搜索数据库与经纬度和长半径使用PHP的MySQL(Search database with lat and long by radius using php mysql)

 我试图开发一个小型站点，通过设置的半径从数据库中检索结果并将它们放置在地图上。 我使用谷歌地图api来获取经纬度，并将其传递给我的php脚本，该脚本查询数据库并将结果集作为JSON对象返回。 
 
 我有一个小的问题，使用json将经纬度和长度都发送到php。 
 
 我的主要问题是，我的sql搜索数据库似乎是错误的，因为它只是检索我的数据库中的前10个结果。 我希望它从搜索点返回半径范围内的前10个结果。 
 
 这是我的ajax代码 
 
function showCarPark(location)
{
    var lat = location.lat();
    var lng = location.lng();
    //alert("Lat: " +lat.toFixed(6));
    //alert("Lng: " +lng.toFixed(6));
    document.getElementById('carParkResults').innerHTML = "";
    var criterion = document.getElementById("address").value;
    var count = 0;
    $.ajax({    
              url: 'process.php',
              type: 'GET',
              data: "lat=" + lat + "&lng=" + lng,
              dataType: 'json',
              success: function(data) 
              {
                  jQuery.each(data, function()
                  {
                    $('<p>').text("Car Park: " + data[count].name).appendTo('#carParkResults');
                    placeCarParks(data[count].postcode,data[count].name, data[count].street, data[count].type);
                    count++;
                  });
              },
              error: function(e) 
              {
                //called when there is an error
                console.log(e.message);
                alert("error" + e.message);
              }
    });
 
 这是我的PHP脚本 
 
$rad = 20;
$lat = $_GET['lat'];
$lng = 1.4681464; //put a temporary number in as it wont pass the lng in the JSON

$sql="SELECT *, (3959 * acos(cos(radians('".$lat."')) * cos(radians(lat)) * cos( radians(long) - radians('".$lng."')) + sin(radians('".$lat."')) * 
sin(radians(lat)))) 
AS distance 
FROM carpark HAVING distance < 15 ORDER BY distance LIMIT 0 , 10";

$result = mysql_query($sql);

while($r = mysql_fetch_assoc($result)) $rows[] = $r;

echo json_encode($rows);
 
 我的表中的列被称为拉特和长任何帮助表示赞赏。 

Im trying to develop a small site which retrieves a results from a database by a set radius and places them on a map. I use Google maps api to get the lat and long from there search criteria, pass these to my php script which queries the database and returns the result set as a JSON object.
 
Im having small a problem sending both the lat and long to the php using json.
 
My main problem is that my sql to search the database appears to be wrong as it just retrieves the first 10 results in my database. I want it to return the first 10 results within the radius from the search point.
 
Here is my ajax code
 
function showCarPark(location)
{
    var lat = location.lat();
    var lng = location.lng();
    //alert("Lat: " +lat.toFixed(6));
    //alert("Lng: " +lng.toFixed(6));
    document.getElementById('carParkResults').innerHTML = "";
    var criterion = document.getElementById("address").value;
    var count = 0;
    $.ajax({    
              url: 'process.php',
              type: 'GET',
              data: "lat=" + lat + "&lng=" + lng,
              dataType: 'json',
              success: function(data) 
              {
                  jQuery.each(data, function()
                  {
                    $('<p>').text("Car Park: " + data[count].name).appendTo('#carParkResults');
                    placeCarParks(data[count].postcode,data[count].name, data[count].street, data[count].type);
                    count++;
                  });
              },
              error: function(e) 
              {
                //called when there is an error
                console.log(e.message);
                alert("error" + e.message);
              }
    });
 
And here is my php script
 
$rad = 20;
$lat = $_GET['lat'];
$lng = 1.4681464; //put a temporary number in as it wont pass the lng in the JSON

$sql="SELECT *, (3959 * acos(cos(radians('".$lat."')) * cos(radians(lat)) * cos( radians(long) - radians('".$lng."')) + sin(radians('".$lat."')) * 
sin(radians(lat)))) 
AS distance 
FROM carpark HAVING distance < 15 ORDER BY distance LIMIT 0 , 10";

$result = mysql_query($sql);

while($r = mysql_fetch_assoc($result)) $rows[] = $r;

echo json_encode($rows);
 
The columns in my table are called lat and long Any help is appreciated.

原文：https://stackoverflow.com/questions/10621345