Re: CREATE TABLE QUERY PHP/MySQL

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answers inline

Ignatius
____________________________________________
----- Original Message ----- 
From: "Jonathan" <dbphp@ftbwebsolutions.com>
To: <php-db@lists.php.net>
Sent: Friday, December 06, 2002 10:39 PM
Subject: RE:  CREATE TABLE QUERY PHP/MySQL


> 
> 
> Granting access makes sense.  
> 
> Now what is the difference between the user information I put into the
> mysql_connect function and "PHP user"?
> 
one and same thing.

> If there is a difference , how would I grant the access via a query and
> how would it affect the new users who will use the application when it
> is given to my client?
> 
> Also, a preview reply mentioned mysql_error().  How would I use function
> to verify that the table was created?  Do I simply write
> 
> If (mysql_error() != "")
> Return false
> 
> 
A typical way to use it is
mysql_***(  *** ) or die( mysql_error() ) ;

> -----Original Message-----
> From: Ignatius Reilly [mailto:ignatius.reilly@free.fr] 
> Sent: Friday, December 06, 2002 3:34 PM
> To: php-db@lists.php.net
> Subject: Re:  CREATE TABLE QUERY PHP/MySQL
> 
> Very likely you have to GRANT the CREATE permission to your PHP user
> account.
> 
> Also investigate the CREATE TEMPORARY TABLE grant status available since
> version 4+. Very convenient.
> 
> Ignatius
> ____________________________________________
> ----- Original Message -----
> From: "Jonathan" <dbphp@ftbwebsolutions.com>
> To: <php-db@lists.php.net>
> Sent: Friday, December 06, 2002 10:00 PM
> Subject:  CREATE TABLE QUERY PHP/MySQL
> 
> 
> > This is the first time I am trying to create a table in MySQL based on
> > user input.
> >
> > Here is the scenario,
> >
> > The user will enter in the name of a stage, e.g. Harley Davidson.
> > I then look for the first space (to indicate the first word), make it
> > into lower case and then do the following:
> >
> > $new_stage_name = "stage_".$var;
> >
> > This will give $new_stage_name the value of "stage_harley".
> >
> > That works fine.
> >
> > Now I want to run the following query:
> >
> > $SQL = "CREATE TABLE ".$new_stage_name." (set values and their
> > properties)";
> >
> > If I run this query using PHPMyAdmin, minus the var of course, the
> table
> > is created.
> >
> > When I run it in my code, nothing happens. I do a die($SQL) to see the
> > final outcome of the query and it's correct and is what I am using to
> > run the query in PHPMyAdmin.
> >
> > This is what my code is doing:
> >
> > I have class RecordSet which makes the db connection and its
> constructor
> > takes the $SQL as an arg.
> >
> > So, $SQL = "what is listed above";
> > $objRecordSet = new RecordSet($SQL);
> > //I used this all the time with all kinds of other Select, Insert,
> > Update, and Delete queries and it works perfect.
> >
> > For one, I don't know how to test whether the table has been created
> or
> > not.  I know I could probably do some sort of mysql_list_tables and
> look
> > for it or whatever but I was looking for something similar to
> >
> > If (mysql_affected_rows() == -1)
> > Because that is very easy.
> >
> > And ideas as to why the table is not being created.
> >
> > ===================
> > Jonathan Villa
> > Application Developer
> > IS Design & Development
> > www.isdesigndev.com
> > 414.429.0327
> > ===================
> >
> >
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
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> >
> >
> 
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