Question

if($inserted=mysqli_affected_rows($conn))
{
    $a=mysqli_query($conn,"select * from c_agent where fps_name='$shop_name'");
    if (mysqli_num_rows($a)>0)
    {
        $b=mysqli_fetch_assoc($a);
            $sh_id=$b['fps_id'];


    }
}
$ss=(string)($sh_id);
$ins="Create table $ss("
        ."'aadhar_id' int not null ,"
        ."'name' varchar(10),"
        ."'wheat' int,"
        ."'rice' int,"
        ."'kerosine'int"
        . " )"
        . " ENGINE = InnoDB;";  

if(mysqli_query($conn,$ins))
{
    echo "Table created successfully";
} else
{
        echo "ERROR: Could not able to execute  ". mysqli_error($conn). mysqli_errno($conn);
}

所以基本上我想用变量创建一个表，它显示我下面的错误。数据已成功插入变量$sh_id。但我仍然无法使用变量名创建表。

错误：无法执行您的SQL语法有错误;查看与您的MariaDB服务器版本对应的手册，以便在＃104; 1047附近使用正确的语法（＆＃39; aadhar_id＆＃39; int not null，＆＃39; name＆＃39; varchar（10），＆＃ 39;小麦＆＃39; int，＆＃39;大米＆＃39; int，＆＃39; kerosi＆＃39;在11064行

无法从php变量创建一个表来存储表中的值

0 个答案: