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Posted by Jerry Stuckle on 06/17/07 11:51
Shutey wrote:
> On 17 Jun, 09:31, Georgi Naumov <gonau...@gmail.com> wrote:
>> Shutey :
>>
>>
>>
>>> I have a strange issue with dropdowns. Using php4, mySQL5, Apache 2 on
>>> a fast XP pro PC, I have a form which requires 5 dropdowns populated
>>> with indentical values. I extract the values using SQL and populate 2
>>> variables and use a for-next loop to create the dropdown. The dropdown
>>> contains some 310 items! It works beautifully if I have 1 or 2
>>> dropdowns but as soon as I add more it partially creates the 3rd and
>>> just stops until it times out!
>>> All 5 work because I have altered their positions but it always stops
>>> part way through creating the 3rd dropdown.
>>> I tried an alternative method by doing away with the variables and
>>> constructing the dropdowns with a 'while' statement reading through
>>> the results of my SQL and got exactly the same problem only 2 and a
>>> bit dropdowns.
>>> Stopping it running after a few seconds or several minutes makes no
>>> difference, it seems to reach a certain point and just stop - no error
>>> messages or anything!
>>> I have tried increasing the resourses in my php.ini file but this also
>>> made no difference.
>>> Can anyone suggest an alternative or a fix?
>>> Thanks
>> .........................................................................................
>> Can you paste same code ?- Hide quoted text -
>>
>> - Show quoted text -
>
> Here's the code I am currently using, there are 4 more identical
> 'name=option2, 3, 4 and 5' but otherwise the same.
>
> $qry = 'SELECT field1, field2 ';
> $qry .= 'FROM database ';
> $qry .= 'ORDER BY field1, field2;';
>
> echo '<table border="0" cellpadding="2" cellspacing="0"><tr>';
> echo "<td>Label 1:</td>";
> echo "<td><select name='option1' size='1'>";
> echo "<option value=''></option>";
> $x = -1;
> $res = mysql_query($qry) or die("Query failed);
> while($row = mysql_fetch_array($res))
> {
> extract($row);
> $x+=1;
> $sel = '';
> if ($field1 == $ref_variable1) $sel = 'SELECTED';
> echo "<option value='" . $field1 . "' $sel>" . $field2 . "</
> option>";
> }
> echo "</select></td>";
> echo "</tr></table>";
>
> Thanks for your interest
>
310 items in a single dropdown is a lot - and 5 dropdowns with the same
items is going to generate a lot of HTML. Not what I would call real
user-friendly, but it should still work.
You've shown the code for one of the dropdowns, which looks good. But
what about the other 4? How do you generate them, also?
You could also try the following - it will be a little more efficient
than your existing code. Not that there's anything wrong with your
existing code, but this might do something a little different:
echo '<table border="0" cellpadding="2" cellspacing="0"><tr>';
echo "<td>Label 1:</td>";
echo "<td><select name='option1' size='1'>";
echo "<option value=''></option>";
$x = -1;
$res = mysql_query($qry) or die("Query failed);
while($row = mysql_fetch_array($res))
{
$x++; // Do you actually need $x?
if ($row['field1'] == $ref_variable1) // Just use the array value
$sel = ' SELECTED';
else
$sel = '';
echo "<option value='" . $row['field1'] . "'$sel>" . $row['field2'] .
"</option>";
}
echo "</select></td>";
echo "</tr></table>";
--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================
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