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using a variable to name the table in the CREATE TABLE function

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MaxMumford MaxMumford is offline

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using a variable to name the table in the CREATE TABLE function

Jun 30th, 2008

Hey,

Im just wondering how i can get something like this to work:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
$tablename = $_POST['tablename']; 
mysql_query(CREATE TABLE `databasename`.`$tablename` (`name` VARCHAR( 200 ) NOT NULL))
$tablename = $_POST['tablename']; 
mysql_query(CREATE TABLE `databasename`.`$tablename` (`name` VARCHAR( 200 ) NOT NULL))

At the moment its coming up with an error because I am using a variable for the table name instead of inputting it normally.

Thanks,

Max

Last edited by MaxMumford; Jun 30th, 2008 at 5:20 pm.

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Re: using a variable to name the table in the CREATE TABLE function

Jun 30th, 2008

you can do that.

try this:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
$tablename = $_POST['tablename'];
$sql = "CREATE TABLE `databasename`.`{$tablename}` (//table fields here)";
mysql_query($sql);
$tablename = $_POST['tablename'];
$sql = "CREATE TABLE `databasename`.`{$tablename}` (//table fields here)";
mysql_query($sql);

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nav33n

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Re: using a variable to name the table in the CREATE TABLE function

Jun 30th, 2008

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$tablename = "test";
$query = "create table ".$tablename." (name varchar(200) not null )";
mysql_query($query);
$tablename = "test";
$query = "create table ".$tablename." (name varchar(200) not null )";
mysql_query($query);

Like this.

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*PM asking for help will be ignored*

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Join Date: Oct 2006

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MaxMumford MaxMumford is offline

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Re: using a variable to name the table in the CREATE TABLE function

Jul 1st, 2008

Thanks for the help

however I don't see how this is not working :s

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
$sql = 'CREATE TABLE `web163-gallery-7`.`{$picurl}` (`name` VARCHAR( 200 ) NOT NULL ,`picurl` VARCHAR( 200 ) NOT NULL ,`group` VARCHAR( 200 ) NOT NULL ,`dateadded` TIMESTAMP( 200 ) NOT NULL DEFAULT CURRENT_TIMESTAMP ,`id` INT( 200 ) NOT NULL ,PRIMARY KEY ( `id` ) ,UNIQUE (`name` ,`picurl`)';
 
mysql_query($sql)
$sql = 'CREATE TABLE `web163-gallery-7`.`{$picurl}` (`name` VARCHAR( 200 ) NOT NULL ,`picurl` VARCHAR( 200 ) NOT NULL ,`group` VARCHAR( 200 ) NOT NULL ,`dateadded` TIMESTAMP( 200 ) NOT NULL DEFAULT CURRENT_TIMESTAMP ,`id` INT( 200 ) NOT NULL ,PRIMARY KEY ( `id` ) ,UNIQUE (`name` ,`picurl`)';

mysql_query($sql)

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nav33n

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Re: using a variable to name the table in the CREATE TABLE function

Jul 1st, 2008

Missing ending )
P.S. group is a keyword. Its not advisable to use keywords as column names.
This works fine for me.

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$sql = "CREATE TABLE test.".$picurl ."
(name VARCHAR(200) NOT NULL ,
picurl VARCHAR( 200 ) NOT NULL ,
group1 VARCHAR( 200 ) NOT NULL ,
dateadded TIMESTAMP( 200 ) NOT NULL DEFAULT CURRENT_TIMESTAMP ,
id INT( 200 ) NOT NULL ,
PRIMARY KEY ( id ) ,UNIQUE (name ,picurl))";
$sql = "CREATE TABLE test.".$picurl ."
(name VARCHAR(200) NOT NULL ,
picurl VARCHAR( 200 ) NOT NULL ,
group1 VARCHAR( 200 ) NOT NULL ,
dateadded TIMESTAMP( 200 ) NOT NULL DEFAULT CURRENT_TIMESTAMP ,
id INT( 200 ) NOT NULL ,
PRIMARY KEY ( id ) ,UNIQUE (name ,picurl))";

Ignorance is definitely not bliss!

*PM asking for help will be ignored*

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mexabet

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Re: using a variable to name the table in the CREATE TABLE function

Jul 1st, 2008

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
# Drop any existing table `posts`
# --------------------------------------------------------
DROP TABLE IF EXISTS `posts`;
 
# --------------------------------------------------------
# Table structure for table `posts`
# --------------------------------------------------------
CREATE TABLE posts (
                    post_id int(11) NOT NULL auto_increment,
					category_id int(11) NOT NULL,
					user_id int(11) NOT NULL,
					title varchar(150) NOT NULL,
					body text NOT NULL,
					posted timestamp,
					PRIMARY KEY (post_id)
					);
# Drop any existing table `posts`
# --------------------------------------------------------
DROP TABLE IF EXISTS `posts`;

# --------------------------------------------------------
# Table structure for table `posts`
# --------------------------------------------------------
CREATE TABLE posts (
                    post_id int(11) NOT NULL auto_increment,
					category_id int(11) NOT NULL,
					user_id int(11) NOT NULL,
					title varchar(150) NOT NULL,
					body text NOT NULL,
					posted timestamp,
					PRIMARY KEY (post_id)
					);

•

Originally Posted by MaxMumford

Hey,

Im just wondering how i can get something like this to work:
Help with Code Tags

PHP Syntax (Toggle Plain Text)
$tablename = $_POST['tablename']; 
mysql_query(CREATE TABLE `databasename`.`$tablename` (`name` VARCHAR( 200 ) NOT NULL))
$tablename = $_POST['tablename']; mysql_query(CREATE TABLE `databasename`.`$tablename` (`name` VARCHAR( 200 ) NOT NULL))
At the moment its coming up with an error because I am using a variable for the table name instead of inputting it normally.

Thanks,

Max

•

Join Date: Oct 2006

Posts: 215

Reputation: MaxMumford is an unknown quantity at this point

Solved Threads: 2

MaxMumford MaxMumford is offline

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Posting Whiz in Training

Re: using a variable to name the table in the CREATE TABLE function

Jul 1st, 2008

Thanks

I thought it wasnt working but that was because i needed to call the table something with .jpg at the end and iv realised i cant do that. No worries iv found a work around.

Max

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