Re: first record not displayed

Paul Bern 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 ($nrows) matches what I get from MySQL and the first record gets > displayed (using PHPMyAdmin). Even if I change the select to pull only > certain records, the very first one does not get displayed. > > Can anyone shed any light on this for me? > > Thanks! > > Paul That sounds like you may somewhere have a mysql_fetch_array() from which you are not using the result; thsat ill fetch the first record, and if you then start using the data from a "while($row = mysql_fetch_array" it will start with the second record. Something looks odd about your for loop, but my brain is a bit tired at the moment and I can't pinpoint it :-) Cheers -- David Robley This library isn't safe - I just stumbled on an idea.