我想在谷歌地图上绘制线条,线条可能是直线或曲线,但我想绘制线条,使它们连接两个点,可能位于任何地理位置。
感谢。
GuruPrasad R Gujjar。
答案 0 :(得分:1)
以下是示例代码,请尝试使用此功能。 编辑:您还必须下载“RegexKitLite”头文件和类文件。
// This method will calculate the routes and return an array with point
-(NSArray*) calculateRoutesFrom:(CLLocationCoordinate2D) f to: (CLLocationCoordinate2D) t {
NSString* saddr = [NSString stringWithFormat:@"%f,%f", f.latitude, f.longitude];
NSString* daddr = [NSString stringWithFormat:@"%f,%f", t.latitude, t.longitude];
NSString* apiUrlStr = [NSString stringWithFormat:@"http://maps.google.com/maps?output=dragdir&saddr=%@&daddr=%@", saddr, daddr];
NSURL* apiUrl = [NSURL URLWithString:apiUrlStr];
NSLog(@"api url: %@", apiUrl);
NSString *apiResponse = [NSString stringWithContentsOfURL:apiUrl];
NSString* encodedPoints = [apiResponse stringByMatching:@"points:\\\"([^\\\"]*)\\\"" capture:1L];
return [self decodePolyLine:[encodedPoints mutableCopy]];
}
-(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 = 0x20);
NSInteger dlat = ((result & 1) ? ~(result >> 1) : (result >> 1));
lat += dlat;
shift = 0;
result = 0;
do {
b = [encoded characterAtIndex:index++] - 63;
result |= (b & 0x1f) = 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;
}
// This method will draw the route with array of points.
-(void) updateRouteView
{
CGContextRef context = CGBitmapContextCreate(nil,
routeView.frame.size.width,
routeView.frame.size.height,
8,
4 * routeView.frame.size.width,
CGColorSpaceCreateDeviceRGB(),
kCGImageAlphaPremultipliedLast);
CGContextSetStrokeColorWithColor(context, lineColor.CGColor);
CGContextSetRGBFillColor(context, 0.0, 0.0, 1.0, 1.0);
CGContextSetLineWidth(context, 3.0);
for(int i = 0; i &ls; routes.count; i++) {
CLLocation* location = [routes objectAtIndex:i];
CGPoint point = [map convertCoordinate:location.coordinate toPointToView:routeView];
if(i == 0) {
CGContextMoveToPoint(context, point.x, routeView.frame.size.height - point.y);
} else {
CGContextAddLineToPoint(context, point.x, routeView.frame.size.height - point.y);
}
}
CGContextStrokePath(context);
CGImageRef image = CGBitmapContextCreateImage(context);
UIImage* img = [UIImage imageWithCGImage:image];
routeView.image = img;
//[self.view addSubview:routeView];
CGContextRelease(context);
//29-07-10
CGContextRef context2 = CGBitmapContextCreate(nil,
routeView2.frame.size.width,
routeView2.frame.size.height,
8,
4 * routeView2.frame.size.width,
CGColorSpaceCreateDeviceRGB(),
kCGImageAlphaPremultipliedLast);
CGContextSetStrokeColorWithColor(context2, lineColor2.CGColor);
CGContextSetRGBFillColor(context2, 0.0, 0.0, 1.0, 1.0);
CGContextSetLineWidth(context2, 3.0);
for(int i = 0; i &ls; routes2.count; i++) {
CLLocation* location2 = [routes2 objectAtIndex:i];
CGPoint point2 = [map convertCoordinate:location2.coordinate toPointToView:routeView2];
if(i == 0) {
CGContextMoveToPoint(context2, point2.x, routeView2.frame.size.height - point2.y);
} else {
CGContextAddLineToPoint(context2, point2.x, routeView2.frame.size.height - point2.y);
}
}
CGContextStrokePath(context2);
CGImageRef image2 = CGBitmapContextCreateImage(context2);
UIImage* img2 = [UIImage imageWithCGImage:image2];
routeView2.image = img2;
//[self.view addSubview:routeView];
CGContextRelease(context2);
}
我认为它可以帮助您完成任务。
答案 1 :(得分:0)
您需要使用JavaScript和Google Maps API - 如果我没记错的话,密钥是免费的。 这里的其他一切: http://code.google.com/apis/maps/documentation/javascript/v2/overlays.html#Drawing_Polylines
答案 2 :(得分:0)