我可以从PHP $ var中的另一个表创建表吗？

Question

As seen below, I' m taking some data in a variable from a table and I' m trying to create a new table in another database (of course in the same host). The table is never created and I dont'even know if it's possible.

Please note that I can connect only with user1 to db1 and with user2 to db2, so that's why I try this stuff.

Any help?

mysql_connect("localhost:3036", "user1", "pass1")or die("cannot connect"); 
mysql_select_db("db1")or die("unable to select db1");// connect to db1 as user1

$takedata="SELECT ID, post_date, post_content, guid 
    FROM `db1`.`wp_posts`"; // in var $takedata , I save the query


$result_new=mysql_query($takedata); //execute the query 

if($result_new){
echo "data from db1 taken sucsessfully";
echo "<BR>";

}

else {
echo "ERROR taking data from db1";
echo "<BR>";
}


mysql_close(); // close connection with db1

mysql_connect("localhost:3036", "user2", "pass2")or die("cannot connect"); 
mysql_select_db("db2")or die("unable to select db2");// connect to db2 as user2

$sql="CREATE TABLE `db2`.`newtable`  AS (`$result_new`)"; //Here is the tricky part
            //I'm trying to create table in db2 with the data
                                   //I took from db1.

$result=mysql_query($sql);


if($result){
echo "Table newtable Created";
echo "<BR>";

}

else {
echo "ERROR Creating Table newtable";
echo "<BR>";
}

mysql_close(); // close connection with db2

Answer 1

You code does not make sense, as I have no idea what $result_new will contain. The create table command requires a certain syntax.

However, I can answer your question :

The table is never created and I dont'even know if it's possible.

The reason why you don't know if you are not properly querying the error. Use the mysql_error() function to do this.

if($result){
   echo "Table newtable Created";
   echo "<BR>";

}

else {
   echo "ERROR Creating Table newtable : ";
   echo mysql_error();
   echo "<BR>";
}