Question

Did a search, didn’t find any issues on this.

My application takes .csv files, switches the data around and spits it out in a way my company finds useful. The issue is that the program seems to require filenames without whitespace. For instance “My data.csv” will not work, but “My_data.csv” will work. Here is what I believe to be the relevant code

 NSArray *allowedTypes = [NSArray arrayWithObjects:@"CSV",@"csv",nil];
openDLG = [NSOpenPanel openPanel];
[openDLG setDelegate:self];
[openDLG setCanChooseFiles:YES];
[openDLG setAllowedFileTypes:allowedTypes];

NSInteger openReturn = [openDLG runModal];
if (openReturn == NSFileHandlingPanelOKButton) {
    NSArray *rawCSVs = [openDLG URLs];
    NSEnumerator *enumerator = [rawCSVs objectEnumerator];
    id object;
    while ((object = [enumerator nextObject])) {
        NSString *tempDump = [NSString stringWithContentsOfURL:object encoding:NSUTF8StringEncoding error:NULL];
        NSArray *bigArray = [tempDump csvRows];
        int total = [bigArray count];
        ....do other things

In debug mode, the “object” comes up as nil for filenames with whitespace and “tempdump” string comes up as @”nil” for filenames with whitespace, but comes up wonderfully for names_with_underscores

Answer 1

An URL does not permit every character, so you will need to do some conversion.

NSString *rawString = [NSString stringWithString:@"/Users/myusername/test 1.txt"];
NSString *escapedString = [rawString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *url = [NSURL URLWithString:escapedString];
NSLog(@"url: %@", url);