (index < len) {
NSInteger b;
NSInteger shift = 0;
NSInteger result = 0;
do {
b = [encoded characterAtIndex:index++] - 63;
result |= (b & 0x1f) << shift;
shift += 5;
} while (b >= 0x20);
NSInteger dlat = ((result & 1) ? ~(result >> 1) : (result >> 1));
lat += dlat;
shift = 0;
result = 0;
do {
b = [encoded characterAtIndex:index++] - 63;
result |= (b & 0x1f) << shift;
shift += 5;
} while (b >= 0x20);
NSInteger dlng = ((result & 1) ? ~(result >> 1) : (result >> 1));
lng += dlng;
NSNumber *latitude = [[[NSNumber alloc] initWithFloat:lat * 1e-5] autorelease];
NSNumber *longitude = [[[NSNumber alloc] initWithFloat:lng * 1e-5] autorelease];
printf("[%f,", [latitude doubleValue]);
printf("%f]", [longitude doubleValue]);
CLLocation *loc = [[[CLLocation alloc] initWithLatitude:[latitude floatValue] longitude:[longitude floatValue]] autorelease];
[array addObject:loc];
}
return array;
}
在文件中包含RegexKitLite.h.
include RegexKitLite.h in your file.
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