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转储表的数据categoria

插入categoria( idcateg, descri) 值 (1, 'Action'), (2, 'Classic'), (3, 'Fight'), (4, 'Others'), (5, 'Puzzles'), (6, '赛车'), (7, '射击'), (8, '运动'), (9, '塔防'), (10, '僵尸');

$varCategorias_GameData = "0";
if (isset($_GET["cat"])) {
$varCategorias_GameData = $_GET["cat"];
}
mysql_select_db($database_gameconnection, $gameconnection);
$query_GameData = sprintf("SELECT * FROM jogos WHERE jogos.intCategoria = %s", GetSQLValueString($varCategorias_GameData, "int"));
$GameData = mysql_query($query_GameData, $gameconnection) or die(mysql_error());
$row_GameData = mysql_fetch_assoc($GameData);
$totalRows_GameData = mysql_num_rows($GameData);

<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>
...

我想编辑与分类列表相关的每个链接,这在我的数据库中,链接的代码在哪里?我怎样才能编辑每一个?

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1 回答 1

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while ($row = mysql_fetch_assoc($result)) {
echo '<a href="link_categoria.php?     cat='.$row['idcateg'].'">'.$row['descri'].'</a>';
 }
于 2013-04-27T04:56:58.207 回答