iOS - 将CLLocations的数组拆分为只是纬度和经度字符串的数组

Question

在当前时刻，NSMutableArray CLLocations CLLocation <-42.86672600,+147.32411407> +/- 0.00m (speed -1.00 mps / course -1.00) @ 12/11/14, 7:51:06 AM Australian Eastern Daylight Time-<-42.86413769,+147.32383317> +/- 0.00m (speed -1.00 mps / course -1.00) @ 12/11/14, 7:52:30 AM Australian Eastern Daylight Time-<-42.86596128,+147.32764541> +/- 0.00m (speed -1.00 mps / course -1.00) @ 12/11/14, 7:52:33 AM Australian Eastern Daylight Time" ，我不需要每个-42.86672600,+147.32411407/-42.86413769,+147.32383317/-42.86596128,+147.32764541。当前时刻的日志是：

{{1}}

如何将阵列缩小，只显示纬度和经度，即阵列如下： {{1}} ...

干杯， SebOH

Answer 1

如果您确定他们只是CLLocation个对象，那么......

for (CLLocation *location in array) {
    NSLog(@"%f, %f", location.coordinate.latitude, location.coordinate.longitude);
}

这将在一行中注销每个lat和long。

如果你想要那个特定的字符串......

NSMutableString *string = [NSMutableString string];

for (CLLocation *location in array) {
    [string appendFormat:@"%f,%f/", location.coordinate.latitude, location.coordinate.longitude];
}

NSLog(@"%@", string);

这将记录您在问题中输入的确切字符串。