Easy mysql dynamic table name.

mamanybono

Hi everybody,
I just want to learn this, I have $country_name variable in my code and if $country_name=turkey, I need to create a table in database as turkey_ip because of this I wrote below code and it works good. But I wonder is there a way to not create $ip variable. I mean I want to use $country_name variable and ip string in mysql query so ı do not need one extra variable.

$_ip=$country_name.'_ip';

	mysql_query("CREATE TABLE $_ip 
(
   id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
   p1 VARCHAR(50),
   p2 VARCHAR(50),
   p3 VARCHAR(50),
   p4 VARCHAR(50),
   p5 VARCHAR(50)
)");

mamanybono

I found the answer 🙂.

mysql_query("CREATE TABLE {$country_name}_ip
(
id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
p1 VARCHAR(50),
p2 VARCHAR(50),
p3 VARCHAR(50),
p4 VARCHAR(50),
p5 VARCHAR(50)
)");

bradgrafelman

A better question might be: Why would you need to create a new DB table based on the value of $country_name?

Either way, the answer to your question would be to either use concatenation or the so-called "complex (curly) syntax"; both methods are described on the PHP manual page for the [man]string[/man] type.

mamanybono

bradgrafelman;11013071 wrote:
A better question might be: Why would you need to create a new DB table based on the value of $country_name?

Either way, the answer to your question would be to either use concatenation or the so-called "complex (curly) syntax"; both methods are described on the PHP manual page for the [man]string[/man] type.

Yes, I thought about database design and I changed it, I am not going to create new table using by $country_name. İnstead of this, I am going to create one table and record all data into this table using by $country_name.