Question

我的代码已经填充了选择下拉列表中的两个不同的值来自mysql列和硬编码。选择tester时，它会正确显示该tester的记录。

现在我想在用户选择其中一个硬编码选项（在mysql列中为列名）时在表中显示一些用户定义的列。我是php的新手，所以请你帮忙。

    <form name="frmdropdown" method="post" action="emp_dd_display.php">
    <h2 align="center">Records</h2>
            <strong> Select Option: </strong> 
            <select id="tester" name="tester" class="tester"> 
               <option value=""> -----------ALL-----------</option>
               <option value="machine_details"> Machine Details </option>
               <option value="tester"> Tester </option>
               <option value="test_event"> Event </option>
            <?php
                 $dd_res=mysqli_query($conn, "Select DISTINCT tester from workflow1");
                 while($r=mysqli_fetch_row($dd_res))
                 { 
                       echo "<option value='$r[0]'".($_POST['tester']==$r[0]?' selected':'')."> $r[0] </option>";
                 }
             ?>
            </select>
     <input type="submit" name="find" value="find"/> 
     <table border="1">
     <?php
       if($_SERVER['REQUEST_METHOD'] == "POST")
     {
         $des=$_POST["tester"]; 
         if($des=="")  // if ALL is selected in Dropdown box, show all columns
         { 
             $res=mysqli_query($conn, "Select * from workflow1");
             //$res=mysqli_query($conn, "Select date1, tester from workflow1");
                 echo "<tr align=center>";
                 echo "<th>Sr. No. </th>";   
                 echo "<th>Date</th>";  
                 echo "<th>Tester</th>";   
                 echo "<th>Machine Details</th>";  
                 echo "<th>Event</th>";
                 echo "</tr>";  
                 while($r=mysqli_fetch_row($res))
         {                  
                 echo "<tr>";
                 echo "<td align=center>$r[0]</td>";
                 echo "<td width=200>$r[1]</td>";
                 echo "<td alig=center width=40> $r[3]</td>";
                 echo "<td align=center width=200>$r[4]</td>";
                 echo "<td width=100 align=center>$r[5]</td>";
                 echo "</tr>";
        }    
         }
         else
         { 
             $res=mysqli_query($conn, "Select date1, tester from workflow1 where tester='".$des."'");
                 echo "<tr align=center>";
                 echo "<th>Emp_Id </th>";   
                 echo "<th>Emp_name </th>";  
                 echo "</tr>";               
         }
           while($r=mysqli_fetch_row($res))
         {
                 echo "<tr>";
                 echo "<td align='center'>$r[0]</td>";
                 echo "<td align='center'>$r[1]</td>";
                 echo "</tr>";
        }
    }
    ?>
   </table>
   </center>
   </form>

期望的输出：

如果未选择任何内容，则设置为“ALL”，

---------------------------------------------------------- | Sr. NO | Date | Tester | Machine Details| Event | |10-05-2017 | testing1| Ganesh | dgrgr | event1| |10-05-2017 | testing1| Manoj | rewtw | event2| ----------------------------------------------------------

如果用户选择其中一个tester（Ganesh），

-------------------- | Date | Tester | |10-05-2017 | Ganesh | --------------------

如果用户选择其中一个列说test_event，

---------------------------------------- | Date | Event | Machine Details| |10-05-2017 | testing1| abcd | |10-05-2017 | testing1| ddfe | ----------------------------------------

SQL Fiddle中的示例表数据：http://sqlfiddle.com/#!9/b8bd9/1

根据php中的select选项显示列或特定记录

0 个答案: