Error: "...supplied argument is not a valid MySQL-Link resource..."

Hi,

I've set up a database using phpMyAdmin and am now trying to access it from
a PHP file but I get the following error.

"Warning: mysql_query(): supplied argument is not a valid MySQL-Link
resource in /files/home3/steventownley/househunting/search_response.php on
line 59"

Here is the relevant code extract. Line 59 is the last one shown.

/* Declare database parameters */
$db_host = "rumpus";
$username = "steventownley";
$password = // removed from this email!
$db_name = "steventownley_pn";

/* Set up database connection */
$db = mysql_connect($db_host, $username, $password, $db_name);
/* Set up query */
$query = 'select * from Houses where 1';
/* Send query to database */
$result = mysql_query($db, $query);

All help much appreciated.

Thanks.

Steve

--
----
Steven Townley
Stockport, England




--
----
Steven Townley
Stockport, England
MSN Messenger: steven@czechbook.freeserve.co.uk

On Thu, 21 Apr 2005 22:49:41 +0100, "Steve T"
< pedallingminstrel@NOSPAMpedallingminstre
l.fsworld.co.uk> wrote:

>I've set up a database using phpMyAdmin and am now trying to access it from
>a PHP file but I get the following error.
>
>"Warning: mysql_query(): supplied argument is not a valid MySQL-Link
>resource in /files/home3/steventownley/househunting/search_response.php on
>line 59"
>
>Here is the relevant code extract. Line 59 is the last one shown.
>
>  /* Declare database parameters */
>  $db_host = "rumpus";
>  $username = "steventownley";
>  $password = // removed from this email!
>  $db_name = "steventownley_pn";
>
>  /* Set up database connection */
>  $db = mysql_connect($db_host, $username, $password, $db_name);

You've not bothered to check for errors here.

>  /* Set up query */
>  $query = 'select * from Houses where 1';
>  /* Send query to database */
>  $result = mysql_query($db, $query);

.. and so you're probably passing 'false' to the function here.

To see why the connect is failing, use mysql_error().

$db = mysql_connect($db_host, $username, $password, $db_name)
or die(mysql_error());

--
Andy Hassall / <andy@andyh.co.uk> / <http://www.andyh.co.uk>
<http://www.andyhsoftware.co.uk/space> Space: disk usage analysis tool

Thanks for that.

In fact it appears that the call to mysql_connect() had too many arguments
(perhaps because I've been reading a book on PHP5 and my ISP is running
PHP4?) I've now changed it to connect and then specify a database in two
stages, like this (below), with the call to mysql_connect taking only three
arguments (not four).

/* Set up database connection */
mysql_connect($db_host, $username, $password);
mysql_select_db($db_name);

This works.

Thanks again

Steve



"Andy Hassall" <andy@andyh.co.uk> wrote in message
 news:ca8g61lrau10mba17u3rrdtm67ghqnhlde@
4ax.com...
> On Thu, 21 Apr 2005 22:49:41 +0100, "Steve T"
> < pedallingminstrel@NOSPAMpedallingminstre
l.fsworld.co.uk> wrote:
> 
>
> You've not bothered to check for errors here.
> 
>
> ... and so you're probably passing 'false' to the function here.
>
> To see why the connect is failing, use mysql_error().
>
> $db = mysql_connect($db_host, $username, $password, $db_name)
>    or die(mysql_error());
>
> --
> Andy Hassall / <andy@andyh.co.uk> / <http://www.andyh.co.uk>
> <http://www.andyhsoftware.co.uk/space> Space: disk usage analysis tool

Steve,

It doesn't look like you have a valid "filter" for your query.

$query = 'select * from Houses where 1';
I think it should relate to one of your database field tables, like

$query = 'select * from Houses where number_bedroom==1';

where "number_bedroom" is a field in your database table.

You should be able to use a simplified result:

$result = mysql_query($query);

Joe G.

