这个是我的数据库连接类,要用换红线部分的方法这是我的index.php页面,调用了这个类文件,和top部分现在问题出在 top.php上面,我用一个函数来生成菜单,但是调用不到query_to_array()这个查询方法了
自己查资料,用new也不行下面是报错的内容Warning: Missing argument 1 for MysqlTool::__construct(), called in E:\Project\2013-8\DaHuangFeng\top.php on line 10 and defined in E:\Project\2013-8\DaHuangFeng\inc\connclass.php on line 7各位大神帮帮忙啊!!!!!!!
PHP类
自己查资料,用new也不行下面是报错的内容Warning: Missing argument 1 for MysqlTool::__construct(), called in E:\Project\2013-8\DaHuangFeng\top.php on line 10 and defined in E:\Project\2013-8\DaHuangFeng\inc\connclass.php on line 7各位大神帮帮忙啊!!!!!!!
PHP类
在类的最下面实例化了,但是在top.php的函数里面取不到这个类的查询方法
这个是数据库类的
<?
$isdebug = false;class MysqlTool
{
public $conn;
public function __construct($server, $name, $pwd, $dbname, $charset='utf8')
{
$this->conn = mysql_connect($server, $name, $pwd);
mysql_query("use $dbname", $this->conn);
mysql_query("set character set $charset", $this->conn);
mysql_query("set names $charset", $this->conn);
} //insert update delete
public function exe($sql)
{
global $isdebug;
$temp = mysql_query($sql, $this->conn);
$err = mysql_error();
if (empty($err) == false && $isdebug == true)
{
print( '<br /><hr style="color:green" /><font color="#0000FF">错误信息:' . mysql_error() . '</font><br /><font color="red">执行SQL:' . $sql . '</font>' );
}
else if (empty($err) == false)
{
die("system error");
}
return $temp;
}
// select *....
public function query_to_array($sql)
{
$temp_array = array();
$result = $this->exe($sql);
while ($row = mysql_fetch_array($result))
{
array_push($temp_array, $row);
}
return $temp_array;
}
// select top 1 id from tb...
public function query_single($sql)
{
$result = $this->exe($sql);
if ($row = mysql_fetch_row($result))
{
return $row[0];
}
}
function __destruct()
{
mysql_close($this->conn);
}
}
$db = new MysqlTool("localhost", "root", "", "db_dahuangfeng");
?> 这是top.php的,在引用top.php的文件里面已经include_once了这个类
<?
function select_menu($p_id,$type_class)
{
$ret_types = $db->query_to_array("select * from tb_types where p_id = $p_id and type_class = $type_class order by type_id desc;");
$html = '';
foreach($ret_types as $k => $v)
{
$html .= '<li><a href="';
if($type_class == 1)
{
$html .= 'about';
}
if($type_class == 2)
{
$html .= 'news';
}
$html .= '.php?t=' . $v['type_id'] . '">' . $v['type_name'] . '</a></li>';
}
return $html;
}
?>
Warning: Missing argument 1 for MysqlTool::__construct(), called in E:\Project\2013-8\DaHuangFeng\top.php on line 10
现在不报错了,但是 菜单没出来,数据库里面是有数据的,新手自学php,求指点啊
function select_menu($p_id,$type_class)
{
$ret_types = $db->query_to_array(".....那这个 $db 不是在外面吗?