Thanks Rick, Now it works.
On Thursday 07 February 2002 03:04 pm, Rick Emery wrote:
> The following means you have not opened a link to your database.
> Concerntrate your efforts there.
>
> Warning: Supplied argument is not a valid MySQL-Link
> resource in /var/www/html/list.php3 on line 26
>
> -----Original Message-----
> From: Ken Thompson [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, February 07, 2002 3:23 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Resource link errors
>
>
> Hello all,
> I'm very new to php and have been beating my head on the wall over this for
> 2
> days. I've searched through the php documentation but I guess I either
> don't
>
> know where or what to look for or don't understand what I'm seeing.
> I think this has to be an array but the original doesn't seem to think
> so.(see below)
> I've tried ($myrow = mysql_fetch_array($result)) and still get the same
> errors.
> The original that I used as an example works OK showing the results of
> Name,
>
> Position. They didn't include a data cell for $myrow[3] ...........
> =========================================================
> THIS WORKS:
> <?php
>
> $title = "Welcome to the Web-Site of"; /*I commented this out and finaly
> removed it cuz I don't want the title echoed in the final page.*/
>
> include("header.inc");
>
> $result = mysql_query("SELECT * FROM employees",$db);
>
> echo "<table border=1>\n";
>
> echo "<tr><td>Name</td><td>Position</tr>\n";
>
> while ($myrow = mysql_fetch_row($result)) {
>
> echo("<tr><td>%s %s</td><td>%s</tr>\n", $myrow[1], $myrow[2],
> $myrow[3]);
>
> }
>
> echo "</table>\n";
> include("footer.inc");
>
> ?>
> ===========================================================
> What I am trying to do is to return the results of the query into a table
> with these items:
> +-----------+-------------+------+-----+---------+----------------+
>
> | Field | Type | Null | Key | Default | Extra |
>
> +-----------+-------------+------+-----+---------+----------------+
>
> | id | tinyint(4) | | PRI | NULL | auto_increment |
> | year | int(4) | YES | | NULL | |
> | make | varchar(20) | YES | | NULL | |
> | model | varchar(20) | YES | | NULL | |
> | engine | varchar(20) | YES | | NULL | |
> | fuel | varchar(8) | YES | | NULL | |
> | trans | varchar(10) | YES | | NULL | |
> | condition | varchar(20) | YES | | NULL | |
> | price | varchar(10) | YES | | 0 | |
> | location | varchar(20) | YES | | NULL | |
>
> +-----------+-------------+------+-----+---------+----------------+
> These are the errors that show up when I run the page in the browser.
> The database is local and I'm on Linux-Mandrake 8.1 using Apache web
> server. I can get the results I want from the command line, that is I can
> make queries and get the proper info returned.
> what have I done to cause it not to work?
>
>
> Warning: Supplied argument is not a valid MySQL-Link
> resource in /var/www/html/list.php3 on line 26
>
> Warning: Supplied argument is not a valid MySQL result
> resource in /var/www/html/list.php3 on line 32
>
> THIS DOESN'T:
> I get the table and the item name e.g. Year, Make etc.
> Once I get the 3 cells returning the proper info, I want to add the rest of
> the cells needed to display all the info in the web page. I don't think it
> should be a problem, comments on this?
>
> <?php
>
> include("header.inc");
>
> //(Line 26 below, everything above "<?php" is plain html.)
> $result = mysql_query("SELECT * FROM autos",$db);
>
> echo "<table border=1>\n";
>
> echo "<tr><td>Year</td><td>Make</td><td>Model</td></tr>\n";
>
> //(Line 32 below)
> while ($myrow = mysql_fetch_row($result)) {
>
> //What does the '%s' mean? I read it someplace in the tutorial
> //but can't find it now.
>
> printf("<tr><td>%s %s</td><td>%s</tr>\n", $myrow[1], $myrow[2],
> $myrow[3]);
>
> }
>
> echo "</table>\n";
> include("footer.inc");
>
> ?>
--
Ken Thompson, North West Antique Autos
Payette, Idaho
Email: [EMAIL PROTECTED]
http://www.nwaa.com
Sales and brokering of antique autos and parts.
Linux- Coming Soon To A Desktop Near You
Registered Linux User #183936
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