谁能解决？。。。。。。。。。

[buhao_md]

批量删除后如何获取所有删除的信息

这只能获取一条的信息，如何能获取多条数据，然后插入数据库中

$_SESSION['adminuser']=$_SESSION['user_login'];//接收用户名
$sql1 = mysql_query(
        "SELECT content,hotel_id 
        FROM `hotel_activities_2` 
        WHERE h_id in($hid)");
$content = mysql_fetch_array($sql1); 
//print_r($content);
if($content){
    $logdata['hotel_id']=$content['hotel_id'];
    $logdata['content']=$content['content'];
    $logdata['operating']="删除";
    writedblog($logdata);
/*********开始删除*******/
$query="delete from hotel_activities_2 where h_id in($hid)";
if (mysql_query($query)){	
	echo "<script>alert('数据删除成功');window.location='chaxun.php';</script>";
}else{
echo "<script>alert('数据删除失败');window.location='chaxun.php';</script>";	
}
}
/********定义函数*************/
function writedblog($data){
    $time=date('Y-m-d H:i:s',time());
    return mysql_query("insert into hotel_rizhi(hotel_id,name,content,operating,time) values 
    ('$data[hotel_id]','$_SESSION[adminuser]','$data[content]','$data[operating]','$time')");
}
/**********关闭数据库***********/
mysql_close($connection);

[CunningBoy]

现做select查询操作，然后再做delete操作就可以了。

[xuzuning]

$sql1 = mysql_query(
"SELECT content,hotel_id
FROM `hotel_activities_2`
WHERE h_id in($hid)");
$content = mysql_fetch_array($sql1);
改为
$sql1 = mysql_query(
"SELECT content,hotel_id
FROM `hotel_activities_2`
WHERE h_id in($hid)");
while($row = mysql_fetch_array($sql1)))
$content[] = $row;