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当我调用Google API获取地址时,我在这里调用API。Google Api与Json一起使用
-(void)searchviews:(NSString*)EditString selector:(SEL)sel
{
NSLog(@"Welcome To Search views");
searchviews=sel;
NSString *path =[NSString stringWithFormat:@"http://maps.google.com/maps/api/geocode/json?address=%@&sensor=false",EditString];
NSURL *url=[NSURL URLWithString:path];
NSLog(@"hiii---%@",url);
ASIFormDataRequest *request=[ASIFormDataRequest requestWithURL:url];
[request setRequestMethod:@"POST"];
[request setDelegate:self];
[request startAsynchronous];
[drkSignUp showWithMessage:nil];
NSLog(@" Complet--------------- ");
和对于请求方法我打电话一样。
- (void)requestFinished:(ASIHTTPRequest *)request {
//NSLog(@"%@",[request responseString]);
NSString *func = [self getFunc:[request url]];
NSLog(@"%@\n%@",func,[request responseString]);
if ([func isEqual:@"json?address=%@&sensor=false"])
{
NSDictionary *resDict = [parser objectWithString:[request responseString] error:nil];
NSLog(@"---- ResData%@",resDict);
NSString *result = [resDict objectForKey:@"successful"];
NSLog(@"hiiiii google api calling............");
[drkSignUp hide];
[self.delegate performSelector:searchviews withObject:[resDict objectForKey:@"results"]];
就是这样,但问题在乐趣中创造。当我打电话
if ([func isEqual:@"json?address=%@&sensor=false"])
它不调用cos它是动态String.So我应该在func中置入%@的位置?
@dipangsheth欢迎您。如果它有用,也不要忘记upvote和/或接受答案:) –