Question

我正在尝试加入2个表并根据id更新表。使用外键连接2个表。 Id对于每一行都是唯一的。

这是桌面设计表1

ID | routeno |名称| PHNO

表2

ID | routeno |拾取|时间|

我正在尝试基于routeno加入table1和table2。并根据id编辑表2。在我的代码中，如果table2中有3条记录，其routeno为1，则只有最后一条记录被编辑，其他记录不会被编辑。代码怎么了？

        <?php

    $result = mysql_query("select b.routeno, r.pickup,r.pickuptime,r.dropback,r.rate  from routedetails AS r INNER JOIN  busroute AS b ON b.routeno = r.routeno AND b.routeno = '$routeno'  ");
    if (mysql_num_rows($result) == 0) { echo "<h3 style=color:#0000cc;text-align:center;>No Routes Yet..!</h3>"; } else {  ?>                       

        <table cellpadding="10px" cellspacing="5" width="1200px" style="line-height:25px;font-size:1em;font-family:calibri" border="1" align="center">
        <tr>
    <td  style="color:#1980bb;font-weight:bold;font-size:15px; width:250px;">Pick Up</td>
    <td  style="color:#1980bb;font-weight:bold;font-size:15px;width:250px;">Pick Up Time</td>
    <td style="color:#1980bb;font-weight:bold;font-size:15px;width:250px;">Drop Up Time</td>    
    <td  style="color:#1980bb;font-weight:bold;font-size:15px;width:250px;">Rate</td>
    </tr></table>

    <?php   

    if(isset($_POST['submit']))
    {
    $id=$_POST['id'];   
    $pickup=$_POST['pickup'];
    $pickuptime=$_POST['pickuptime'];
    $dropback=$_POST['dropback'];
    $rate=$_POST['rate'];           
    $query3=mysql_query("update routedetails set id='$id', pickup='$pickup',pickuptime='$pickuptime',dropback='$dropback',rate='$rate' where  id='$id'");
    if($query3)
    {
    echo "<script>window.open('editroute.php?id='.$id.','_self')</script>";
    }
    }
    $query2=mysql_query("select r.id,b.routeno,r.pickup,r.pickuptime,r.dropback,r.rate  from routedetails AS r INNER JOIN  busroute AS b ON b.routeno = r.routeno AND b.routeno = '$routeno'  " );

while ($r1 = mysql_fetch_assoc($query2)) {
        ?>
    </div>

        <table>
    <tr>
    <td><input type="text" name="id" value="<?php echo $r1['id']  ?>"/></td>
    <td><input type="text" name="pickup" value="<?php echo $r1['pickup']  ?>"/></td>
    <td><input type="text" name="pickuptime" value="<?php echo $r1['pickuptime']?>"/></td>
    <td><input type="text" name="dropback" value="<?php echo $r1['dropback']  ?>"/></td>
    <td><input type="text" name="rate" value="<?php echo $r1['rate']  ?>"/></td>

    </tr>
    </table>    

    <?php } ?>  




    <?php }} 
    ?>
    <input type="submit" value="Edit"name="submit" class="nextbutton1">

    </form>

    </td>
    </tr>

    </table>

加入2个表并在php中编辑表

0 个答案: