Php 为特定员工从到表中选择数据

我尝试选择特定员工的请求数据，但没有得到任何结果。请帮帮我

这是第一页的代码。我从列表中选择了员工

<select name="emp_id">
  <?php
  $emp_id = $_SESSION['emp_id'];
    $query= "SELECT emp_id FROM request";   
    $result = mysql_query($query);
if(!$result)
        {die("Query got problem").(mysql_error());}
    else  
    { 
while($row = mysql_fetch_assoc( $result )) {
    echo "<option>".$row['emp_id']."</option>";}
    echo "</select>";

}

?>

这是第二页的代码

<?php
   $con=mysql_connect("localhost","root","");
   mysql_select_db("employee_transfare",$con);
$query="select employee.f_name, employee.emp_id, request.from request.to, request.description, request.date from employee,equest where employee.emp_id=$_POST[emp_id] and employee.emp_id= request.emp_id"  ;
        $result = mysql_query($query); 
    {
        echo "<table border='1'>";
        echo "<tr> <th>Employee Name</th><th>ID</th><th>Current Department</th> <th>Requested Department</th> <th>Reason</th> <th>date</th>  </tr>";
        while($row = mysql_fetch_array( $result )) {
        echo "<tr><td>"; 
        echo $row['f_name'];
        echo "</td><td>"; 
        echo $row['emp_id'];
        echo "</td><td>"; 
        echo $row['from'];
        echo "</td><td>"; 
        echo $row['to'];
        echo "</td><td>"; 
        echo $row['description'];
        echo "</td><td>";
         echo $row['date'];
        echo "</td></tr>"; 
    } 
    echo "</table>";
    } 

?>

我看到的第一个错误：

echo "<option>".$row['emp_id']."</option>";

应该是：

echo '<option value="'.$row['emp_id'].'">'.$row['emp_id'].'</option>';

$query='select employee.f_name, employee.emp_id, request.from request.to, request.description, request.date
    from employee,request
    where employee.emp_id='.intval($_REQUEST['emp_id'], 10).' and employee.emp_id=request.emp_id';

仅针对调试错误，您还可以更改：

$result = mysql_query($query);

到

---

当（$row=mysql\u fetch\u array（$result））{警告：mysql\u fetch\u array（）希望参数1是资源，布尔值在第57行的C:\xampp\htdocs\project\project\employee\u request.php中给出

$result = mysql_query($query);

$result = mysql_query($query) or die(mysql_error());