php - 检查 MYSQL 查询是否为空

Question

我想根据查询是否为 null 来判断条件并打印两个不同的表。以下是我的代码。

$number_of_date= "SELECT SUM(number_of_date) AS number_of_date
                    FROM emp_leaves
                    WHERE emp_id='$userID'
                      AND leave_category='Annual'
                      AND apply_year='$year'";
$number_date = mysql_query ($number_of_date);

if (($number_date == 'NULL'))
{

    $result2 = "SELECT * FROM leave_info
                  WHERE employment_type='permanent'
                    AND leave_type='annual'";
    $sql2 = mysql_query ($result2);

    while($row = mysql_fetch_array($sql2)) {
       ?>
    <tr>
      <td><?php echo $row["leave_type"]?></td>
      <td><?php echo $row["leave_count"]?></td>
      </td></tr><?php }
}
else
{
    $sql = "SELECT * FROM emp_leaves
              WHERE emp_id='$userID'
                AND leave_category='Annual'
              ORDER BY leave_id DESC limit 1";
    $result = mysql_query ($sql);

    while($row = mysql_fetch_array($result)) {
    ?>
    <tr>
        <td><?php echo $row["leave_category"]?></td>
        <?php }?><?php
    while($row11 = mysql_fetch_array($number_date)) {
            ?>
            <td><?php echo $row11["number_of_date"]?></td><?php } } ?>
    </tr>

但是，“如果”条件不起作用。使该条件正确的方法是什么。任何人都请帮助我!

Answer 1

你可以使用isset()或 empty() php函数。

if(isset($number_date)) 或 if(empty($number_date))

编辑

看看HERE

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success or FALSE on error.

所以请试试这个。

if (mysql_query ($number_of_date)) 或在您的代码中这样编辑。

if (($number_date == true))