jeudi 23 octobre 2014

Create new table with the name as a resultant from a variable is failing


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I want it so that every single time a new person registers on my site it creates a new table specifically for them. This is for a file system i am implementing. However, when trying to use a variable as the name of the table. It seems to give me a error. I was wondering if anyone knew why.


The following code is my create table code, I have already connected and selected a database earlier in the script so that is not the issue



$newuser = $_POST['username'];

$create = "CREATE TABLE".$newuser." (
column_one varchar (20) NOT NULL,
column_two int NOT NULL auto_increment PRIMARY KEY,
column_three int NOT NULL,
column_four varchar (15) NOT NULL,
column_five year
)";

$results = mysql_query($create) or die (mysql_error());


I then get the error...



You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'TABLEAdministrator9 ( column_one varchar (20) NOT NULL, column' at line 1


However, when just naming it something like "potato" it will go ahead and create it. So I am wondering if it is an issue to the syntax i am using when wanting it to use the resultant from the variable $newuser.


I am brand new to back end programming.


Thanks



asked 10 mins ago


1 Answer



Vote count:

1




You are missing a space:



$create = "CREATE TABLE ".$newuser." (
-----------------------^
column_one varchar (20) NOT NULL,
column_two int NOT NULL auto_increment PRIMARY KEY,
column_three int NOT NULL,
column_four varchar (15) NOT NULL,
column_five year
)";


answered 5 mins ago






Create new table with the name as a resultant from a variable is failing

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