假设我有一个NSURL
?无论它是否已经有一个空的查询字符串,如何向NSURL
的query
添加一个或多个参数?也就是说,有没有人知道这个函数的实现?
- (NSURL *)URLByAppendingQueryString:(NSString *)queryString
这样它就可以满足这个NSURL+AdditionsSpec.h
文件:
#import "NSURL+Additions.h"
#import "Kiwi.h"
SPEC_BEGIN(NSURL_AdditionsSpec)
describe(@"NSURL+Additions", ^{
__block NSURL *aURL;
beforeEach(^{
aURL = [[NSURL alloc] initWithString:@"http://www.example.com"];
aURLWithQuery = [[NSURL alloc] initWithString:@"http://www.example.com?key=value"];
});
afterEach(^{
[aURL release];
[aURLWithQuery release];
});
describe(@"-URLByAppendingQueryString:", ^{
it(@"adds to plain URL", ^{
[[[[aURL URLByAppendingQueryString:@"key=value&key2=value2"] query] should]
equal:@"key=value&key2=value2"];
});
it(@"appends to the existing query sting", ^{
[[[[aURLWithQuery URLByAppendingQueryString:@"key2=value2&key3=value3"] query] should]
equal:@"key=value&key2=value2&key3=value3"];
});
});
});
SPEC_END
发布于 2017-05-18 15:59:46
正如其他人所提到的,您可以使用NSURLComponents
来构造URL。
@implementation NSURL (Additions)
- (NSURL *)URLByAppendingQueryParameters:(NSDictionary *)queryParameters
{
NSURLComponents *components = [[NSURLComponents alloc] initWithURL:self resolvingAgainstBaseURL:NO];
NSMutableArray *queryItems = [NSMutableArray array:components.queryItems];
for (NSString *key in [queryParameters allKeys]) {
NSURLQueryItem *queryItem = [[NSURLQueryItem alloc] initWithName:key value:queryParameters[key]];
[queryItems addObject:queryItem];
}
components.queryItems = queryItems;
return [components URL];
}
@end
发布于 2011-06-11 05:16:06
下面是一个实现,它通过了您的规范:
@implementation NSURL (Additions)
- (NSURL *)URLByAppendingQueryString:(NSString *)queryString {
if (![queryString length]) {
return self;
}
NSString *URLString = [[NSString alloc] initWithFormat:@"%@%@%@", [self absoluteString],
[self query] ? @"&" : @"?", queryString];
NSURL *theURL = [NSURL URLWithString:URLString];
[URLString release];
return theURL;
}
@end
下面是NSString
的一个实现
@implementation NSString (Additions)
- (NSURL *)URLByAppendingQueryString:(NSString *)queryString {
if (![queryString length]) {
return [NSURL URLWithString:self];
}
NSString *URLString = [[NSString alloc] initWithFormat:@"%@%@%@", self,
[self rangeOfString:@"?"].length > 0 ? @"&" : @"?", queryString];
NSURL *theURL = [NSURL URLWithString:URLString];
[URLString release];
return theURL;
}
// Or:
- (NSString *)URLStringByAppendingQueryString:(NSString *)queryString {
if (![queryString length]) {
return self;
}
return [NSString stringWithFormat:@"%@%@%@", self,
[self rangeOfString:@"?"].length > 0 ? @"&" : @"?", queryString];
}
@end
发布于 2015-11-06 17:05:26
对于那些不想在用NSURLComponents
构建NSURL
时编写样板代码的人来说,这只是一个友好的帖子。
自从iOS8以来,我们有了NSURLQueryItem
,它可以帮助我们快速构建URL请求。
我写了一个方便的类别来简化工作,你可以在这里获取:URLQueryBuilder
下面的示例说明了使用它是多么容易:
NSString *baseURL = @"https://google.com/search";
NSDictionary *items = @{
@"q" : @"arsenkin.com",
@"hl" : @"en_US",
@"lr" : @"lang_en"
};
NSURL *URL = [NSURL ars_queryWithString:baseURL queryElements:items];
// https://google.com/search?q=arsenkin.com&hl=en_US&lr=lang_en
https://stackoverflow.com/questions/6309698
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