I dont know why but when I am calling the following URL
it gives me BAD REQUEST - INVALID URL, although this URL is working fine on safari browser and other browsers as well
http://www.ysl.com/wx/shop-product/women/top-handles#{"ytosQuery":"true","department":"handbags_tophandle_w","gender":"D","brand":"","macro":"","micro":"","season":"A,P,E","color":"","size":"","site":"","section":"","sortRule":"","yurirulename":"searchwithdepartment","microcolor":"","agerange":"","macroMarchio":"","page":"2","productsPerPage":"50","modelnames":"","look":"","washtype":"","fabric":"","prints":"","suggestion":"false","suggestionValue":"","material":"","occasion":"","weight":"","gal
I am using following code:
NSString *str = [NSString stringWithFormat:#"%#",[payload stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]]];
NSString* webStringURL = [str stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
webStringURL = [str stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *payload = [NSURL URLWithString:webStringURL];
Any Help Guys? What is that I am doing wrong?
Don't use stringByReplacingPercentEscapesUsingEncoding: on the hole URL, but just on the GET parameters.
Now http:// will also be escaped, thus becoming http%3A%2F%2F which is not valid as an URL.
Related
I have a strange issue with Xcode6.1
_mainURl is my ' ServerLink '
service is ' GetUserById '
NSString* str = [NSString stringWithFormat:#"%#%#",_mainURl,service];
NSURL *url = [NSURL URLWithString:str];
When I append two strings and Create a url with NSURL, url getting 'null'
But, When I directly given the server link followed by serviceName I can generate URL.
NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:#"ServerLink/GetUserById"]];
Based on what you have written, the string you format would be "ServerLinkGetUserById". That is different to the string you enter manually of #"ServerLink/GetUserById".
Try updating your format to be:
NSString* str = [NSString stringWithFormat:#"%#/%#",_mainURl,service];
NSString* str = [NSString stringWithFormat:#"%#/%#",_mainURl,service];
NSURL *url = [NSURL URLWithString:str];
don't you try like this ??
i think i am not sure exact what issue having you but as your bellow description, i think you left this part
I have url that I am going to request data from. Here is the code.
urlStr = [urlStr stringByAddingPercentEscapesUsingEncoding:
NSASCIIStringEncoding];
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:[NSURL URLWithString:urlStr]];
This works except when I have certain characters such as ş. The urlstr will return null. Is there a way around this to except certain character types? Any tips or suggestions are appreciated.
I've used the following method with success in many applications:
- (NSString *)urlEncode:(NSString *)str {
return (NSString *)CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(NULL, (CFStringRef)str, NULL, CFSTR("!*'();:#&=+$,/?%#[]"), kCFStringEncodingUTF8));
}
Note that I use this on only the params of the URL, so the following would work (notice I added the ş, which seemed to work, although I'm not familiar with that character):
NSString *baseURL = #"http://www.google.com";
NSString *paramsString = #"testKey=test value_with some (weirdness)!___ş";
NSString *resultingURLString = [NSString stringWithFormat:#"%#?%#", baseURL, [self urlEncode:paramsString]];
Which produces the result:
http://www.google.com?testKey%3Dtest%20value_with%20some%20%28weirdness%29%21___%C5%9F
I've got a small problem that seems a little bit odd to me. I often used NSString or NSLog while adding NSNumbers into several places:
NSNumber *categoryId = [[NSNumber alloc]initWithInt:0];
NSURL *url = [NSURL URLWithString:#"http://shop.rs/api/json.php?action=getCategoryByCategory&category=%i",[categoryId integerValue]];
Now xcode tells me that I'm too many arguments. What am I doing wrong? Setting up an NSNumber into NSStrings or NSLogs works as I did it above.
