sending NSDictionary with openURL - iphone

Question

I want to open an other application using
[[UIApplication sharedApplication]openURL:[[NSURL alloc]initWithString:myString]];
as log as myString is like
NSString *myString=[NSString stringWithFormat:#"testHandleOpenUrl://?%#",#"123"];
it works fine but if I try to use an NSDictionary like
NSString *myString=[NSString stringWithFormat:#"testHandleOpenUrl://?%#",userInfo];
it fails without an error
Hope you can help me.

Answer 1

Try enumerating all the elements in your dictionary and appending those to your URL.
NSMutableString *params = [[[NSMutableString alloc] init] autorelease];
NSEnumerator *keys = [userInfo keyEnumerator];
NSString *name = [keys nextObject];
while (nil != name) {
[params appendString: name];
[params appendString: #"="];
[params appendString: [userInfo objectForKey:name]];
name = [keys nextObject];
if (nil != name) {
[params appendString: #"&"];
}
}
NSString *myString=[NSString stringWithFormat:#"testHandleOpenUrl://?%#",params];

Answer 2

When you use an NSDictionary in the format string, you get a URL that looks like this:
testHandleOpenUrl://?{
bar = foo;
}
The result of calling -description on userInfo is simply substituted for the %#. Presumably, you want to pass parameters contained in the dictionary in the URL. Probably something like this:
NSDictionary *userInfo = [NSDictionary dictionaryWithObjectsAndKeys:#"bar", #"foo", nil];
NSString *myString=[NSString stringWithFormat:#"testHandleOpenUrl://?foo=%#", [userInfo objectForKey:#"foo"]];
NSLog(#"myString: %#", myString); // prints "myString: testHandleOpenUrl://?foo=bar"

Answer 3

NSString * paramString = #"";
int i = 0;
for(NSString * key in [userInfo allKeys]){
NSString * value = (NSString*)[userInfo objectForKey:key];
NSString * valueParam = [NSString stringWithFormat:#"%#%#=%#",(i==0)?#"?":#"&",key,value];
paramString = [paramString stringByAppendingString:valueParam];
i++;
}
NSString *myString=[NSString stringWithFormat:#"testHandleOpenUrl://%#", paramString];