Question

我有一些代码可以获取任何地址并返回lat和long作为php变量。

if ($_SESSION['where']) {
    $where = stripslashes($_SESSION['where']);
    $whereurl = urlencode($where);

$location = file("http://maps.google.com/maps/geo?q=new+york+New+york
&output=csv&key=xxxxxxxxxxxxxxxxxxxxxxxx");
        list ($stat,$acc,$north,$east) = explode(",",$location[0]);
        $html = "Information for ".$where."<br>";
        $html .= "North: $north, East: $east<br>";
        $html .= "Accuracy: $acc, Status: $stat<br>";
} else {
        $html = "Varibles Undefined";
}

?>


     <?php $_SESSION['lat'] = "$html"; $_SESSION['lon'] = "$whereurl"; echo"<strong>"?><?php echo "$north";?>&deg; North, <?php echo "$east";?>&deg; East</strong>

我知道它有效，因为当我进入

http://maps.google.com/maps/geo?q=$whereurl
    &output=csv&key=xxxxxxxxxxxxxxxxxxxx

手动进入浏览器，返回

200,4,40.7143528,-74.0059731

这是我需要保存的PHP变量。然而，它并没有将其保存为$ north或$ east。对于如何解决这个问题，有任何的建议吗？提前谢谢。

这是它给我的东西：

{“name”：“new york new york”，“Status”：{“code”：200，“request”：“geocode”}，“Placemark”：[{“id”：“p1”，“地址“：”New York，NY，USA“，”AddressDetails“：{”Accuracy“：4，”Country“：{”AdministrativeArea“：{”AdministrativeAreaName“：”NY“，”SubAdministrativeArea“：{”Locality“： {“LocalityName”：“New York”}，“SubAdministrativeAreaName”：“New York”}}，“CountryName”：“USA”，“CountryNameCode”：“US”}}，“ExtendedData”：{“LatLonBox”：{ “north”：40.8495342，“south”：40.5788964，“east”： - 73.7498543，“west”： - 74.2620919}}，“Point”：{“coordinates”：[ - 74.0059731,40.7143528,0]}}，{“ id“：”p2“，”address“：”Manhattan，New York，NY，USA“，”AddressDetails“：{”Accuracy“：4，”Country“：{”AdministrativeArea“：{”AdministrativeAreaName“：”NY“ ，“SubAdministrativeArea”：{“Locality”：{“DependentLocality”：{“DependentLocalityName”：“Manhattan”}，“LocalityName”：“New York”}，“SubAdministrativeAreaName”：“New York”}}，“CountryName”： “USA”，“CountryNameCode”： “US”}}，“ExtendedData”：{“LatLonBox”：{“north”：40.8200450，“south”：40.6980780，“east”： - 73.9033130，“west”： - 74.0351490}}，“Point”：{“坐标“：[ - 73.9662495,40.7834345,0]}}]}

Answer 1

使用file_get_contents（返回一个字符串）。

简短示例：

$location = file_get_contents("http://maps.google.com/maps/geo?q=$whereurl&output=csv");
$test = explode(",",$location);
print_r($test);

完整解决方案：

http://www.ece.uvic.ca/~casualt/random.php?location=Bermuda

来源：

$theLocation = "new+york";
$locIn = $_GET['location'];
if($locIn)
{
    $theLocation = urlencode($locIn);   //Just to be safe (even though this got passed in via get
}
$location = file_get_contents('http://maps.google.com/maps/geo?q='.$theLocation.'&output=csv&key=AIzaSyAQtiJHUt5KiL-aGONLlACQQ3OdwpbvFzI');
$test = explode(",",$location);
//print_r($test);
$lat = $test[2];
$long = $test[3];
echo '<h1>Magical Location of '.$theLocation.':</h1>';
echo '<ul>';
echo '<li>Longitude: '.$long.'</li>';
echo '<li>Latitude: '.$lat.'</li>';
echo '</ul>';

那应该为你做的伎俩。：）

谷歌地图API麻烦

1 个答案:

http://www.ece.uvic.ca/~casualt/random.php?location=Bermuda