Question

我有一个iPhone应用程序使用JSON形式的NSData对象发出发布请求（下面的代码供参考）。我正在从CLLocation发送GPS坐标，但是想要访问PHP中的各个元素，这样我就可以将它们存储在某个db中（稍后会这样做）。

由于某些原因，我无法找到一种方法来访问PHP中的各个字段，而我或者得到整个数组，或者什么都没有。

Objective-C代码供参考：

  NSDictionary *locationDic = [[NSDictionary alloc] initWithObjectsAndKeys:
      [NSNumber numberWithDouble:location.coordinate.latitude], @"lat",
      [NSNumber numberWithDouble:location.coordinate.longitude], @"long",
      [NSNumber numberWithDouble:location.speed], @"speed",
                       nil];


  NSError *error;
  NSData *jsonData = [NSJSONSerialization dataWithJSONObject:locationDic
      options:NSJSONWritingPrettyPrinted // Pass 0 if you don't care about the readability of the generated string
      error:&error];

  NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"http://localhost/location.php"]];
  [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
  [request addValue:@"postValues" forHTTPHeaderField:@"HTTP_METHOD"];
  [request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
  [request setHTTPMethod:@"POST"];    
  [request setHTTPBody:jsonData];

PHP：

$handle = fopen("php://input", "rb");
$http_raw_post_data = '';
while (!feof($handle)) {
    $http_raw_post_data .= fread($handle, 8192);
}
fclose($handle);

$post_data = json_decode($http_raw_post_data,true);

if (is_array($post_data))
{
    $response = array("status" => "ok", "code" => 0, "original request" => $post_data);
}
else
{
    $response = array("status" => "error", "code" => -1, "original_request" => $post_data);
}

$processed = json_encode($response);

echo $processed;
echo $processed[0];

知道我在这里可能缺少什么吗？

Answer 1

我比你更简单地获得有效载荷：

  $content = file_get_contents('php://input');
  $ary = json_decode($content , true);

并在JSON格式良好时获取关联数组。

访问从iPhone应用程序发送的PHP中的数组对象

1 个答案: