Objective-C URL string contains single percent sign

Question

I have a string which ends with percent sign(%),

this string is prepared for an URL request as a parameter:

NSString *parameter = @"/param=%";
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL urlWithString:[NSString stringWithFormat:@"http://www.whatev%@",parameter]]];

The request returns nil.

I've tried:

NSString *parameter = @"/param=\uFF05";
//request returns nil

and

NSString *parameter =  @"/param=%";
NSString *newParameter = [parameter stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
//request returns /param=%25 ...where does 25 come from?!

How could I have only one % converted to a request url?

Any advice would be appreciated.

used this http://stackoverflow.com/questions/739682/how-to-add-percent-sign-to-nsstring — Chirag Shah, Apr 20 '15 at 10:49
http://stackoverflow.com/questions/21403323/url-encoding-a-string/21404487#21404487 — Sport, Apr 20 '15 at 10:51
%25 is for %... That is because of Encoding... Like space is converted to %20 — Fahim Parkar, Apr 20 '15 at 11:01

score 0 · Accepted Answer · answered Apr 20 '15 at 10:51

0

The percent sign has a special purpose in URL's and is used to encode special characters of all kinds. For example a space ( ) is %20 and the percent sign itself is %25.

http://en.wikipedia.org/wiki/Percent-encoding

The last one should be correct, I assume you have problems using it as a request?

answered Apr 20 '15 at 10:51

Covalence

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u're right the last works fine ! I was just confused about the 25 and forgot to give it a try lol... – bluenowhere Apr 20 '15 at 13:09

score 0 · Answer 2 · answered Apr 20 '15 at 10:52

0

escape sequence for % is %%, so @"/param=%%" should solve the problem

answered Apr 20 '15 at 10:52

thorb65

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Objective-C URL string contains single percent sign

2 Answers2