Objective-C/iPhone - String replacing

Question

How would I go about removing the "&expires_in=87131" within the following NSString:

Obviously the value 87131 may change in the future and it's always the last part of the string as well.

NSString *accessToken = @"136369349714439%7C2.nNIKZW8Z7Yw_aaaffqKv7lVFYJg__.86400.1301824800-705896566%7Cp-z9A68pJqTDNjEMj0TrHogc2bw&expires_in=87131";
NSString *newStr = [accessToken stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

Is it always the last part of the string?

Georg Schölly
– Georg Schölly

2011-04-02 10:29:59 +00:00
Commented Apr 2, 2011 at 10:29 — Georg Schölly
– Georg Schölly, Commented Apr 2, 2011 at 10:29
Correct, it's always the last part of the string.

gotnull
– gotnull

2011-04-02 10:31:47 +00:00
Commented Apr 2, 2011 at 10:31 — gotnull
– gotnull, Commented Apr 2, 2011 at 10:31

user94896 · Accepted Answer · 2011-04-02 10:44:56Z

2

If the string is guaranteed to only have one ampersand in it:

NSArray *components = [accessToken componentsSeparatedByString:@"&"];
accessToken = [components objectAtIndex:0];

answered Apr 2, 2011 at 10:44

user94896

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3 Comments

gotnull Over a year ago

I think this will do, I'm pretty certain it will be the first ampersand before that particular string value.

Kalle Over a year ago

If not, consider componentsSeparatedByString:@"&expires_in" instead, in case you have a stray & sign in the actual access token (don't believe it can happen but you never know).

gotnull Over a year ago

Added Kalle's suggestion, thanks for all your help guys. Much much appreciated.

Jim Blackler · Accepted Answer · 2011-04-02 09:55:25Z

2

For 'fixing the %7C characters' see NSString stringByReplacingPercentEscapesUsingEncoding

http://developer.apple.com/library/ios/#documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html#//apple_ref/occ/instm/NSString/stringByReplacingPercentEscapesUsingEncoding

answered Apr 2, 2011 at 9:55

Jim Blackler

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1 Comment

Jim Blackler Over a year ago

It's not easy actually with the default libraries. A regular expression would be error-prone. Your best bet is to decompose the URL into parameters, remove the one you don't want, then rebuild the URL. You really want a URL parser. There's discussion of one here. stackoverflow.com/questions/2225814/…

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