Thanks for your response, I did finally sove it by changing mysql_fetch_row($result); to mysql_fetch_array($result); and assigning my variable as $somevar = $list['fieldname'];
Now I get my real value. Thanks again. Alp "Raditha Dissanayake" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > Alp wrote: > > >Hi there, > > > >I've been fighting with PHP/MySQL to get proper results but without success. > >For example: > > $sql = "SELECT * FROM prod WHERE shortname='Myprog"; > > $result = mysql_query($sql); >>> returns Resource #7 > > $list = mysql_fetch_row($result); >>> returns Array > > > >and actually the data in the table is a four digit number! The query returns > >what is expected in MySQL but refuses to do so under php. What is it (or > >could be) that I am doing wrong/missing? > > > > > PHP is behaving exactly as it should. The fact that your table has one > column only is of no relevence you still have to retrieve the first > element of the array that is returned mysql_fetch_row($result); > > -- > Raditha Dissanayake. > ------------------------------------------------------------------ > http://www.radinks.com/print/card-designer/ | Card Designer Applet > http://www.radinks.com/upload/ | Drag and Drop Upload -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
