如何使一个链接到一个PHP的cookie，但不同的输出

Question

即时通讯使用mysql数据库。我想做一个链接到一个PHP页面的链接。

<?php 
$connection = mysql_connect("localhost","root",""); 
mysql_select_db("scratchdisk",$connection); 
$sql = "SELECT * FROM note"; 
$result = mysql_query($sql); 
echo "<table>"; 
echo "<tr><td>#</td><td>Title</td><td>Action</td></tr>"; 

while($row=mysql_fetch_assoc($result)){ 
echo "<tr><td>".$row['#']."</td><td>".$row['Title']."</td><td><a href='view.php' id='view'>View</a></td></tr>"; } 
echo "</table>"; 
?> 

我要显示我$row['#']但同样view.php文件中的下一个页面基地。

Answer 1

您可以使用此代码：

<?php 
$connection = mysql_connect("localhost","root",""); 
mysql_select_db("scratchdisk",$connection); 
$sql = "SELECT * FROM note"; 
$result = mysql_query($sql); 
echo "<table>"; 
echo "<tr><td>#</td><td>Title</td><td>Action</td></tr>"; 

while($row=mysql_fetch_assoc($result)){ 
echo "<tr><td>".$row['#']."</td><td>".$row['Title']."</td><td><a href='view.php?id=" . $row['#'] . "' id='view'>View</a></td></tr>"; } 
echo "</table>"; 
?> 

在view.php

if(isset($_GET['id']) && !empty($_GET['id'])){ 
    //Write logic here 
    $id = $_GET['id']; 
}