Question

关于如何简化下面的PHP脚本的任何建议？这是我之前的问题：How to check if a checkbox/ radio button is checked in php 与此相关的，我在这里要做的是根据选中的复选框输出数据。但是我的代码并不是很好，如果2个结果满足条件，它会显示2个表。正如您在下面的代码中看到的，有关如何简化此操作的任何建议吗？

$id =  mysql_real_escape_string($_POST['idnum']);

if ( $_POST['yr'] == 'year'  and $_POST['sec'] == 'section' ){
    $result2 = mysql_query("SELECT * FROM student WHERE IDNO='$id'");

    echo "<table border='1'>
        <tr>
        <th>IDNO</th>
        <th>YEAR</th>
        <th>SECTION</th>

        </tr>";

    while($row = mysql_fetch_array($result2))
    {
        echo "<tr>";
        echo "<td>" . $row['IDNO'] . "</td>";
        echo "<td>" . $row['YEAR'] . "</td>";
        echo "<td>" . $row['SECTION'] . "</td>";

        echo "</tr>";
    }
    echo "</table>";
}

if ( $_POST['yr'] == 'year'  and $_POST['sec'] == 'section' and  $_POST['lname'] == 'lastname'){
    $result3 = mysql_query("SELECT * FROM student WHERE IDNO='$id'");

    echo "<table border='1'>
        <tr>
        <th>IDNO</th>
        <th>YEAR</th>
        <th>SECTION</th>
        <th>LASTNAME</th>

        </tr>";

    while($row = mysql_fetch_array($result3))
    {
        echo "<tr>";
        echo "<td>" . $row['IDNO'] . "</td>";
        echo "<td>" . $row['YEAR'] . "</td>";
        echo "<td>" . $row['SECTION'] . "</td>";
        echo "<td>" . $row['LASTNAME'] . "</td>";
        echo "</tr>";
    }
    echo "</table>";
}

mysql_close($con);
?>

Answer 1

虽然不是PHP人员，但我会根据何时在表格中包含姓氏列来做一个逻辑。这样的事情会有所帮助并保持简化......

$ShowLastName = ( $_POST['yr'] == 'year'  
                 and $_POST['sec'] == 'section' 
                 and  $_POST['lname'] == 'lastname');

$id =  mysql_real_escape_string($_POST['idnum']); 
$result2 = mysql_query("SELECT * FROM student WHERE IDNO='$id'"); 


echo "<table border='1'> 
        <tr> 
        <th>IDNO</th> 
        <th>YEAR</th> 
        <th>SECTION</th> ";

if( $ShowLastName )
        echo "<th>LASTNAME</th> ";

echo "</tr>"; 

while($row = mysql_fetch_array($result2)) 
{ 
     echo "<tr>"; 
     echo "<td>" . $row['IDNO'] . "</td>"; 
     echo "<td>" . $row['YEAR'] . "</td>"; 
     echo "<td>" . $row['SECTION'] . "</td>"; 

     if( $ShowLastName )
         echo "<td>" . $row['LASTNAME'] . "</td>"; 

     echo "</tr>"; 
} 
echo "</table>"; 

mysql_close($con); 
?>

Answer 2

你应该做几件事：

首先，您应该从程序逻辑中拆分显示代码。你真的应该使用像smarty

这样的模板引擎

在第二步中，您应该将数据库代码移动到单独的类中。

通过这种分离，您可以获得更清洁，更好地阅读源代码。现在，您可以更好地对不同的业务案例做出反应。使用被调用的PHP文件作为Controller，它充当数据检索和输出之间的代理。

Answer 3

您应该将PHP与HTML一起使用

<?PHP $id =  mysql_real_escape_string($_POST['idnum']);

if ( $_POST['yr'] == 'year'  and $_POST['sec'] == 'section' ){
    $result2 = mysql_query("SELECT * FROM student WHERE IDNO='$id'");
?>

    <table border='1'>
        <tr>
        <th>IDNO</th>
        <th>YEAR</th>
        <th>SECTION</th>

        </tr>
<?PHP
    while($row = mysql_fetch_array($result2))
    {
        <tr>
        <td> <?PHP  echo $row['IDNO'] ?>
        <td> <?PHP  echo $row['YEAR'] ?></td>
        <td> <?PHP  echo $row['SECTION'] ?></td>

        </tr>
<?PHP   }  ?>
    </table>

<?PHP }

if ( $_POST['yr'] == 'year'  and $_POST['sec'] == 'section' and  $_POST['lname'] == 'lastname'){
    $result3 = mysql_query("SELECT * FROM student WHERE IDNO='$id'");
 ?>
    <table border='1'>
        <tr>
        <th>IDNO</th>
        <th>YEAR</th>
        <th>SECTION</th>
        <th>LASTNAME</th>

        </tr>
<?PHP }
    while($row = mysql_fetch_array($result3))
    {
 ?>
        <tr>
        <td> <?PHP  echo $row['IDNO']  ?></td>
        <td> <?PHP echo $row['YEAR']  ?></td>
        <td> <?PHP echo $row['SECTION']  ?> </td>
        <td> <?PHP echo $row['LASTNAME']  ?> </td>
        </tr>";
<?PHP    }   ?>
    </table>
<?PHP }  ?>
<?PHP mysql_close($con); ?>

已编辑您可以更加简化

<?PHP $id =  mysql_real_escape_string($_POST['idnum']);

if ( $_POST['yr'] == 'year'  and $_POST['sec'] == 'section' ){
    $result2 = mysql_query("SELECT * FROM student WHERE IDNO='$id'");
?>

    <table border='1'>
        <tr>
          <th>IDNO</th>
          <th>YEAR</th>
          <th>SECTION</th>
          <?PHP    if ($_POST['lname'] == 'lastname')?>   <th>LASTNAME</th> <?PHP  }   ?>
        </tr>
    <?PHP  while($row = mysql_fetch_array($result2)) {  ?>
        <tr>
          <td> <?PHP  echo $row['IDNO'] ?>
          <td> <?PHP  echo $row['YEAR'] ?></td>
          <td> <?PHP  echo $row['SECTION'] ?></td>
          <?PHP    if ($_POST['lname'] == 'lastname')?>   <th><?PHP echo $row['LASTNAME']  ?></th> <?PHP  }   ?>
        </tr>
    <?PHP   }  ?>
    </table>

<?PHP } ?>

<?PHP mysql_close($con); ?>

我如何简化这个PHP脚本

3 个答案: