Question

我有这个选择：

<?php
$stmt2 = $mysqli->prepare("SELECT `country`, COUNT(`country`) AS `numar` FROM logs GROUP BY country where username = '$username'");
$stmt2->execute();
mysqli_stmt_bind_result($stmt2, $country, $numar);
$stmt2->store_result();
if($stmt2->num_rows > 0)  //To check if the row exists
  {
      while (mysqli_stmt_fetch($stmt2))
{
  echo "<tr>                

                    <tr>
                    <td class=\"nn\" style=\"text-align:center;\" align=center>$country</td>
                    <td class=\"nn\" style=\"text-align:center;\" align=center>$numar</td>

                    </tr>
                    ";

        }           

} else {
echo"<tr>
<td colspan=\"5\" class=nn><div style=\"text-align:center;\">Not have websites in your account! Click <a href=\"addwebsite\">here</a> to add a website!</div></td>
</tr>";
}   
?>

运行此选择时，返回此错误：致命错误：在第115行的/nginx/html/site/reports.php中的非对象上调用成员函数execute（），第115行与select一致！我尝试使用and选择但同样的错误！哪里有问题？

Answer 1

如下所述纠正sql

$stmt2 = $mysqli->prepare("SELECT `country`, COUNT(`country`) AS `numar` FROM logs where username = '$username' GROUP BY country ");

mysqli选择错误

1 个答案: