URLWithString为资源路径返回nil - iphone

Question

由于某种原因在获取资源的URL时遇到问题:此代码在viewDidLoad中,并且在其他应用程序中有效,但由于某些原因不在此处:

NSString* audioString = [[NSBundle mainBundle] pathForResource:@"sound" ofType:@"wav"];
NSLog(@"AUDIO STRING: %@" , audioString);

NSURL* audioURL = [NSURL URLWithString:audioString];
NSLog(@"AUDIO URL: %d" , audioURL);

NSError* playererror;
audioPlayer = [[AVAudioPlayer alloc] initWithContentsOfURL:audioURL error:&playererror];
[audioPlayer prepareToPlay];    

NSLog(@"Error %@", playererror);

日志输出:

音频字符串:/var/mobile/Applications/D9FA0569-45FF-4287-8448-7EA21E92EADC/SoundApp.app/sound.wav

音频网址:0

错误错误域= NSOSStatusErrorDomain代码= -50"无法完成操作.(OSStatus error -50.)"

Answer 1

你的字符串没有协议,所以它是一个无效的网址.试试这个...

NSString* expandedPath = [audioString stringByExpandingTildeInPath];
NSURL* audioUrl = [NSURL fileURLWithPath:expandedPath];

Answer 2

只需将一行更改为:

    NSURL* audioURL = [NSURL fileURLWithPath:audioString];