IOS not encoding plus(+) sign in x-www-form-urlencoded POST request?

Question

I am using the following code,

NSString *jsonD = [NSString stringWithFormat:@"rawJson=%@",[fbUserInfo jsonUTF8String]];
NSData *myRequestData = [jsonD dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];

NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url
                                                           cachePolicy:policy timeoutInterval:20.0];

[ request setHTTPMethod: @"POST" ];
[request setValue:[NSString stringWithFormat:@"%d", [myRequestData length]] forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[ request setHTTPBody: myRequestData ];

In this code I have the date 2013-06-29T18:33:17+0000. The problem is that my server receiving date as 2013-06-29T18:33:17 0000. See the space. I don't know why this is happening.

Krumelur · Accepted Answer · 2013-07-03T10:07:48.033

5

Just because you specify the encoding header does not mean the NSURLRequest will actually perform said encoding.

The CFURLCreateStringByAddingPercentEscapes function can URL-encode your strings for you.

If you don't care about the encoding and just want to send simple strings, I suggest you skip the encoding entirely and use the "text/plain" content type.

Another option would be to use JSON instead of form encoding ("application/json"). JSON is much easier to use, both on the server side and on the client side. Cocoa has builtin support for JSON.

EDIT: Now that I read your code again, I see that you name your variable suggesting that the data is already JSON. If so, the solution is to use the "application/json" content type and make sure the server understands and decodes this properly.

edited Jul 03 '13 at 10:07

answered Jul 03 '13 at 10:00

Krumelur

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I need x-www-form-urlencoded encoding? Is this same? I will try and let you know – Imran Qadir Baksh - Baloch Jul 03 '13 at 10:01
Not necessarily, if you need to form-encode objects (i.e. arrays, dictionaries, and whatnot) you need to encode them as key-value pairs as well. For simple strings the escaping should be enough. – Krumelur Jul 03 '13 at 10:04
This is not a simple JSON. It coming from 3rd part side. I need to correctly encode this as x-www-form-urlencoded – Imran Qadir Baksh - Baloch Jul 03 '13 at 10:06
Thanks. CFURLCreateStringByAddingPercentEscapes works for now. – Imran Qadir Baksh - Baloch Jul 03 '13 at 10:10
Yep, this is the right answer. Either use JSON data with `application/json` header, or URL encoded data with `application/x-www-form-urlencoded` header. Mixing the two will result in no good. – Guillaume Algis Jul 03 '13 at 10:10

score 2 · Answer 2 · edited May 23 '17 at 11:59

2

If you need to have the ' ' -> '+' substitution, you need to do it on your own.

See: Objective-c iPhone percent encode a string? for the code to do the conversion.

Update

OK, based on your comment, here is some additional information. Based on old handling of query strings, '+' characters are turned into ' ' characters.

It was enshrined into old JavaScript, CGI, and PHP handing of URLs.

Microsoft even has a buggy example demonstrating it.

You can avoid the problem by always converting '+' to '%2B' and ' ' to '%20'.

edited May 23 '17 at 11:59

Community

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answered Jul 03 '13 at 10:03

Jeffery Thomas

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I am sending plus from client. I need to get + on server. I need simply this – Imran Qadir Baksh - Baloch Jul 03 '13 at 10:04
@user960567 I updated my post with some information as to why this happens. – Jeffery Thomas Jul 03 '13 at 10:44

score 1 · Answer 3 · answered Jul 03 '13 at 10:04

1

The AFNetworking library solves it thusly:

static NSString * AFPercentEscapedQueryStringPairMemberFromStringWithEncoding(NSString *string, NSStringEncoding encoding) {
    static NSString * const kAFCharactersToBeEscaped = @":/?&=;+!@#$()~',*";
    static NSString * const kAFCharactersToLeaveUnescaped = @"[].";

    return (__bridge_transfer  NSString *)CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, (__bridge CFStringRef)string, (__bridge CFStringRef)kAFCharactersToLeaveUnescaped, (__bridge CFStringRef)kAFCharactersToBeEscaped, CFStringConvertNSStringEncodingToEncoding(encoding));
}

answered Jul 03 '13 at 10:04

FluffulousChimp

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Would you care to elaborate? – FluffulousChimp Jul 03 '13 at 10:16
Yes, sorry that was kind of mean. My point is that a third party lib is not the solution to all problems. This does solves (I guess) the problem, but by avoiding it altogether. I didn't downvote though, as it does provide a solution to the problem :) – Guillaume Algis Jul 03 '13 at 10:21
I agree; I was showing the implementation only as an example of how others have solved it, not to advocate for using that library _in toto_. – FluffulousChimp Jul 03 '13 at 10:58

score 1 · Answer 4 · answered Jun 29 '15 at 10:33

As @Krumelur suggested,The CFURLCreateStringByAddingPercentEscapes function can URL-encode your strings for you. And Here is the code.

static NSString * CTPercentEscapedQueryStringValueFromStringWithEncoding(NSString *string, NSStringEncoding encoding) {
    return (__bridge_transfer  NSString *)CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, (__bridge CFStringRef)string, NULL, (__bridge CFStringRef)kCharactersToBeEscapedInQueryString, CFStringConvertNSStringEncodingToEncoding(encoding));
}

And Complete solution : To get encoded url from parameters use below code.

/**
 get parameterized url from url and query parameters.
 */
+(NSString *)getParameterizedUrl:(NSString *)url withParameters:(NSDictionary *)queryDictionary
{
    NSMutableArray *mutablePairs = [NSMutableArray array];
    for (NSString *key in queryDictionary) {
        [mutablePairs addObject:[NSString stringWithFormat:@"%@=%@", CTPercentEscapedQueryStringKeyFromStringWithEncoding(key, NSUTF8StringEncoding), CTPercentEscapedQueryStringValueFromStringWithEncoding(queryDictionary[key], NSUTF8StringEncoding)]];
    }

    return [[NSString alloc]initWithFormat:@"%@?%@",url,[mutablePairs componentsJoinedByString:@"&"]];
}

static NSString * const kCharactersToBeEscapedInQueryString = @":/?&=;+!@#$()',*";

static NSString * CTPercentEscapedQueryStringKeyFromStringWithEncoding(NSString *string, NSStringEncoding encoding) {
    static NSString * const kCharactersToLeaveUnescapedInQueryStringPairKey = @"[].";

    return (__bridge_transfer  NSString *)CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, (__bridge CFStringRef)string, (__bridge CFStringRef)kCharactersToLeaveUnescapedInQueryStringPairKey, (__bridge CFStringRef)kCharactersToBeEscapedInQueryString, CFStringConvertNSStringEncodingToEncoding(encoding));
}

static NSString * CTPercentEscapedQueryStringValueFromStringWithEncoding(NSString *string, NSStringEncoding encoding) {
    return (__bridge_transfer  NSString *)CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault, (__bridge CFStringRef)string, NULL, (__bridge CFStringRef)kCharactersToBeEscapedInQueryString, CFStringConvertNSStringEncodingToEncoding(encoding));
}

And use in your code as

NSString *url = [self getParameterizedUrl:@"http://www.example.com" withParameters:self.params];

Note : params here is dictionary.

IOS not encoding plus(+) sign in x-www-form-urlencoded POST request?

4 Answers4