Re: [PHP-DB] letting a table name be a constant

Ben Galin wrote:

> I want to define() TABLE as a constant instead of using the variable $tabl
e
>
> [PHP code]
>   // This runs without an error but does *not* insert a new row
>   $name="name";
>   define("TABLE", "mytable");
>   $mysql_connect("localhost","","") or die("Error: ".mysql_error());
>   $mysql_select_db("mydb") or die("Error: ".mysql_error());
>   $mysql_query("INSERT INTO TABLE (`id`,`name`) VALUES ('','$name')");
> [/PHP code]

You can't have a constant in a string. You'd do it like this:

mysql_query("INSERT INTO " . TABLE . " (`id`,`name`) VALUES ('','$name')");

The other error you have, and I don't know how you think this code runs,
are the dollar signs before mysql_connect(), mysql_select_db() and
mysql_query() are not needed.

--

---John Holmes...

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On Aug 23, 2004, at 8:12 PM, John Holmes wrote:

> You can't have a constant in a string. You'd do it like this:
>
> mysql_query("INSERT INTO " . TABLE . " (`id`,`name`) VALUES
> ('','$name')");

Thanks, John and Justin.  Wasn't thinking about that.

> The other error you have, and I don't know how you think this code
> runs, are the dollar signs before mysql_connect(), mysql_select_db()
> and mysql_query() are not needed.

You're right.  It doesn't run.  I retyped instead of copy-and-paste'd;
that's a typo.

Cheers,
Ben

On Tuesday 24 August 2004 08:18, Ben Galin wrote:

> You're right.  It doesn't run.  I retyped instead of copy-and-paste'd;
> that's a typo.

Please, when you post code make sure they're verbatim by using copy & paste.
That way people can focus on your actual problem rather than on your typos.

--
Jason Wong -> Gremlins Associates -> www.gremlins.biz
Open Source Software Systems Integrators
* Web Design & Hosting * Internet & Intranet Applications Development *
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