Question

我目前正在学习一门非常基本的网络编程主题，但我遇到了完成项目的问题。

下面的代码显示了我创建的所有超链接，当我点击链接时，它会根据我想要显示的信息显示不同的信息（代码在hyerlink代码之后）。我使用过if-else命令，但无论如何我可以使用Array来存储信息吗？我已经尝试过这么多，但我似乎没有为Array做正确的事。

此代码是超链接页面。

<body>
<h1 style="text-align: center">Welcome to School of IIT</h1>
<p>Please choose a diploma you are interested in:</p>

<ul>
<li><a href="dipListShow.php?dip_id=1">Diploma in IT</a></li>
<li><a href="dipListShow.php?dip_id=2">Diploma in Internet and Multimedia Design</a></li>
<li><a href="dipListShow.php?dip_id=3">Diploma in Mobile and Wireless</a></li>
<li><a href="dipListShow.php?dip_id=4">Diploma in Financial Business Informatics</a></li>
<li><a href="dipListShow.php?dip_id=5">Diploma in Cyber and Digital Security</a></li>
<li><a href="dipListShow.php?dip_id=6">Diploma in Gaming and    Entertainment Technology</a></li>
</ul>
</body>

此代码是php页面：

<body>
<h1 style="text-align: center">Welcome to School of Informatics and IT School</h1>
<?php
$dip_id = $_GET['dip_id'];
if ($dip_id == "1")
{
 $dipname = "Diploma in Information Technology";
 $imgfile= "dip_it.jpg";
 echo "<font color = red>You have selected $dipname</font></p>";
}
else if($dip_id == "2")
{
  $dipname = "Diploma in Interactive Multimedia Informatics";
  $imgfile= "dip_imd.jpg";
  echo "<font color = green>You have selected $dipname</font></p>";
}
else if($dip_id == "3")
{
  $dipname = "Diploma in Mobile and Wireless Technology";
  $imgfile= "dip_mwc.jpg";
  echo "<font color = blue>You have selected $dipname</font></p>";
}
else if($dip_id == "4")
{
  $dipname = "Diploma in Financial Business Informatics";
  $imgfile= "dip_fbi.jpg";
  echo "<font color = purple>You have selected $dipname</font></p>";
}
else if($dip_id == "5")
{
  $dipname = "Diploma in Cyber and Digital Security";
  $imgfile= "dip_cds.jpg";
  echo "<font color = orange>You have selected $dipname</font></p>";
}
else if($dip_id == "6")
{
  $dipname = "Diploma in Gaming and Entertainment Technology";
  $imgfile= "dip_get.jpg";
  echo "<font color = grey>You have selected $dipname</font></p>";
}
echo "<img src='images/$imgfile'/>";
?>
</body>

请指导我如何在Array中完成此操作，而不是使用if-else以及如何在两个页面之间进行链接。谢谢:)）

Answer 1

以下代码仅在您的毕业证书是连续的时才有效。

$data = array (
        array (
                'dipName' => 'Diploma in Information Technology',
                'img' => 'dip_it.jpg',
                'color' => 'red' 
        ),
        array (
                'dipName' => 'Diploma in Interactive Multimedia Informatics',
                'img' => 'dip_imd.jpg',
                'color' => 'green'
        ),
        array (
                'dipName' => 'Diploma in Mobile and Wireless Technology',
                'img' => 'dip_mwc.jpg',
                'color' => 'blue'
        ),
        array (
                'dipName' => 'Diploma in Interactive Multimedia Informatics',
                'img' => 'dip_imd.jpg',
                'color' => 'red'
        )
);


echo '<font color = "'.$data[$dip_id]['color'].'">You have selected '.$data[$dip_id]['dipName'];
echo '<img src="images/'.$data[$dip_id]['img'].'"/>';

Answer 2

有很多方法可以做到这一点。如果你想要一个简单的数组，只需要这样做：

$dip_ids = array(
    '1' => array(
        'dipname' => "Diploma in Information Technology",
        'imgfile' => "dip_it.jpg",
        'msg' =>"<font color = red>You have selected $dipname</font></p>"
        ),
    '2' => array(
        'dipname' => "Diploma in Interactive Multimedia Informatics",
        'imgfile' => "dip_imd.jpg",
        'msg' =>"<font color = green>You have selected $dipname</font></p>"
        ),
    '3' => array(
        'dipname' => "Diploma in Mobile and Wireless Technology",
        'imgfile' => "dip_mwc.jpg",
        'msg' =>"<font color = blue>You have selected $dipname</font></p>"
        ),
    '4' => array(
        'dipname' => "Diploma in Financial Business Informatics",
        'imgfile' => "dip_fbi.jpg",
        'msg' =>"<font color = purple>You have selected $dipname</font></p>"
        ),
    '5' => array(
        'dipname' => "Diploma in Cyber and Digital Security",
        'imgfile' => "dip_cds.jpg",
        'msg' =>"<font color = orange>You have selected $dipname</font></p>"
        ),
    '6' => array(
        'dipname' => "Diploma in Gaming and Entertainment Technology",
        'imgfile' => "dip_get.jpg",
        'msg' =>"<font color = grey>You have selected $dipname</font></p>",
        )
    );

然后只需访问$dip_ids[$dip_id]['dipname']等数据

Answer 3

您应该使用database.Store在表中存储此信息并通过其id检索它。数据库结构应该是这样的：

table_name
 -id
 -tipname
 -image

或使用Array（这不是好习惯）：
现场演示： https://eval.in/84787
在相应的键中设置正确的值。我有插入测试值

 <body>
    <h1 style="text-align: center">Welcome to School of Informatics and IT School</h1>
    <?php
     $dip_id = isset($_GET['dip_id']) ? $_GET['dip_id']: '';
     // set your $dip_id for testing purpose
     //$dip_id = 3;
        $arrData = array("1" => array( "dpname" =>"Diploma in Information Technology","image" =>"dip_it.jpg","color"=>"color1"),"2" => array( "dpname" =>"Diploma in Interactive Multimedia Informatics","image" =>"dip_imd.jpg","color"=>"color1"),"3" => array( "dpname" =>"Diploma in Mobile and Wireless Technology","image" =>"dip_it.jpg","color"=>"color1"),"4" => array( "dpname" =>"set value for 4","image" =>"set value for 4","color"=>"color1"),"5" => array( "dpname" =>"set value for 5","image" =>"set value for 5","color"=>"color5"),"6" => array( "dpname" =>"set value for 4","image" =>"set value for 4","color"=>"color6"));

        if(isset($dip_id) || $dip_id){
         echo "<font color = '{$arrData[$dip_id]['color']}'>You have selected    {$arrData[$dip_id]['dpname']}</font></p>";
         $img = $arrData[$dip_id]['image'];

         echo "<img src='Assignment_1/$img'/>";

    }

    ?>
    </body>

如何使用Array以ID作为索引显示正确的信息？

3 个答案: