echo 上次从数据库 创建的表名.在这里,我无法呼应这个名字


echo last created table name from database . here i am not able to echo that name

我收到警告 --> mysql_fetch_array() 期望参数 1 是资源,布尔值

enter code here
$db // database name
$sql1="SELECT TABLE_NAME FROM INFORMATION_SCHEMA.TABLES WHERE TABLE_TYPE = 'BASE TABLE' AND       TABLE_SCHEMA=".$db." order by create_time desc limit 1";
$res1 = mysql_query($sql1);
if($row = mysql_fetch_array($res1))
echo $row['TABLE_NAME'];
else echo "unable to fetch table name";
您需要在

包含$db变量时加上引号。更改此行:

$sql1="SELECT TABLE_NAME FROM INFORMATION_SCHEMA.TABLES WHERE TABLE_TYPE = 'BASE TABLE' AND       TABLE_SCHEMA=".$db." order by create_time desc limit 1";

自:

$sql1="SELECT TABLE_NAME FROM INFORMATION_SCHEMA.TABLES WHERE TABLE_TYPE = 'BASE TABLE' AND       TABLE_SCHEMA='".$db."' order by create_time desc limit 1";

请注意,mysql_* 函数已弃用。将 PDO 或 mysqli_* 与预处理语句结合使用,以避免应用程序中的安全风险。