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@xxxxxxxxxxxxxxxx> 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.
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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]);-- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
