使用openURL发送NSDictionary

Question

我想使用以下命令打开另一个应用程序

[[UIApplication sharedApplication]openURL:[[NSURL alloc]initWithString:myString]];

与myString一样的对数

NSString *myString=[NSString stringWithFormat:@"testHandleOpenUrl://?%@",@"123"];

它工作得很好，但是如果我尝试使用NSDictionary，比如

NSString *myString=[NSString stringWithFormat:@"testHandleOpenUrl://?%@",userInfo];

它会失败，不会出现错误

希望你能帮助我。

Answer 1

尝试枚举字典中的所有元素，并将这些元素附加到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

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];

Answer 3

当您在格式字符串中使用NSDictionary时，您会得到如下所示的URL：

testHandleOpenUrl://?{
    bar = foo;
}

在userInfo上调用-description的结果将简单地替换为%@。您可能希望在URL中传递字典中包含的参数。可能是这样的：

NSDictionary *userInfo = [NSDictionary dictionaryWithObjectsAndKeys:@"bar", @"foo", nil];
NSString *myString=[NSString stringWithFormat:@"testHandleOpenUrl://?foo=%@", [userInfo objectForKey:@"foo"]];
NSLog(@"myString: %@", myString); // prints "myString: testHandleOpenUrl://?foo=bar"