"Steve T" < pedallingminstrel@NOSPAMpedallingminstre
l.fsworld.co.uk> wrote in
message news:42681ff2$0$539$ed2619ec@ptn-nntp-reader03.plus.net...
> Hi,
>
> I've set up a database using phpMyAdmin and am now trying to access it
> from
> a PHP file but I get the following error.
>
> "Warning: mysql_query(): supplied argument is not a valid MySQL-Link
> resource in /files/home3/steventownley/househunting/search_response.php on
> line 59"
>
> Here is the relevant code extract. Line 59 is the last one shown.
>
>  /* Declare database parameters */
>  $db_host = "rumpus";
>  $username = "steventownley";
>  $password = // removed from this email!
>  $db_name = "steventownley_pn";
>
>  /* Set up database connection */
>  $db = mysql_connect($db_host, $username, $password, $db_name);
>  /* Set up query */
>  $query = 'select * from Houses where 1';
>  /* Send query to database */
>  $result = mysql_query($db, $query);
>
> All help much appreciated.
>
> Thanks.
>
> Steve
>
> --
> ----
> Steven Townley
> Stockport, England
>
>
>
>
> --
> ----
> Steven Townley
> Stockport, England
> MSN Messenger: steven@czechbook.freeserve.co.uk
>

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Joe Gazda wrote:
> Steve,
>
> It doesn't look like you have a valid "filter" for your query.
>
> $query = 'select * from Houses where 1';
>  I think it should relate to one of your database field tables, like
>
> $query = 'select * from Houses where number_bedroom==1';
>
> where "number_bedroom" is a field in your database table.
>
> You should be able to use a simplified result:
>
> $result = mysql_query($query);
>
> Joe G.
>
> "Steve T" < pedallingminstrel@NOSPAMpedallingminstre
l.fsworld.co.uk> wrote 
in
> message news:42681ff2$0$539$ed2619ec@ptn-nntp-reader03.plus.net...
> 
>
>
>

References: <42681ff2$0$539$ed2619ec@ptn-nntp-reader03.plus.net> <Mu6dnc-Vwt
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*Sorry, I posted the wrong copy - I somehow opened two :-(*

Joe Gazda wrote:
> It doesn't look like you have a valid "filter" for your query.
>
> $query = 'select * from Houses where 1';

Hi Steve,
The SQL is valid, it will simply return every record in the database

>  I think it should relate to one of your database field tables, like
>
> $query = 'select * from Houses where number_bedroom==1';
>
> where "number_bedroom" is a field in your database table.
>
> $query = 'select * from Houses where number_bedroom==1';
Should that not read "..... number_bedroom = 1" ?? Note the single = sign.
<snip>

> "Steve T" < pedallingminstrel@NOSPAMpedallingminstre
l.fsworld.co.uk> wrote 
in
> message news:42681ff2$0$539$ed2619ec@ptn-nntp-reader03.plus.net...
 
This is usually a symptom of a previous statement (your connection, or
your database selection) failing.
 

The line above looks like your problem - the type for the fourth
parameter should be boolean (True, or False), you are passing a String.
You will need to use mysql_connect() then mysql_select_db() *but*
after each call you should check the return value from the function...
if it is false, the function has failed.  You can use a simple die......
if you wish, but it is often nicer to handle the error more gracefully.


/* Declare database parameters */
$db_host = "rumpus";
$username = "steventownley";
$password = // removed from this email!
$db_name = "steventownley_pn";

/* Set up database connection */
$connection = mysql_connect($db_host, $username, $password);

if($connection == false)
{
// Handle your error here, use the functions
// to tell you what is wrong.
echo "There was an error : ".mysql_errno()." : ".mysql_error();
// Then do something useful before ending the script
exit();
}

// Or using die which will terminate the script instantly
$db = mysql_select_db($db_name) or die ("There was an error : "
.mysql_errno()." : ".mysql_error());

Also use the error checking after your call to mysql_query... etc.

HTH, regards,

Andy