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Posted by Janet Valade on 10/16/36 11:14
Mark Sargent 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
This means that $result is not a valid result resource. You can
var_dump($result) to see that this is true. It probably contains false.
You get false when there's a problem with your query. If you echo
mysql_error() after the query, you will get more information about what
is wrong.
Janet
> *
> 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.
>
--
Janet Valade -- janet.valade.com
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