PHP mysqli_fetch_array選擇最小值

Question

嘿傢伙，謝謝你的幫助。 我想要做的是使用用戶的緯度和經度，並將其與我在我的數據庫中保存的經度和緯度與相關城市名稱進行比較。我現在的代碼工作正常，但問題是半徑30英里內可能有多行。看看我的代碼，如果可以的話，幫助我。

<?php function distance($lat1, $lon1, $lat2, $lon2) { 

    $pi80 = M_PI/180; 
    $lat1 *= $pi80; 
    $lon1 *= $pi80; 
    $lat2 *= $pi80; 
    $lon2 *= $pi80; 

    $r = 6372.797; // mean radius of Earth in km 
    $dlat = $lat2 - $lat1; 
    $dlon = $lon2 - $lon1; 
    $a = sin($dlat/2) * sin($dlat/2) + cos($lat1) * cos($lat2) * sin($dlon/2) * sin($dlon/2); 
    $c = 2 * atan2(sqrt($a), sqrt(1 - $a)); 
    $km = $r * $c; 

    //echo '<br/>'.$km; 
    $km = $km * 0.621371; 
    return $km; 
} 
    include_once("dbconnect.php"); 
    /* check connection */ 
    if ($mysqli->connect_errno) { 
     printf("Connect failed: %s\n", $mysqli->connect_error); 
     exit(); 
    } 
    $grabCities = "SELECT * FROM cities"; 
    $result = $mysqli->query($grabCities); 
    while($row = mysqli_fetch_array($result)){ 
    $distance = distance(41.08866717,-81.4780426,$row['lat'],$row['longitude']); 
    if ($distance < 30){ 
     echo $myCity = $row['City'] . "<BR>"; 
    } 

    } 
    $mysqli->close(); 
?> 

Answer 1

您可以使用MYSQL做到這一點，1結果返回這是你想要的。

sprintf(' 
    SELECT 
     *, 
     (((acos(sin((%1$d*pi()/180)) * 
      sin((`lat`*pi()/180))+cos((%1$d*pi()/180)) * 
      cos((`lat`*pi()/180))*cos(((%2$d-`longitude`) * 
      pi()/180))))*180/pi()) 
     ) as distance 
     FROM `cities` 
    ORDER BY 
     distance ASC 
    LIMIT 1 
    ', 
    $latitude, 
    $longitude 
); 

Answer 2

是這樣的嗎？未經測試。

$smallestDistance = null; 

while($row = mysqli_fetch_array($result)){ 
    $distance = distance(41.08866717,-81.4780426,$row['lat'],$row['longitude']); 
    if ($distance < $smallestDistance){ 
     $myCity = $row['City']; 
    } 
    $smallestDistance = $distance; 
} 

echo $myCity;