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請在下面的代碼中提供幫助。發生的事情是,當我點擊一個按鈕時,我得到了一個地方名稱的經緯度。但是,最近它不起作用。它打印出「地址x失敗到地理編碼。接收狀態」請注意,x是給定的地址,並且沒有給出狀態碼。使用v3進行Google地理編碼不會顯示錯誤
$id=$_REQUEST['id'];
define("MAPS_HOST", "maps.googleapis.com");
define("KEY", "xxxxxxx");
$query = "SELECT * FROM markers WHERE mid = $id LIMIT 1";
$result = mysql_query($query);
if (!$result) {
die("Invalid query: " . mysql_error());
}
// Initialize delay in geocode speed
$delay = 0;
$base_url = "http://" . MAPS_HOST . "/maps/api/geocode/xml?address=";
if($row = @mysql_fetch_assoc($result)){
$geocode_pending = true;
$address = $row["address"];
while ($geocode_pending) {
$request_url = $base_url . urlencode($address) . "&sensor=false&key=" . KEY;
$xml = simplexml_load_file($request_url) or die("url not loading");
$status = $xml->Response->Status->code;
if (strcmp($status, "200") == 0) {
// Successful geocode
$geocode_pending = false;
$coordinates = $xml->Response->Placemark->Point->coordinates;
//$coordinatesSplit = split(",", $coordinates);
// Format: Longitude, Latitude, Altitude
//$lat = $coordinatesSplit[1];
//$lng = $coordinatesSplit[0];
list($lat,$lng) = explode(",",$coordinates);
$query = sprintf("UPDATE markers " .
" SET lat = '%s', lng = '%s' " .
" WHERE mid = '%s' LIMIT 1;",
mysql_real_escape_string($lng),
mysql_real_escape_string($lat),
mysql_real_escape_string($id));
$update_result = mysql_query($query);
if (!$update_result) {
die("Invalid query: " . mysql_error());
}else{
echo "$lat;$lng";
}
} else if (strcmp($status, "620") == 0) {
// sent geocodes too fast
$delay += 100000;
} else {
// failure to geocode
$geocode_pending = false;
echo "Address " . $address . " failed to geocode. ";
echo "Received status " . $status . "
\n";
}
usleep($delay);
}
}
非常感謝Dr.Molle。我刪除密鑰後,它就像一個魅力。 – user1149162