How do i specify the query component of a URL?

Suppose i have a URL like:

NSString *urlString = @"";

And i want to specify the query component of this URL, so i did this:

NSURLComponents *components = [[NSURLComponents alloc] initWithString:urlString];
[components setQuery:@"guid=56d962c7e4b0b00655a669a2"];
NSLog(@"%@", components.query);

The first log displays which is not what i want, however the second log displays guid=56d962c7e4b0b00655a669a2. I don't understand the output of first log, if the second log displays that the query component was added, why does the first log not print it ? What i actually want to do is make a NSMutableRequest to the URL but when i do that after the above code and log the result of NSMutableRequest, it says that the required parameter is missing. How do i proceed about this issue. Any help will be appreciated. Thank you.


If you want to get the URL from the NSURLComponents, use the "URL" property of NSURLComponents.

NSString *urlString = @"";
NSURLComponents *components = [[NSURLComponents alloc] initWithString:urlString];
[components setQuery:@"guid=56d962c7e4b0b00655a669a2"];
NSLog(@"%@", components.URL); // Should log: ""
NSLog(@"%@", components.query);

NSURL *url = components.URL;

Alternatively, you could just append your queryString onto the end of your string and then turn the whole thing into an NSURL:

NSString *queryString = @"?guid=56d962c7e4b0b00655a669a2";
NSString *urlString = [@"" stringByAppendingString:queryString];
NSURL *url = [NSURL URLWithString:urlString];
NSLog(@"%@", url.absoluteString);

Why don't you try it this way?

NSString *query = @"guid=56d962c7e4b0b00655a669a2"
NSString *urlString = [NSString stringWithFormat:@"",query];
NSURL *url = [NSURL URLWithString:urlString];
NSURLRequest *request = [NSURLRequest requestWithURL:url];

So every time you want to change the query, you can change the query string and then the request will change too.

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