Re: Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource

Hegedűs Balázs wrote:

Hi all, ever tried to test if your connection to the database succeeded? maybe that's why it shouted that "...not a valid result resource..." msg. Balazs Hegedus Mark Sargent wrote:

Drewcore wrote:

just put a @ symbol before the function calll, eg while ($myrow = @mysql_fetch_row($result)) { etc On 4/25/05, Mark Sargent <powderkeg@snow.email.ne.jp> wrote:

Hi All, get the following error when calling data from mysql, *Warning*: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in */var/www/html/phpmysqltable.php* on line *36 * Below are snippets of my code, Line 22: $result = mysql_query("SELECT product_name, product_model_number, product_serial_number FROM Products",$db); Line 36: while ($myrow = mysql_fetch_row($result)) {

                         printf("<tr><td>%s %s</td><td>%s</td><td>%s

%</td></tr>\n",

                         $myrow[1], $myrow[2], $myrow[3]);

I'm following a webmonkey.com tut for php/mysql. http://webmonkey.wired.com/webmonkey/99/21/index3a.html Cheers. Mark Sargent. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php

Hi All, ok, that fixed the error message, thanx, but, no data is being displayed, only the table/row/cells. Database table, Products definitely has at least 1 entry in it. Cheers. Mark Sargent. printf("<tr><td>%s %s</td><td>%s</td><td>%s</td></tr>\n",

               $myrow[1], $myrow[2], $myrow[3]);

Hi All,

yes, succeeded in connecting to the database earlier, with the below code, <html> <body> <?php $db = mysql_connect("localhost", "root", "pwd omitted"); mysql_select_db("status",$db); $result = mysql_query("SELECT * FROM Products",$db); printf("Product Name: %s<br>\n", mysql_result($result,0,"product_name")); printf("Product Price: %s<br>\n", mysql_result($result,0,"product_price")); printf("Product Serial Number: %s<br>\n", mysql_result($result,0,"product_serial_number")); printf("Product Model Number: %s<br>\n", mysql_result($result,0,"product_model_number")); ?> </body> </html> Cheers. Mark Sargent.