選擇並插入一個查詢SQL

Question

我有一個網頁，用戶單擊地圖，緯度和經度將傳遞到文本字段，並且連同其他一些詳細信息一起將數據提交到數據庫。 我有一個計算最近的城鎮的公式，還有一些城鎮的經緯度樣本表。

目前，我可以顯示最近的城鎮和距離，但無法將其發送到數據庫。 我想我需要將兩個查詢放在一起，但我不知道如何。 請客氣，我知道我的代碼很糟糕，並且有很多安全問題，稍后我將嘗試解決。 謝謝！

<?php

 $conn    = Connect();

/**
 * Use the Haversine Formula to display the 100 closest matches to $origLat, $origLon
 * Only search the MySQL table $tableName for matches within a 10 mile ($dist) radius.
 */

$origLat = $conn->real_escape_string($_POST['lat']);
$origLon = $conn->real_escape_string($_POST['lng']);

// This is the maximum distance (in miles) away from $origLat, $origLon in which to search
$dist = 50; 

$query = "SELECT Townland, lat, lng, 3956 * 2 * 
          ASIN(SQRT( POWER(SIN(($origLat - lat)*pi()/180/2),2)
          +COS($origLat*pi()/180 )*COS(lat*pi()/180)
          *POWER(SIN(($origLon-lng)*pi()/180/2),2))) 
          as distance FROM townland WHERE 
          lng between ($origLon-$dist/cos(radians($origLat))*69) 
          and ($origLon+$dist/cos(radians($origLat))*69) 
          and lat between ($origLat-($dist/69)) 
          and ($origLat+($dist/69)) 
          having distance < $dist ORDER BY distance limit 1"; 

$result = mysqli_query($conn, $query) or die(mysql_error());
while($row = mysqli_fetch_assoc($result)) {
    echo $row['Townland']." > ".$row['distance']."<BR>";
}


$User_ID = $conn->real_escape_string($_POST['User_ID']);
$Species_Name = $conn->real_escape_string($_POST['Species_Name']);
$No_Of_Birds = $conn->real_escape_string($_POST['No_Of_Birds']);
$newdate = $conn->real_escape_string($_POST['Sighting_Date']);
//Converts date to 'yyyy-mm-dd' acceptable to mysql
$newdate=date('Y-m-d',strtotime($newdate));    
$Sighting_Details = $conn->real_escape_string($_POST['Sighting_Details']);
$lat = $conn->real_escape_string($_POST['lat']);
$lng = $conn->real_escape_string($_POST['lng']);
$townland = $row['Townland'];

$query = "INSERT into sighting 
                (User_ID, Species_Name, No_Of_Birds, 
                 Sighting_Date, Sighting_Details, 
                 lat, lng, Townland) 
           VALUES('" . $User_ID . "','" . $Species_Name . "','" . $No_Of_Birds . "','"
             . $newdate . "','" . $Sighting_Details . "','" 
             . $lat . "','" . $lng . "','" . $townland . "')";

$success = $conn->query($query);

if (!$success) {
    die("Couldn't enter data: ".$conn->error);
} 

header("Location: ../View/thankYouSubmitSighting.php");

$conn->close();
exit();

while($row = mysqli_fetch_assoc($result)) {
    echo $row['Townland']." > ".$row['distance']."<BR>";
}
mysqli_close($conn); 

?>

Answer 1

如果您需要將查詢的結果插入表中，例如：my_table可能是您需要插入選擇查詢（由兩部分組成的單個sql命令）

 "INSERT  into my_table (Townland, lat, lng, distance)
 SELECT Townland, lat, lng, 3956 * 2 * 
        ASIN(SQRT( POWER(SIN(($origLat - lat)*pi()/180/2),2)
        +COS($origLat*pi()/180 )*COS(lat*pi()/180)
        *POWER(SIN(($origLon-lng)*pi()/180/2),2))) 
        as distance FROM townland WHERE 
        lng between ($origLon-$dist/cos(radians($origLat))*69) 
        and ($origLon+$dist/cos(radians($origLat))*69) 
        and lat between ($origLat-($dist/69)) 
        and ($origLat+($dist/69)) 
        having distance < $dist ORDER BY distance limit 1";