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@NogDogJul 31.2006 — #This line...
...is reading the first row of the result set [i]and moving the pointer to the next row.[/i] Either delete this line if you aren't actually doing anything with it (in your code snippet you don't), or else do this before your loop to reset the pointer to the first row:
php mysql -while($row=mysql_fetch_array($result))
I have a straightforward query on a straightforward table as …
$validity=”Y”;
$query=”SELECT * FROM employees WHERE office=’$asking’ AND validity=’$validity'”;
$result=mysql_query($query) or die(mysql_error());
$num=mysql_num_rows($result);
$row=mysql_fetch_array($result,MYSQL_ASSOC);
if($num>0){
while($row=mysql_fetch_array($result))
{
echo'<tr>
<td>’,$row[reference],’ ‘,$num,'</td>
</tr>’;}
I know there are three valid rows and $num confirms this but I am only getting the last two out following while($row=………….
If I change $row=mysql_fetch_array($result,MYSQL_ASSOC); to be
$row=mysql_fetch_row($result,MYSQL_ASSOC);
Then I get all three rows but they are accompanied by a WARNING as
Warning: Wrong parameter count for mysql_fetch_row() in /home/cauli/public_html/malcolm/delete-agentoffice2.php on line 84
Have to say its driving me …..
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@NogDogJul 31.2006 — #This line...[code=php]$row=mysql_fetch_array($result,MYSQL_ASSOC);[/code]...is reading the first row of the result set [i]and moving the pointer to the next row.[/i] Either delete this line if you aren't actually doing anything with it (in your code snippet you don't), or else do this before your loop to reset the pointer to the first row:
[code=php]
mysql_data_seek($result, 0);
[/code]