從特定員工的表中選擇數據

Question

我嘗試選擇特定員工的請求數據，但我沒有收到任何東西。請幫助我。

這是第一頁的代碼。我從列表中選擇了員工。

<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>"; 

} 

?> 

這是第二頁的代碼

<?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>"; 
    } 

?> 

Answer 1

第一個錯誤我看到：

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

應該是：

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

下一個可能出現的問題：

$query="select employee.f_name, employee.emp_id, request.from request.to, request.description, request.date from employee,equest where employee.emp_id=$_POST[emp_id] and employee.emp_id= request.emp_id" ; 

應該是：

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

只是爲了調試錯誤你也可以改變：

$result = mysql_query($query); 

到

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