php - php错误传递给google map api的地址坐标

Question

我收到以下错误

    Notice: Trying to get property of non-object in xxx/index.php on line 22

当我使用此代码时:

    $lon = $_GET['lon'];
    $lat = $_GET['lat'];
    $address = $_GET['address'];
    $range = $_GET['range'];

    if($address != 'false') {
      $address = str_replace(" ", "+", $address); // replace all the white space with         "+" sign to match with google search pattern
      $url = "http://maps.google.com/maps/api/geocode/json?sensor=false&address=$address";
      $response = file_get_contents($url);
      $json = json_decode($response,TRUE); //generate array object from the response from the web

      //these are the latitude and longitude definitions
      $lat = $response->results->geometry->location->lat;
      $lon = $response->results->geometry->location->lng;
    }

我很不熟悉php。谁能给我一些指导吗？
谢谢您的帮助

**编辑(显示完整代码):
2010年1月由TECFA的Daniel K. Schneider制作。这是免费软件。
将连接到MySQL数据库，执行SQL语句并
然后将结果作为有效的XML返回。
*/
error_reporting(E_ALL);

      // ---------------------------  Configuration section
      $lon = $_GET['lon'];
      $lat = $_GET['lat'];
      $address = $_GET['address'];
      $range = $_GET['range'];

      if($address != 'false') {
        $address = str_replace(" ", "+", $address); // replace all the white space with           "+" sign to match with google search pattern
        $url = "http://maps.google.com/maps/api/geocode/json?sensor=false&address=$address";
        $response = file_get_contents($url);
        $json = json_decode($response); //generate array object from the response from the web

        //these are the latitude and longitude definitions
        $lat = $json->results->geometry->location->lat;
        $lon = $json->results->geometry->location->lng;
      }
      // Fill in according to your db settings (example is for a local and locked "play" WAMP server)
      // Make sure to keep the ; and the ""
      $host       = "x";
      $user       = "x";
      $pass       = "x";
      $database   = "x";

      // Replace by a query that matches your database
      $SQL_query =
      "SELECT 3956 * 2 * ASIN(SQRT(POWER(SIN(('".$lat."' -abs(lat)) * pi()/180 / 2),2) +           COS('".$lat."' * pi()/180 ) * COS(abs(lat) *pi()/180) * POWER(SIN(('".$lon."' - `long`) *pi()/180 / 2), 2) ))as distance FROM wp_places_locator dest having distance < '".$range."' ORDER BY distance limit 10;";

      // Optional: add the name of XSLT file.
      // $xslt_file = "mysql-result.xsl"; 

      // -------------------------- no changes needed below

      $DB_link = mysql_connect($host, $user, $pass) or die ("Could not find or access the database.");
      mysql_select_db($database, $DB_link) or die ("Could not find or access the database.");
      $result = mysql_query($SQL_query, $DB_link) or die ("Data not found. Your SQL           query didn't work... ");

      // we produce XML
      header("Content-type: text/xml");
      $XML = "<?xml version=\"1.0\"?>\n";
      //if ($xslt_file) $XML .= "<?xml-stylesheet href=\"$xslt_file\" type=\"text/xsl\"  
      // root node
      $XML .= "<result>\n";
      // rows
      while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {    
        $XML .= "\t<row>\n"; 
        $i = 0;
        // cells
        foreach ($row as $cell) {
          // Escaping illegal characters - not tested actually ;)
          $cell = str_replace("&", "&amp;", $cell);
          $cell = str_replace("<", "&lt;", $cell);
          $cell = str_replace(">", "&gt;", $cell);
          $cell = str_replace("\"", "&quot;", $cell);
          $col_name = mysql_field_name($result,$i);
          // creates the "<tag>contents</tag>" representing the column
          $XML .= "\t\t<" . $col_name . ">" . $cell . "</" . $col_name . ">\n";
          $i++;
        }
        $XML .= "\t</row>\n"; 
       }
      $XML .= "</result>\n";

      // output the whole XML string
      echo $XML;
      ?>

Answer 1

您收到错误消息是因为您将json编码为一个关联数组(您将true作为第二个参数传递)，并且试图使用另一个变量名将其作为一个对象进行访问，因此请删除第二个参数

//$json = json_decode($response,TRUE); //generate array object from the response from the web
//Don't pass true as second parameter
$json = json_decode($response); 

并使用正确的对象变量

$lat = $json->results->geometry->location->lat;
$lon = $json->results->geometry->location->lng;