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 <[EMAIL PROTECTED]> 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]);
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.
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