php - mysql从任何距离参数返回相同的结果

Question

我有一个很奇怪的问题。 http://meineanzeige24.de/androidApp/ilanImages/0/10/0.9148%2C32.86342555这个 url 是我用来根据他们的 lat，long 用户 lat，long 和距离在 db 上搜索项目的其余 url。

我的查询被 url 更改了，例如当我将 10 更改为 1000 时，搜索查询尝试搜索距离 < 1000。

可以在链接上看到查询结果 http://meineanzeige24.de/androidApp/ilanImages/0/10/0.9148%2C32.86342555

问题是，当我更改距离并将查询粘贴到 phpmyadmin 并运行查询时，查询有效

但是

在页面上，当我将距离更改为任何整数值时，在网络上的结果总是相同的。此外，距离<10 显示所有项目。您可以通过播放查询来尝试。但正如我所说，当我粘贴到phpmyadmin.

php 代码在这里，我使用 mysqlliDB.php 作为 mysqlconnection等待您的帮助。

        function getIlanImages($catId,$rangeKM,$location) {
    global $host,$user,$password,$db_name;
    global $db;

    $loc = explode(',', $location);

    if($catId>0)
    {


        $q2 = "Select lat,lon,Id,resim1,kategoriId,(     DEGREES(ACOS(COS(RADIANS($loc[0])) * COS(RADIANS(lat)) *             COS(RADIANS($loc[1]) - RADIANS(lon)) +             SIN(RADIANS($loc[0])) * SIN(RADIANS(lat))))    )*111.045 as distance from (SELECT SUBSTRING( map, 2, INSTR(map,',')-2) AS lat, SUBSTRING( map, INSTR(map,',')-length(map),length(map)-INSTR(map,',')-4) AS lon,Id,resim1,kategoriId FROM  firmalar   ) as tbl  WHERE kategoriId=$catId  HAVING distance <= $rangeKM ORDER BY distance";
    }
    else
    {


        $q2 = "Select distance,Id,resim1 from (Select lat,lon,Id,resim1,(     DEGREES(ACOS(COS(RADIANS($loc[0])) * COS(RADIANS(lat)) *             COS(RADIANS($loc[1]) - RADIANS(lon)) +             SIN(RADIANS($loc[0])) * SIN(RADIANS(lat))))    )*111.045 as distance from (SELECT SUBSTRING( map, 2, INSTR(map,',')-2) AS lat, SUBSTRING( map, INSTR(map,',')-length(map),length(map)-INSTR(map,',')-4) AS lon,Id,resim1 FROM  firmalar   ) as tbl) as d WHERE d.distance <= $rangeKM ORDER BY distance"; 
        $result = $db->rawQuery($q2, null); 
    }
    echo $q2; 

   if ($db->count == 0) {
        return "herhangi bir mesaj yok";
    }
    return json_encode($result);

}

Answer 1

问题是关于 mysqldb.php 类 https://github.com/joshcam/PHP-MySQLi-Database-Classrawquery 函数不返回正确的结果我使用核心 php mysql 函数