Best Regards
What is wrong is on
NSURL *url = [NSURL URLWithString:#"http://shop.rs/api/json.php?action=getCategoryByCategory&category=%i",[categoryId integerValue]];
you are calling URLWithString: and then pass in a string that is not being formatted correctly. If you want to do it all on one line then you need to be using stringWithFormat: like
NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:#"http://shop.rs/api/json.php?action=getCategoryByCategory&category=%i",[categoryId integerValue]]];
Because it is adding a parameter you can't just create a string like you normally would with #"some text" you need to format it using the stringWithFormat: which will return an NSString * with the text held within #"" and the paramters you pass in. So [NSString stringWithFormat:#"My String will come with %#", #"Apples"]; this would provide an NSString with "My String will come with Apples". For more information check out the Apple Documentation for NSString and stringWithFormat:
Try this :
NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:#"http://shop.rs/api/json.phpaction=getCategoryByCategory&category=%i", [categoryId integerValue]]];
Initially code was wrong because of : "categoryId integerValue]" (I forgot a '[').
You can use NSString to form your NSURL. You can then pass it to your URLWithString like below:
NSNumber *categoryId = [NSNumber numberWithInteger:0];
NSString *urlString = [NSString stringWithFormat:#"http://shop.rs/api/json.php?action=getCategoryByCategory&category=%i",[categoryId integerValue]];
NSURL *url = [NSURL URLWithString:urlString];
I have a UISearchBar from which I'm extracting the text that represents an address and building from it a JSON URL for Google Geocoder.
NSString* address = searchbar.text;
NSString* url = [NSString stringWithFormat:#"http://maps.google.com/maps/api/geocode/json?address=%#&sensor=false", address];
If I copy & paste the url from debug window to the browser it works like a charm but when I try to convert it to NSURL i get nil
NSURL* theUrl = [NSURL URLWithString:url];
Any ideas?
I added the encoding to the address prior to concatenating the url and that fixed the problem so now the code looks like this:
NSString* address = searchbar.text;
NSString *encodedString = (NSString *)CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(NULL, CFBridgingRetain(address), NULL, (CFStringRef)#"!*'();:#&=+$,/?%#[]", kCFStringEncodingUTF8));
NSString* url = [NSString stringWithFormat:#"http://maps.google.com/maps/api/geocode/json?address=%#&sensor=false", encodedString];
I used this code from the Stack Overflow question: URLWithString: returns nil:
//localisationName is a arbitrary string here
NSString* webName = [localisationName stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSString* stringURL = [NSString stringWithFormat:#"http://maps.google.com/maps/geo?q=%#,Montréal,Communauté-Urbaine-de-Montréal,Québec,Canadae&output=csv&oe=utf8&sensor=false", webName];
NSString* webStringURL = [stringURL stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL* url = [NSURL URLWithString:webStringURL];
When I copied it into my code, there wasn't any issue but when I modified it to use my url, I got this issue:
Data argument not used by format string.
But it works fine. In my project:
.h:
NSString *localisationName;
.m:
NSString* webName = [localisationName stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSString* stringURL = [NSString stringWithFormat:#"http://en.wikipedia.org/wiki/Hősök_tere", webName];
NSString* webStringURL = [stringURL stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL* url = [NSURL URLWithString:webStringURL];
[_webView loadRequest:[NSURLRequest requestWithURL:url]];
How can I solve this? Anything missing from my code?
The # in the original string is used as a placeholder where the value of webName is inserted. In your code, you have no such placeholder, so you are telling it to put webName into your string, but you aren't saying where.
If you don't want to insert webName into the string, then half your code is redundant. All you need is:
NSString* stringURL = #"http://en.wikipedia.org/wiki/Hősök_tere";
NSString* webStringURL = [stringURL stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL* url = [NSURL URLWithString:webStringURL];
[_webView loadRequest:[NSURLRequest requestWithURL:url]];
The +stringWithFormat: method will return a string created by using a given format string as a template into which the remaining argument values are substituted. And in the first code block, %# will be replaced by value of webName.
In your modified version, the format parameter, which is #"http://en.wikipedia.org/wiki/Hősök_tere", does not contain any format specifiers, so
NSString* stringURL = [NSString stringWithFormat:#"http://en.wikipedia.org/wiki/Hősök_tere", webName];
just runs like this (with the warning Data argument not used by format string.):
NSString* stringURL = #"http://en.wikipedia.org/wiki/Hősök_tere";