Question

所以我尝试使用多个变量进行半径类型搜索，我已经为我的数据库中的每个城市存储了Lat / Long，但我注意到我想要做的大多数解决方案都要求我查询整个数据库（最终可能超过50,000行）使用函数中存储的lat / long的值。

无论如何都要使用这个存储的数据库值而不先从查询中获取它，并在查询期间使用它。

这就是我现在的工作方式，但我认为这需要很多，特别是一旦数据库达到100,000 +行，可能需要一些时间才能完成每一行，如果我能检查距离会更好我认为这样的查询可以更好地过滤结果，而无需检查每一行。

function getLocation($address) {
    if (!empty($address)) {

        //Formatted address
        $formattedAddr = str_replace(' ', '+', $address);

        //Send request and receive json data by address
        $geocodeFromAddr = file_get_contents('http://maps.googleapis.com/maps/api/geocode/json?address=' . $formattedAddr . '&sensor=true_or_false');
        $output1 = json_decode($geocodeFromAddr);

        //Get latitude and longitute from json data
        $latitude = $output1->results[0]->geometry->location->lat;
        $longitude = $output1->results[0]->geometry->location->lng;

        return "$latitude,$longitude";
    } 
    else {
        return false;
    }
}

function GetDrivingDistance($lat1, $lat2, $long1, $long2) {
    $url = "https://maps.googleapis.com/maps/api/distancematrix/json?origins=" . $lat1 . "," . $long1 . "&destinations=" . $lat2 . "," . $long2 . "&mode=driving&units=imperial";
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch, CURLOPT_PROXYPORT, 3128);
    curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0);
    curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
    $response = curl_exec($ch);
    curl_close($ch);
    $response_a = json_decode($response, true);
    $dist = $response_a['rows'][0]['elements'][0]['distance']['text'];
    $time = $response_a['rows'][0]['elements'][0]['duration']['text'];

    return $dist;

    //return array('distance' => $dist, 'time' => $time);
}

$sql = "SELECT * FROM mytable WHERE `date` >= '$selected_date'";

if ($length && !empty($length)) {
    $sql.= " AND `length`<='$length'";
}

$sql.= " ORDER BY date_created DESC";

$CITY = "DALLAS, TX";
 // FOR TESTING
$SLAT = getLocation($CITY);
$oparr = split("\,", $SLAT);
$FLAT = $oparr[0];
$FLONG = $oparr[1];

while ($row = $result->fetch_assoc()) {
    $id = $row['id'];
    $city = $row['city'];
    $state = $row['state'];
    $lat = $row['lats'];
    $long = $row['longs'];
    $distance = GetDrivingDistance($FLAT, $lat, $FLONG, $long);

    if ($distance <= 200) {

        // SHOW TABLE RESULTS

    } 
    else {
        echo "No results";
    }
}

Answer 1

基于来自Fastest Way to Find Distance Between Two Lat/Long Points的查询，我发现您可以计算MySQL中的距离，从而减少在调用Google Maps API之前从数据库中读取记录的需要。请参阅下面的代码示例。

看一下表中有3条记录的this SQLFiddle，其中两条距离德克萨斯州达拉斯200英里（另一条在休斯顿附近）。

//get location for Dallas
//getLocation('Dallas, TX') //
$lat = 32.7856717;
$lng = -96.77326000000001;
    $query = 'SELECT city, ( 3959 * acos( cos( radians('.$lat.') ) * cos( radians( mytable.lats ) ) 
    * cos( radians(mytable.longs) - radians('.$lng.')) + sin(radians('.$lat.')) 
    * sin( radians(mytable.lats)))) AS distance  FROM mytable having distance < 200';

//$connection initialized earlier via mysqli_connect()
$results = $connection->query($query);
while($row = $results->fetch_object()) {
    //show table results
    echo 'record: '.$row->city.' - distance: '.$row->distance.'<br />';
}

您可能还考虑在构建中添加this plugin，以便可以使用用户定义的函数（UDF） hasrsine_distance（） - 然后您的查询可以简化为：< / p>

$query = 'SELECT city, haversine_distance('.$lat.','.$lng.',mytable.lats,mytable.longs) AS distance  FROM mytable having distance < 200';

Answer 2

$ sql =“ SELECT caid，months，visa，tickets，package，income，expenses cash按日期（现金月）分组； $ result = $ conn->查询（$ sql）;

  $result=mysqli_query($conn, $sql);

 $rows=mysqli_fetch_array($result);
//      $d = $data["expenses"];
  $res=mysqli_query($conn, $sql);
   $data=mysqli_fetch_array($res);



  if ($result-> num_rows >0){
      while($row = $result-> fetch_assoc()){
          //$expenses =  $row["expenses"];
         // echo $expenses;




     echo  $row["visa"]."</br>" ;

      }
     }

如何在php查询中使用sql数据库值

2 个答案: