Re: first record not displayed

Paul, Is $link relevant? Have you run a repair on the table? The for loop seems odd to me, but I use the while construct exclusively. Adapted to one of my tables, this worked correctly: $result = mysql_query("SELECT * FROM photos "); while($row = mysql_fetch_array($result)){ echo "id={$row["p_id"]} seq={$row["p_desc"]} <br>\n"; } On 1/23/06, Paul Bern <Phbern@syr.edu> wrote: > Hi, > > I have the feeling I'm missing something very obvious here, but it's driving me nuts! > Using this select statement: > > $result = mysql_query("SELECT * FROM latlong ",$link); > $nrows = mysql_num_rows($result); > > > and either of these to display the results: > > for ($j=0; $rec=mysql_fetch_array($result); $j++){ > print "j=$j id={$rec["id"]} seq={$rec["seq"]}\n"; > } > > > while($row = mysql_fetch_array($result)){ > echo "id={$row["id"]} seq={$row["seq"]} \n"; > } > > The very first record gets dropped/not displayed. The number of records returned ($nrows) > matches what I get from MySQL and the first record gets displayed (using PHPMyAdmin). Even if I > change the select to pull only certain records, the very first one does not get displayed. > > Can anyone shed any light on this for me? > > Thanks! > > Paul > > > > > ******************************************************************************************************** > Paul H. Bern, Ph.D. > Numeric Data Services Librarian The only thing worse than not > 352 Bird Library getting what you wanted > 222 Waverly Ave. is getting what you asked for. > Syracuse, NY 13244 > 315-443-1352 > http://library.syr.edu/information/mgi/nds > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > -- Gerry http://portal.danen.org/