我正在使用CodeIgniter,我想使用从MySQL表获取的信息在模型页面中测试条件。
我的问题是,如何回声从"评论" MySQL表中获取的值总数。结果为3-在
中找到的所有数据在这里我的代码在模型页面
中$query = "
SELECT 3 - COUNT(*) AS value FROM `reviews`
INNER JOIN users ON user_id = users.uid
WHERE value = 1 AND user_id = '" . $logged_in_user_via_session . "'
";
$query = $this->db->query($query);
// how to I echo the total number of values fetched from the "reviews" mysql table.
// so I basically want to know if the result is 0 or 1 or 2 so i can continue with additional conditions
预先感谢
最终干净答案
滴答的答案不错,还有所有其他答案。但是,CodeIgniter确实允许您将类CI_DB_MYSQLI_RESULT的对象添加到数字中,如 $ rescorme-> num_rows()-3; 中所示。它建议将其添加到记录后的变量中,如下所示
$query = "
SELECT COUNT(*) AS value FROM `reviews`
INNER JOIN users ON user_id = users.uid
WHERE value = 1 AND user_id = '" . $logged_in_user_via_session . "'
";
$records = $this->db->query($query);
$countedRows=$records->num_rows();
$countedRows;
// subtracting three (3)
$total_minus_three = $countedRows - 3;
echo $total_minus_three;
$query = "
SELECT COUNT(*) AS value FROM `reviews`
INNER JOIN users ON user_id = users.uid
WHERE value = 1 AND user_id = '" . $logged_in_user_via_session . "'
";
$records = $this->db->query($query);
echo "Total Records ->".$records->num_rows();
to print all records use print_r($records->row_array());
编辑
当我看到您的评论时,您想从结果中减去3。
echo '-3 result is ->'. $records->num_rows() -3 ;
您应该尝试以这种方式尝试
$query = "
SELECT COUNT(*) AS value FROM `reviews`
INNER JOIN users ON user_id = users.uid
WHERE value = 1 AND user_id = '" . $logged_in_user_via_session . "'
";
$records = $this->db->query($query);
$countedRows=$records->num_rows();
echo "Total number".$countedRows;
echo "What you want is".$countedRows-3;
您可以使用此命令找到记录总数。
$query = "
SELECT 3 - COUNT(*) AS value FROM `reviews`
INNER JOIN users ON user_id = users.uid
WHERE value = 1 AND user_id = '" . $logged_in_user_via_session . "'
";
echo $query->num_rows(); // for total record
或
$this->db->from("table name")->count_all_results()
或
echo "Total Records ->".$records->row_array();