Cannot create SQL table with query variable as table name in PHP

Question

I am passing a query variable in PHP. I would like to make it the table name, but I know there is probably a SQL syntax error. When I print the statement, the variable is passed meaning it works, but the database simply isn't created.

Here is my code for the creation of the database:

$DBName = "database_name";  
$sql = "CREATE TABLE '$DBName'.'$login' (  
    ClientID int NOT NULL AUTO_INCREMENT,   
    AgentClients varchar(15),  
    ClientTotal int  
    )";  
mysql_query($sql,$link);
where `$login = $_POST['login'];

Also, I'm not worried about security breaches at the moment, so don't worry about that.

Any insight would be greatly appreciated.

You're mixing "table" and "database" a lot. Has the database been created before this code? (I'm not even sure if that matters) But you might need to run CREATE DATABASE '$DBName' first? — Ian
– Ian, Commented Oct 29, 2012 at 19:26
Run the printed query in phpMyAdmin or any other mysql client and you will get all the errors. — air4x
– air4x, Commented Oct 29, 2012 at 19:26
table for each login, is not great structure, and i rally hope you are sanitising the post value. — user557846
– user557846, Commented Oct 29, 2012 at 19:27
Replace ' by ` and add PRIMARY KEY after AUTO_INCREMENT — air4x
– air4x, Commented Oct 29, 2012 at 19:30

JvdBerg · Accepted Answer · 2012-10-29 19:32:17Z

6

You must use backticks for your tablename, and not quotes:

$sql = "CREATE TABLE `$DBName`.`$login` (  
  ClientID int NOT NULL AUTO_INCREMENT,   
  AgentClients varchar(15),  
  ClientTotal int,
  PRIMARY KEY (`ClientID`) 
)";

edited Oct 29, 2012 at 19:32

answered Oct 29, 2012 at 19:26

JvdBerg

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2 Comments

Query Over a year ago

or omit them altogether and use no back ticks.

Thomas Kelley Over a year ago

He's also using an AUTO_INCREMENT without defining a key. That will fail, right?

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