Proper URL path escaping in Cocoa
Don't do this:
NSString *escaped = [path stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];Do this:
CFStringRef escaped = CFURLCreateStringByAddingPercentEscapes(NULL,
(CFStringRef)path,
NULL,
CFSTR(";?#"),
kCFStringEncodingUTF8);Written by Adam Yanalunas
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Because stringByAddingPercentEscapesUsingEncoding alone does not strictly conform to what can be escaped in arguments and you may end up with skipped, unencoded characters in your string. That would result in a malformed URL with incorrect parameters.
That would result in great sadness. And probably sadness as well.
over 1 year ago
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