Question

我想从SQL数据库获取地址，将它们放入地理编码器函数中以获取每个位置的经度和纬度，然后在将它们绘制到地图上之前以XML格式显示位置信息。然而，经度和纬度变得空白。我做错了什么？

 <? $sqlLink = mysqli_connect($host, $sqlusername, $sqlpassword, $db_name);
    $sqlLinkError = mysqli_connect($host, $sqlusername, $sqlpassword, $db_name);
    $mysqli = new mysqli($host, $sqlusername, $sqlpassword, $db_name);
    if ($mysqli->connect_errno)
    {
        print "Error Connecting to Database";
        exit(); 
    }
    if (mysqli_connect_errno())
    {
        print "Error Connecting to Database";
        exit();
    }


function parseToXML($htmlStr)
{
$xmlStr=str_replace('<','&lt;',$htmlStr);
$xmlStr=str_replace('>','&gt;',$xmlStr);
$xmlStr=str_replace('"','&quot;',$xmlStr);
$xmlStr=str_replace("'",'&#39;',$xmlStr);
$xmlStr=str_replace("&",'&amp;',$xmlStr);
return $xmlStr;
}

// Opens a connection to a MySQL server
$connection=mysql_connect ('localhost', $sqlusername, $sqlpassword);
if (!$connection) {
  die('Not connected : ' . mysql_error());
}

// Set the active MySQL database
$db_selected = mysql_select_db($db_name, $connection);
if (!$db_selected) {
  die ('Can\'t use db : ' . mysql_error());
}

// Select all the rows in the markers table
$query = "SELECT id, address1, city, state, zipcode FROM client_information LIMIT 0,12";
$result = mysql_query($query);
if (!$result) {
  die('Invalid query: ' . mysql_error());
}

header("Content-type: text/xml");

// Start XML file, echo parent node
echo '<markers>';

// Iterate through the rows, printing XML nodes for each
while ($row = @mysql_fetch_assoc($result)){

    $data_arr = geocode($row['address1']);

    if($data_arr){

        $latitude = $data_arr[0];
        $longitude = $data_arr[1];
        $formatted_address = $data_arr[2];
    }
    else{
        echo "Error";
    }

  // ADD TO XML DOCUMENT NODE
  echo '<marker ';
   echo 'address="' . parseToXML($row['address1']) . '" ';
  echo 'lat="' . $latitude . '" ';
  echo 'lng="' . $longitude . '" ';
  echo '/>';




}

// End XML file
echo '</markers>';
?>

<?
// function to geocode address, it will return false if unable to geocode address
function geocode($address){

    // url encode the address
    $address = urlencode($address);

    // google map geocode api url
    $url = "http://maps.google.com/maps/api/geocode/json?address={$address}";

    // get the json response
    $resp_json = file_get_contents($url);

    // decode the json
    $resp = json_decode($resp_json, true);

    // response status will be 'OK', if able to geocode given address 
    if($resp['status']=='OK'){

        // get the important data
        $lati = $resp['results'][0]['geometry']['location']['lat'];
        $longi = $resp['results'][0]['geometry']['location']['lng'];
        $formatted_address = $resp['results'][0]['formatted_address'];

        // verify if data is complete
        if($lati && $longi && $formatted_address){

            // put the data in the array
            $data_arr = array();            

            array_push(
                $data_arr, 
                    $lati, 
                    $longi, 
                    $formatted_address
                );

            return $data_arr;

        }else{
            return false;
        }

    }else{
        return false;
    }
}
?>

Answer 1

行：$url = "http://maps.google.com/maps/api/geocode/json?address={$address}";

我正在使用我认为现在已经过时了，我将其修改为现在使用的那个

$url = "https://maps.googleapis.com/maps/api/geocode/json?address={$address}&key="MY API KEY";

它现在有效。

Answer 2

其他人因为他们有类似问题而阅读此内容，我将分享我的解决方案，因为我在使用类似技术时遇到类似的问题，连接到Google Geocode时。事实证明，如果您使用的是PHP 5.4版，则$resp_json = file_get_contents($url);将不会返回对象。它会抛出错误而不是返回false。升级到5.6版为我修复了它。

php geocode功能不返回经度和纬度

2 个答案: