Php 我的密码有问题吗？

我试图显示某个员工的休假历史记录，但当我选择某个月时，例如：我选择显示一月的历史记录，而不是显示一月的历史记录，它显示今年的所有历史记录。我的代码有什么问题吗？下面是我的代码：

<?php
    echo 'View the application history by :<select name="date">
            <option value="january" selected="selected">January</option>
            <option value="february" >February</option>
            <option value="march">March</option>
            <option value="april">April</option>
            <option value="may">May</option>
            <option value="june">June</option>
            <option value="july">July</option>
            <option value="august">August</option>
            <option value="september">September</option>
            <option value="october">October</option>
            <option value="november">November</option>
            <option value="december">December</option>';
    echo'</select>';
            $value = 'value';
            if($value=='january')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-01-01' and Date_Apply<'2013-01-31'"); 
            }
            else if($value=='february')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-02-01' and Date_Apply<'2013-02-28'");
            }
                else if($value=='march')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-03-01' and Date_Apply<'2013-03-31'");
            }
                else if($value=='april')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-04-01' and Date_Apply<'2013-04-30'");
            }
                else if($value=='may')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-05-01' and Date_Apply<'2013-05-31'");
            }
                else if($value=='june')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-06-01' and Date_Apply<'2013-06-30'");
            }
                else if($value=='july')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-07-01' and Date_Apply<'2013-07-31'");
            }
                else if($value=='august')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-08-01' and Date_Apply<'2013-08-31'");
            }
                else if($value=='september')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-09-01' and Date_Apply<'2013-09-30'");
            }
                else if($value=='october')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-10-01' and Date_Apply<'2013-10-31'");
            }
                else if($value=='november')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-11-01' and Date_Apply<'2013-11-30'");
            }
                else if($value=='december')
            {
                 $check_user=mysql_query("select*from employee e, `leave` l where e.emp_id = l.emp_id and l.Emp_ID='".$aid."' and Date_Apply >='2013-12-01' and Date_Apply<'2013-12-31'");
            }

        echo "<table border='1'>";

        echo "<tr>";
        echo "<th>Leave No</th>";
        echo "<th>Leave Start</th>";
        echo "<th>Leave End</th>";
        echo "<th>Date Apply</th>";
        echo "<th>Duration</th>";
        echo "<th>Leave Type</th>";
        echo "<th>Leave Reason</th>";
        echo "<th>Status</th>";


    $counter = 0;
    while($rows = mysql_fetch_assoc($check_user))
    {
        echo"<tr>";
        echo"<td>" . $rows['Leave_ID'] . "</td>";
        echo"<td>" . $rows['Leave_Start'] . "</td>";
        echo"<td>" . $rows['Leave_End'] . "</td>";
        echo"<td>" . $rows['Date_Apply'] . "</td>";
        echo"<td>" . $rows['Duration'] . "</td>";
        echo"<td>" . $rows['Leave_Type'] . "</td>";
        echo"<td>" . $rows['Leave_Reason'] . "</td>";
        echo"<td>" . $rows['Status'] . "</td>";
        $counter++;
    }
    echo "</table>";
?>

是。将$value的值设置为“value”。然后检查该变量是否包含月份的名称。那是不可能的。另外：月份的名称将在名为$date的变量中。。。如果您从选择框中读取它（并将其放入名为$date的变量中）。除此之外，选择框不会做任何事情，因为它需要位于表单标记内部。（和submit按钮或其他东西来提交表单..然后读出它（使用GET）并将其放入$date变量..尽管我会使用不同的名称，以确保名称是唯一的..

“我的代码有什么错误吗？”-这个问题最适合于，看到他没有得到期望的输出，他问他的代码有什么问题

我的代码有什么问题吗？

有！它不可读，多个层混淆，使用了不推荐的数据库API，使用if/else代替switch，可能还有这个列表。我的经理可以节省很多钱获得免费的代码审查。谢谢，我很感激：）