Iphone 如何从非标准URL NSString对象提取参数?
我正在进行iOS开发,我的iPhone应用程序有一个定制的URL方案,看起来像Iphone 如何从非标准URL NSString对象提取参数?,iphone,objective-c,ios,Iphone,Objective C,Ios,我正在进行iOS开发,我的iPhone应用程序有一个定制的URL方案,看起来像myApp://?q=200。我有以下代码来获取查询参数 NSString *urlString = [url absoluteString]; NSString *query = [urlString stringByReplacingOccurrencesOfString:@"myApp://?q=" withString:@""]; …但如果我添加更多参数,我想让它更能证明未来。如何以更安全的方式提取“q”参数
myApp://?q=200
。我有以下代码来获取查询参数
NSString *urlString = [url absoluteString];
NSString *query = [urlString stringByReplacingOccurrencesOfString:@"myApp://?q=" withString:@""];
…但如果我添加更多参数,我想让它更能证明未来。如何以更安全的方式提取“q”参数
提前感谢您的智慧 您可以将从URL返回的查询按
&
和=
进行拆分,并将它们放入字典中
NSURL *url = [NSURL URLWithString:@"myApp://?q=200"];
NSArray *query = [[url query] componentsSeparatedByString:@"&"];
NSMutableDictionary *parameters = [NSMutableDictionary dictionaryWithCapacity:[query count]];
for(NSString *parameter in query)
{
NSArray *kv = [parameter componentsSeparatedByString:@"="];
[parameters setObject:[kv count] > 1 ? [[kv objectAtIndex:1] stringByReplacingPercentEscapesUsingEncoding:NSISOLatin1StringEncoding] : [NSNull null]
forKey:[[kv objectAtIndex:0] stringByReplacingPercentEscapesUsingEncoding:NSISOLatin1StringEncoding]];
}
NSLog(@"Parameters: %@", parameters);
NSLog(@"q = %@", [parameters objectForKey:@"q"]);
在本例中,如果参数没有值,我只需将其设置为
NSNull
。这意味着您需要检查NSNull
,或者更改逻辑以跳过带有值的键,或者将它们设置为空字符串。这可以从我的头顶开始,但还不包括错误检查输入
-(NSDictionary*) parameterDictionaryFromString: (NSURL*) url {
//input can be something like: "myApp://?q=one&q2=two&q3=three"
NSString *requestString = [url query];
//now we have q=one&q2=two&q3=three
NSArray *requests = [requestString componentsSeparatedByString: @"&"];
NSMutableDictionary *resultDictionary = [NSMutableDictionary dictionary];
for (NSString *singleParameter in requests) {
NSArray *keyValuePair = [singleParameter componentsSeparatedByString: @"="];
[resultDictionary setObject: [keyValuePair objectAtIndex: 1] forKey: [keyValuePair objectAtIndex: 0]];
}
NSURL *u = [NSURL URLWithString: @"myApp://something?q=1&check=yes"];
NSLog(@"paramStr = %@", [u parameterString]);
return [resultDictionary copy];
}
关键是利用NSArray类而不是StringReplace。检查我们的
//Your Example:
//@"myApp://?q=200"
//Break:
NSArray *queryParts = [urlString componentsSeparatedByString:@"?q="];
//Assure Content:
if ([[array objectAtIndex:1] length]>0) {
//Setter:
NSString *queryString = [array objectAtIndex:1];
//... Use away...
}