|
|
Posted by Mark Sargent on 10/13/37 11:14
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]);
Navigation:
[Reply to this message]
|