Creating Database in PHP code

Question

I am a very beginner in PHP & mysql. I was searching for the syntax for creating a new database and found the following: (( Assuming I am trying to create "employee" database))

$result= mysql_create_db("employee", $DBconnect);

$SQLString = "CREATE DATABASE $employee";
$QueryResult = @mysql_query($SQLString, $DBconnect);

$sql = 'CREATE DATABASE 'employee';

$createdb=mysql_create_db(employee, $dbconnect);

Can anyone help me to figure the differences between all of those?? PLZ let me know if there is any syntax mistakes!!

Syntax mistakes: $SQLString = "CREATE DATABASE `employee`"; // was "CREATE DATABASE $employee"; — Emilio
– Emilio, Commented Mar 22, 2014 at 16:23
You should not be using mysql_* functions as the extension has been deprecated. Use PDO or mysqli instead! — Patrick Geyer
– Patrick Geyer, Commented Mar 22, 2014 at 16:24

Ry- · Accepted Answer · 2014-03-22 16:24:53Z

You generally shouldn’t be creating databases dynamically. What’s this for?

Anyways, this:

$result = mysql_create_db("employee", $DBconnect);

uses the deprecated mysql_ extension, and it shouldn’t be used. It also creates a database named “employee”.

This:

$SQLString = "CREATE DATABASE $employee";
$QueryResult = @mysql_query($SQLString, $DBconnect);

squashes errors and uses the deprecated mysql_ extension, and it shouldn’t be used. It creates a database with the name of whatever’s in $employee.

This:

$sql = 'CREATE DATABASE 'employee';

$createdb=mysql_create_db(employee, $dbconnect);

is syntactically invalid, doesn’t use the $sql variable, uses a deprecated extension, uses $dbconnect instead of $DBConnect, and attempts to use the constant employee and probably fails over to the literal 'employee'. Shouldn’t be used.

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