Question

我有一个名为categoria的表，有10行：

(idcateg, descri) VALUES 
(1, 'Action'), 
(2, 'Classic'), 
(3, 'Fight'), 
(4, 'Others'), 
(5, 'Puzzles'), 
(6, 'Racing'), 
(7, 'Shooting'), 
(8, 'Sports'), 
(9, 'Tower Defense'), 
(10, 'Zombie');

并且它都链接到名为link_categoria.php的页面

<a href="link_categoria.php?cat=1">Action</a>
<a href="link_categoria.php?cat=2">Classic</a>
<a href="link_categoria.php?cat=3">Fight</a>
<a href="link_categoria.php?cat=4">Others</a>
...

我想知道的是，像这个网站：http://www.gameonline.org 我想点击类别时更改每个页面的信息。我该怎么做？

由于

mysql_select_db($database_gameconnection, $gameconnection);
$query_Recordset1 = "SELECT * FROM categoria ORDER BY categoria.descri";
$Recordset1 = mysql_query($query_Recordset1, $gameconnection) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);

while ($row_Recordset1 = mysql_fetch_assoc($Recordset1))
{
echo '<a href="link_categoria.php?cat='.
   $row_Recordset1['idcateg'] .'">'.
   $row_Recordset1['descri']. '</a><br>';
}

Answer 1

在link_categoria.php上，您可以使用$_GET['cat']

获取类别

// Check if $_GET['cat'] is set [ie. ?cat=##] and is numeric
if(isset($_GET['cat']) && is_numeric($_GET['cat'])){

  // Get categoria and escape it before using in the query
  $cat = mysql_real_escape_string($_GET['cat']);

  // Query to get categoria title
  $sql_categoria = "SELECT * FROM categoria WHERE idcateg = $cat";
  $query_categoria = mysql_query($sql_categoria, $gameconnection) or die(mysql_error()); 
  $categoria = mysql_fetch_assoc($query_categoria);

  // Echo the categoria title
  echo '<h1>'.$categoria['descri']. '</h1>';

} // ends if(isset($_GET['cat']) && is_numeric($_GET['cat']))

假设您有按照列表列出的游戏表

gameid | idcateg | game
  1         3      game1
  2         5      game2
....

您可以执行类似的查询以获取该类别中的游戏列表

// Check if $_GET['cat'] is set [ie. ?cat=##] and is numeric
if(isset($_GET['cat']) && is_numeric($_GET['cat'])){

  // Get categoria and escape it before using in the query
  $cat = mysql_real_escape_string($_GET['cat']);

  $query_Recordset1 = "SELECT * FROM games WHERE idcateg = $cat";
  $Recordset1 = mysql_query($query_Recordset1, $gameconnection) or die(mysql_error());
  $totalRows_Recordset1 = mysql_num_rows($Recordset1);

  while ($row_Recordset1 = mysql_fetch_assoc($Recordset1))
  {
  echo '<a href="game.php?gameid='.
     $row_Recordset1['gameid'] .'">'.
     $row_Recordset1['game']. '</a><br>';
  }
}

请注意，从文档开始 - 自PHP 5.5.0起，mysql_*扩展名已弃用，将来将被删除。相反，应使用MySQLi或PDO_MySQL扩展名。有关详细信息，另请参阅MySQL: choosing an API指南和related FAQ。

更改类别页面的内容

1 个答案: