如何从数据库行创建有效的 JSON

我不知道如何让PHP与我的JSON JavaScript代码一起使用。我假设我需要它将每一行解析为一个主数组，该数组为每一行分配一个从 1 到无穷大的唯一值键。但是，老实说，我已经看过了，我不知道该怎么做。另外，如果您发现映射添加标记有任何问题，请告诉我。

PHP代码：

 <?php
 include 'dbconnect.php';
 $result = mysql_query("SELECT * FROM coords ORDER BY name DESC") or die ("Could 
 not query");
 while($row = mysql_fetch_array($result)) {
 $r[] = array(
  "name" => $row['name'],
  "lat" => $row['lat'],
  "lng" => $row['lng'],
  "speed" => $row['speed'],
  "altitude" => $row['altitude'],
   "distance" => $row['distance']
);
}
$encoded = json_encode($r);
echo $encoded;
exit($encoded);
mysql_close($conn);
?> 

JAVASCRIPT 代码：

  var usermarker;
    var markloc;
    function deleteUserOverlay() {
        if (usermarker) {
            usermarker.setMap(null);
        }
    }
    function calluserlocation(){
     console.log('calluserlocation fires');
$.ajax( { 
url: "getdata.php",
type: "GET", 
dataType: "json", 
success: function(data) { for (var i = 0; i < data.length; i++) { markloc = new google.maps.LatLng(data[i].b, data[i].c); adddata(markloc); } }, error: function(data) { console.log( "error" ); } });
console.log("sucessful run of function");
}
   function adddata(markloc){ 
    marker = new google.maps.Marker({
    position: markloc,
     icon: 'http://www.wolfdoginfo.net/app/cropcircles.png',
    map: map
  });
deleteUserOverlay();
usermarker = marker;  
   }

我在控制台中收到错误，我爸爸现在输出如下

[{"name"："test2"，"lat"："39.8441792"，"lng"："-105.104921"，"speed"："bad"，"altitude"："dontcare"，"distance"："whatever"}，{"name"："test"，"lat"："39.729431999999996"，"LNG"："-104.831919"，"speed"："speed"，"altitude"："altitude"，"distance"}，{"name"："grant3"，"lat"：">

39.729431999999996"，"LNG"："-104.831919"，"speed"："speed"，"altitude"："altitude"，"distance"："distance"}，{"name"："grant2"，"lat"："test34"，"lng"："test34"，"speed"："speed"，"altitude"："height"，"距离"："距离"}，{"名称"："格兰特"，"纬度"："39.729431999999996"，"LNG"："-104.831919"，"速度"："速度"，"高度"："高度"，"距离"："距离"}，{"名称"："，"纬度"："39.75198511"，"LNG"："-104.85021166"，"速度"："速度"，"高度"，"距离"："距离"}][{"名称"："test2"，"lat"："39.8441792"，"LNG"："-105.104921"，"速度"："坏"，"高度"："dontcare"，"距离"："随便"}，{"名称"："测试"，"lat"："39.729431999999996"，"LNG"："-104.831919"，"速度"："速度"，"高度"："高度"，"距离"："距离"}，{"名称"："Grant3"，"lat"："39.729431999999996"，"LNG"："-104.831919"，"速度"："速度"，"高度"："高度"，"距离"："距离"}，{"名称"："Grant2"，"lat"："test34"，"LNG"："Test34"，"速度"："速度"，"高度"："高度"，"距离"："距离"}，{"名称"："格兰特"，"纬度"："39.729431999999996"，"LNG"："-104.831919"，"速度"："速度"，"高度"："高度"，"距离"："距离"}，{"名称"："，"纬度"："39.75198511"，"液化天然气"："-104.85021166"，"速度"："速度"，"高度"："高度"，"距离"："距离"}]

所以现在我的 PHP 由于某种原因只正确输出了两次但是我的 JSON 代码不起作用。我需要填充每个条目的标记。

缺点说：

[01:57:52.768] GET http://wolfdoginfo.net/app/show/getdata.php [HTTP/1.1 200 OK 86ms]
[01:57:52.705] "calluserlocation fires"
[01:57:52.705] "sucessful run of function"
[01:57:52.826] "error"

while($row = mysql_fetch_array($query)) {
    echo json_encode(array( "a" => $row['name'], "b" => $row['lat'], "c" => $row['lng'], "d" => $row['speed'], "e" => $row['altitude'], "f" => $row['distance']));
}
// this code will return this JSON
// { a:?, b:?, c:?, ... }
// { a:?, b:?, c:?, ... }
// ...

$return = array();
while($row = mysql_fetch_array($query)) {
    $return [] = array( "a" => $row['name'], "b" => $row['lat'], "c" => $row['lng'], "d" => $row['speed'], "e" => $row['altitude'], "f" => $row['distance']));
}
echo json_encode($return);
// this code will return
// [
//    { a:?, b:?, c:?, ... },
//    { a:?, b:?, c:?, ... },
//    ...
// ]

$details_array = array();
while($row = mysql_fetch_array($query))
{
array_push($details_array, $row);//pushes all the fetched row data into $details_array
}
echo json_encode($details_array);