我想從一個表中選擇一個隨機地址,並在其他表中的列的一個補充,但它口口聲聲說「不數據庫中選擇」的MySQL從一個表插入到另一個
if ($_SESSION[address] == "")
{
$db = @mysql_select_db($db_bitcoins,$connection)
or die(mysql_error());
$sql = "SELECT Count(*) FROM address";
$result = @mysql_query($sql, $connection) or die(mysql_error());
$rnum = mysql_num_rows($result);
$rrr=rand(1,rnum);
$sql = "SELECT * FROM address WHERE id = '$rrr'";
$result = @mysql_query($sql, $connection) or die(mysql_error());
while ($sql = mysql_fetch_object($result)) {
$_SESSION[address] = $sql -> ads;
$db = @mysql_select_db($db_name,$connection)
or die(mysql_error());
$sql = "UPDATE $table_name SET bitaddress = $_SESSION[address] WHERE username = '$user' and password = password('$pass')";
$result = @mysql_query($sql, $connection) or die(mysql_error());
}
}
謝謝提前!
你有$ table_name的更新查詢嗎? – 2012-04-17 11:34:39
我敢打賭我的錢在數據庫選擇的一些錯誤,你忽略@ – 2012-04-17 11:38:26
檢查[PDO](http://php.net/manual/en/book.pdo.php),所以你可以防止SQL注入。並處理你的錯誤,不要忽視他們,真的 – 2012-04-17 11:56:03