谷歌將REQUEST_DENIED恐怖

Question

我想使用谷歌Places API的，但我不斷收到一個REQUEST_DENIED。我進入了Google API控制檯，並開啓了Google Places API。我的代碼是這樣的：

NSString *searchString = [NSString stringWithFormat:@"Beirut"]; 
    NSString *urlString = [NSString stringWithFormat:@"https://maps.googleapis.com/maps/api/place/textsearch/json?query=%@&sensor=true&key=%@!",searchString, kGOOGLE_API_KEY]; 

    NSURL *requestURL = [NSURL URLWithString:urlString]; 
    NSURLRequest *request = [NSURLRequest requestWithURL:(requestURL)]; 

    //response 
    NSData *response = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil]; 
    NSError *jsonParsingError = nil; 
    NSDictionary *locationResults = [NSJSONSerialization JSONObjectWithData:response options:0 error:&jsonParsingError]; 

    NSLog(@"%@",locationResults); 

我翻了一番檢查，我使用了正確的API密鑰和我收到此輸出：

{ 
    "debug_info" =  (
    ); 
    "html_attributions" =  (
    ); 
    results =  (
    ); 
    status = "REQUEST_DENIED"; 
} 

我再次進入到谷歌API控制檯，點擊「嘗試「通過谷歌地方開/關開關旁邊的」問號「信息按鈕，我直接到另一個具有相同輸出的標籤。我該如何解決這個問題？

編輯：多類型添加到URL會導致錯誤

NSString *urlString = [NSString stringWithFormat:@"https://maps.googleapis.com/maps/api/place/search/json?location=%f,%f&radius=%@&types=food|bar&sensor=true&key=%@", currentLocation.coordinate.latitude, currentLocation.coordinate.longitude, [NSString stringWithFormat:@"%i", 1000], kGOOGLE_API_KEY]; 

Answer 1

您有一個！在urlString的末尾。我刪除了它，測試代碼並將其返回結果

NSString *searchString = [NSString stringWithFormat:@"NewYork"]; 
    NSString *urlString = [NSString stringWithFormat:@"https://maps.googleapis.com/maps/api/place/textsearch/json?query=%@&sensor=true&key=%@",searchString, kGOOGLE_API_KEY]; 

    NSURL *requestURL = [NSURL URLWithString:urlString]; 
    NSURLRequest *request = [NSURLRequest requestWithURL:requestURL]; 

    //response 
    NSData *response = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil]; 
    if(response){ 
     NSError *jsonParsingError = nil; 
     NSDictionary *locationResults = [NSJSONSerialization JSONObjectWithData:response options:0 error:&jsonParsingError]; 

     NSLog(@"%@",locationResults); 
    }else{ 
     NSLog(@"nil"); 
    }