Question

我有数据库记录了访问我网站的用户的IP地址和会话ID。我想显示前10个IP地址的HTML表以及与每个IP地址关联的不同会话数。

所以我从这个查询开始：

$sql="SELECT ip, country, city, COUNT(*) as count FROM Hits GROUP BY ip ORDER BY count DESC LIMIT 10";

然后，对于每个产生的10行，我使用while循环查询同一个数据库，以提取与每个地址相关的不同会话ID的数量：

$sql="SELECT DISTINCT Session FROM Hits WHERE ip='$ipaddress'"

我的挑战是在我通过while循环时构建一个数组（与MySQLi查询结果具有相同的结构），然后我可以使用foreach循环输出HTML表。这是我的完整代码：

$con = mysqli_connect("localhost","user","password","dbase");
$sql="SELECT ip, country, city, COUNT(*) as count FROM Hits GROUP BY ip ORDER BY count DESC LIMIT 10";
$result = mysqli_query($con,$sql);
$data = array();
while ($row = mysqli_fetch_assoc($result)) {
     $ipaddress = $row["ip"];
     $sql="SELECT DISTINCT Session FROM Hits WHERE ip='$ipaddress'";
     $res = mysqli_query($con,$sql);
     $distinctsessions = mysqli_num_rows($res);
     $data[] =$ipaddress.",".$distinctsessions;
};
$colNames = array_keys(reset($data));
foreach($data as $row) {
    echo "<tr>";
    foreach($colNames as $colName) {
    echo "<td>".$row[$colName]."</td>";
    }   
echo "</tr>";
}
echo "</table>";
mysqli_free_result($result);
mysqli_free_result($res);
mysqli_close($con);

我意识到我的错误在于：

$data[] =$ipaddress.",".$distinctsessions;

并且这可能不是正确的结构，但我无法解决问题。感谢你的建议。

Answer 1

无需使用foreach使用以下代码

$con = mysqli_connect("localhost","user","password","dbase");
$sql="SELECT ip, country, city, COUNT(*) as count FROM Hits GROUP BY ip ORDER BY count DESC LIMIT 10";
$result = mysqli_query($con,$sql);
$data = array();
while ($row = mysqli_fetch_assoc($result)) {
     $ipaddress = $row["ip"];
     $sql="SELECT DISTINCT Session FROM Hits WHERE ip='$ipaddress'";
     $res = mysqli_query($con,$sql);
     $distinctsessions = mysqli_num_rows($res);
 ?>
     <tr>
          <td><?php echo $ipaddress; ?></td>
          <td><?php echo $distinctsessions; ?></td>
     </tr>
  <?php }
         echo "</table>";
         mysqli_free_result($result);
         mysqli_free_result($res);
         mysqli_close($con); 
  ?>

创建一个PHP数组以在HTML表格中显示

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