Problem with mysql_select_db()

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I m making a site, something like PhpMyAdmin, that will be hosted on my server.
I have index.php with one form input ("NameOfDatabase") where the user should wrote the name of the database that wanna to create, and ofcourse a submit button, that will link to another page called detail.php.
For now that is ok because it's working.
Next thing is detail.php with the form inputs for table name and other atributs for create a table in mysql, but in the top of the page i must wrote the code for linking the database.
Here is the code
$connection = mysql_connect("localhost", "user", "pass");
this code is ok, but
mysql_select_db("$_GET['NameOfDatabase']", $connection);
the second code is my problem, ithe error says : No database selected.
I think there is a problem because the database in the second page is not defined. Any help???

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You are trying to select database which doesn't exists, right?

First of all, you have to make database, and THEN select it. There is no way to select ANYTHING which doesn't exists...

Hope this helps...

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i think u dont understand me and maybe i dont explain very good.
I repair that, but now i have other problem

$_GET['bracket1'] == "("; $_GET['bracket2'] == ")"; $queryCreateTable = "create table ".$_GET['NameOfTable']." ( ".$_GET['NameOfColumn']." ".$_GET['Type']." ".$_GET['bracket1']." ".$_GET['Value']." ".$_GET['bracket2']." ".$_GET['atribut']." ".$_GET['null']."  ".$_GET['default']." ".$_GET['increment'].")  ";

if i type the name of the table, name of the column, name of the type of variable, value of the variable, (the rest attribut, nill, default ,increment are not important) it says:
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 ') )' at line

the error is because there is )) (two brackets) one by one
this is similar like this query "create table customer(Name varchar(32) );"
and if u notice there is two bracket one by one, and if i write it in mysql command prompt, it says that is ok, there is no error.

Damn sometimes i cant understand myself
sorry is not explained very good, but still try to understand the point

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try
mysql_select_db($_GET['NameOfDatabase'], $connection);

but, this can be security problem, because someone can simply select table that should not be selected...

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Give us a
print $queryCreateTable;
before query

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Peca ::try
mysql_select_db($_GET['NameOfDatabase'], $connection);

but, this can be security problem, because someone can simply select table that should not be selected...

i have that code posted in my first post.
But why security problem? He cant select a table that he doesnt created.


igor86 ::Give us a
print $queryCreateTable;
before query

i dont understand u....if u mean on the query for creating the table in mysql here is it.

create table customer(Name varchar(32) );