如何將圖像放入文件夾並從文檔目錄獲取它們

Question

在我的應用程序中，不同文件夾中有多張圖像（共5張）。 如何將這些文件夾放入文檔目錄？

然后，如何從特定圖像文件夾中獲取計數和圖像名稱？

Answer 1

檢查Working_with_Directories鏈接。

編輯 ： Copy item請參考此鏈接

有關更多詳細信息，請參考NSFileManager ：

Answer 2

UIImage *dimage = [UIImage imagedNamed:@"yourImage.png"]; //get the image name here
 NSString *docDir = [NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES) objectAtIndex:0];

 NSString *imageName = @"myImageName.png";
 NSArray *patharr=[ImagePath componentsSeparatedByString:@"."];
 NSString* Ext=[NSString stringWithFormat: @"%@",[patharr objectAtIndex:1]];
 NSString *FilePath = [NSString stringWithFormat: [NSString stringWithFormat:@"%@/%@",docDir,ImagePath]];


 NSData *data = [NSData dataWithData:UIImagePNGRepresentation(dimage)];
 [data writeToFile:FilePath atomically:YES];

進一步獲取文件目錄文件名的內容...

 NSArray *paths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
    NSString *documentsPath = [paths objectAtIndex:0]; 
    NSArray *dirContents = [[NSFileManager defaultManager] contentsOfDirectoryAtPath:documentsPath error:nil];
    NSMutableArray *array=[[NSMutableArray alloc]init];
    for (NSString *dir in dirContents) {
        if ([dir hasSuffix:@".png"]) {
            NSRange range = [dir rangeOfString:@"."];
            NSString *name = [dir substringToIndex:range.location];
            if (![name isEqualToString:@""]) {

                [array addObject:name];
            }
        }
    } 
NSLog(@"document folder content list %@ ",array);

Answer 3

這是從目錄中獲取圖像的方法

array = [[NSBundle mainBundle] pathsForResourcesOfType:@"png" inDirectory:@"folderName"];

Answer 4

迅速
文檔目錄和商店圖像的路徑

func getDocumentsDirectory() -> URL {
    let paths = FileManager.default.urls(for: .documentDirectory, in: .userDomainMask)
    let documentsDirectory = paths[0]
    return documentsDirectory
}

if let image = UIImage(named: "example.png") {
    if let data = UIImagePNGRepresentation() {
        let filename = getDocumentsDirectory().appendingPathComponent("copy.png")
        try? data.write(to: filename)
    }
}