傳遞網址參數

我傳遞的網址，在這個url一些字符串值和NSInteger價值，但是當我把breakpoint就這個問題和我選項卡上的痕跡，然後它告訴我這個消息exc-bad-access在url我給出大膽的，請看到「大膽'我想通過價值：
[ NSInteger day,NSInteger day1,NSString *fromDate1, NSString *fromDate1,NSString *OriginCode,NSString *DestinCode]。傳遞網址參數

當我把breakpoint放到url上，但是當我進入breakpoint時，我的應用程序崩潰了，爲什麼它會崩潰？幫我。我錯在哪裏？

-(void)sendRequest 

{  








    stringWithFormat:@"http://www.google.com?AvailabilitySearchInputFRSearchView%24ButtonSubmit=Search%20For%20Flights%20&AvailabilitySeast=",day,day1,DestinCode,"2011-09","2011-09",OriginCode]; 
    NSString *urlString = [NSString stringWithFormat:@"http://www.bookairways tickt.com/Sales/FRSearch.aspx?AvailabilitySearchInputFRSearchView%24ButtonSubmit=Search%20For%20Flights%20&AvailabilitySearchInputFRSearchView%24DropDownListMarketDay1=**%i**&AvailabilitySearchInputFRSearchView%24DropDownListMarketDay2=**%i**&AvailabilitySearchInputFRSearchView%24DropDownListMarketDestination1=**%@**&AvailabilitySearchInputFRSearchView%24DropDownListMarketMonth1=**%@**&AvailabilitySearchInputFRSearchView%24DropDownListMarketMonth2=**%@**&AvailabilitySearchInputFRSearchView%24DropDownListMarketOrigin1=**%@**&AvailabilitySearchInputFRSearchView%24DropDownListPassengerType_ADT=1&AvailabilitySearchInputFRSearchView%24DropDownListPassengerType_CHD=0&AvailabilitySearchInputFRSearchView%24DropDownListPassengerType_INFANT=0&AvailabilitySearchInputFRSearchView%24RadioButtonFlowSelector=FlightAndCar&AvailabilitySearchInputFRSearchView%24RadioButtonMarketStructure=RoundTrip&AvailabilitySearchInputFRSearchView%24discountPax=0&__EVENTARGUMENT=&__EVENTTARGET=&__VIEWSTATE=%2FwEPDwUBMGRkg4UKvNNb1NbM14%2F2n9zUxhNQ%2B%2BA%3D&errorlist=",day,day1,DestinCode,fromDate1,fromDate2,OriginCode]; 
    //urlString=[urlString stringByAddingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; 
    NSURL *url = [NSURL URLWithString:urlString]; 
    NSLog(@"************url:%@",url); 

    NSURLRequest *theRequest=[NSURLRequest requestWithURL:url 
            cachePolicy:NSURLRequestUseProtocolCachePolicy 
           timeoutInterval:60.0]; 
    NSURLConnection *theConnection=[[NSURLConnection alloc] initWithRequest:theRequest delegate:self]; 
    if (theConnection) { 
     webData = [[NSMutableData data] retain]; 
     NSLog(@"%@",webData); 
    } else { 

    } 

}

來源

2011-09-22 Harish

哈里什這將幫助ü在創建URL檢查這個我們http://wiki.akosma.com/IPhone_URL_Schemes 這樣

NSString *template = @"appigotodo://com.example.xyzapp/import?name=%@&note=%@&due-date=%@&priority=%@&repeat=%@"; 
NSString *name = @"Buy%20some%20milk"; 
NSString *note = @"Stop%20on%20the%20way%20home%20from%20work."; 
NSString *dueDate = @"2009-07-16"; 
NSString *priority = @"1"; 
NSString *repeat = @"101"; 
NSString *stringURL = [NSString stringWithFormat:template, name, note, dueDate, priority, repeat]; 
NSURL *url = [NSURL URLWithString:stringURL];

[UIApplication的sharedApplication]的OpenURL：URL]。

來源

2011-10-03 03:36:26 Ajay

使你正確的網址是這樣的： -

NSURL *url = [NSURL URLWithString:[*yourstring* stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

來源

2011-09-22 05:34:04 Gypsa

兩件事情：

的網址中有沒有被用作佔位符許多％符號。您代碼中不在'**'之間的％符號需要像這樣轉義：%%。換句話說，SearchInputFRSearchView%24Button應該是SearchInputFRSearchView%%24Button。
您正在使用％i將整數放入您的字符串中。您應該使用％d代替。

來源

2011-09-22 05:40:23

再次檢查上面我再次改變鏈接，但甚至沒有得到解決。我該如何做到這一點 – Harish

傳遞網址參數

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