Objective-c – Parse NSURL query property

iphonensurlobjective c

I have a URL like myApp://action/1?parameter=2&secondparameter=3

With the property query I get following part of my URL

parameter=2&secondparameter=3

Is there any way easy to put this in a NSDictionary or an Array?

Thx a lot

Best Answer

You can use queryItems in URLComponents.

When you get this property’s value, the NSURLComponents class parses the query string and returns an array of NSURLQueryItem objects, each of which represents a single key-value pair, in the order in which they appear in the original query string.

Swift

let url = "http://example.com?param1=value1&param2=param2"
let queryItems = URLComponents(string: url)?.queryItems
let param1 = queryItems?.filter({$0.name == "param1"}).first
print(param1?.value)

Alternatively, you can add an extension on URL to make things easier.

extension URL {
    var queryParameters: QueryParameters { return QueryParameters(url: self) }
}

class QueryParameters {
    let queryItems: [URLQueryItem]
    init(url: URL?) {
        queryItems = URLComponents(string: url?.absoluteString ?? "")?.queryItems ?? []
        print(queryItems)
    }
    subscript(name: String) -> String? {
        return queryItems.first(where: { $0.name == name })?.value
    }
}

You can then access the parameter by its name.

let url = "http://example.com?param1=value1&param2=param2"
print(url.queryParameters["param1"])

Related Solutions

Objective-c – NSString property: copy or retain

For attributes whose type is an immutable value class that conforms to the NSCopying protocol, you almost always should specify copy in your @property declaration. Specifying retain is something you almost never want in such a situation.

Here's why you want to do that:

NSMutableString *someName = [NSMutableString stringWithString:@"Chris"];

Person *p = [[[Person alloc] init] autorelease];
p.name = someName;

[someName setString:@"Debajit"];

The current value of the Person.name property will be different depending on whether the property is declared retain or copy — it will be @"Debajit" if the property is marked retain, but @"Chris" if the property is marked copy.

Since in almost all cases you want to prevent mutating an object's attributes behind its back, you should mark the properties representing them copy. (And if you write the setter yourself instead of using @synthesize you should remember to actually use copy instead of retain in it.)

Objective-c – Creating URL query parameters from NSDictionary objects in ObjectiveC

Introduced in iOS8 and OS X 10.10 is NSURLQueryItem, which can be used to build queries. From the docs on NSURLQueryItem:

An NSURLQueryItem object represents a single name/value pair for an item in the query portion of a URL. You use query items with the queryItems property of an NSURLComponents object.

To create one use the designated initializer queryItemWithName:value: and then add them to NSURLComponents to generate an NSURL. For example:

NSURLComponents *components = [NSURLComponents componentsWithString:@"http://stackoverflow.com"];
NSURLQueryItem *search = [NSURLQueryItem queryItemWithName:@"q" value:@"ios"];
NSURLQueryItem *count = [NSURLQueryItem queryItemWithName:@"count" value:@"10"];
components.queryItems = @[ search, count ];
NSURL *url = components.URL; // http://stackoverflow.com?q=ios&count=10

Notice that the question mark and ampersand are automatically handled. Creating an NSURL from a dictionary of parameters is as simple as:

NSDictionary *queryDictionary = @{ @"q": @"ios", @"count": @"10" };
NSMutableArray *queryItems = [NSMutableArray array];
for (NSString *key in queryDictionary) {
    [queryItems addObject:[NSURLQueryItem queryItemWithName:key value:queryDictionary[key]]];
}
components.queryItems = queryItems;

I've also written a blog post on how to build URLs with NSURLComponents and NSURLQueryItems.

Best Answer

Swift

Related Solutions

Objective-c – NSString property: copy or retain

Objective-c – Creating URL query parameters from NSDictionary objects in ObjectiveC

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