Re: [PHP] form handling
Chris Stinemetz chrisstinem...@gmail.com wrote: I would bet it's the quotes screwing up the js. Can / are you escaping that variable when ajaxing it back? Bastien Koert 905-904-0334 I found a way to make the ajax work with one form. I removed the table and the ajax worked just fine. Aparently you can't embed div containers within a table without affecting the whole table. At least that is what I found out tonight. Please correct me if I am wrong. Chris -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Without seeing your code, its hard to figure out what is happening. Post it onto pastebin or something (large code excerpts are very hard to read on this mailing list). Often, running the code through validator.w3.org will find issues that are affecting your layout. Thanks, Ash http://www.ashleysheridan.co.uk -- Sent from my Android phone with K-9 Mail. Please excuse my brevity. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] form handling
it's (pretty) easy to send two forms at once with jquery nowadays, you just
get all the input of the 2 forms and post them!
function submit2Forms(form1DomId,form2DomId){
$datas={};
$(form1DomId).find(':input').each(function(){
if(($(this).attr('name')
$(this).attr('type')!='checkbox') || ($(this).attr('type')=='checkbox'
$(this).is(':checked')))
$datas[$(this).attr('name')]=$(this).val();
});
$(form2DomId).find(':input').each(function(){
if(($(this).attr('name')
$(this).attr('type')!='checkbox') || ($(this).attr('type')=='checkbox'
$(this).is(':checked')))
$datas[$(this).attr('name')]=$(this).val();
});
$.post(URL,function(html) {
$('body').html(html);
})
return false;
});
it's just a small example so you can see how you can do it!
On 12 August 2011 08:48, Ashley Sheridan a...@ashleysheridan.co.uk wrote:
Chris Stinemetz chrisstinem...@gmail.com wrote:
I would bet it's the quotes screwing up the js. Can / are you
escaping that variable when ajaxing it back?
Bastien Koert
905-904-0334
I found a way to make the ajax work with one form. I removed the table
and the ajax worked just fine. Aparently you can't embed div
containers within a table without affecting the whole table. At least
that is what I found out tonight. Please correct me if I am wrong.
Chris
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Without seeing your code, its hard to figure out what is happening. Post it
onto pastebin or something (large code excerpts are very hard to read on
this mailing list). Often, running the code through validator.w3.org will
find issues that are affecting your layout.
Thanks,
Ash
http://www.ashleysheridan.co.uk
--
Sent from my Android phone with K-9 Mail. Please excuse my brevity.
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[PHP] form validation
I have a select menu created by a foreach loop. I am trying to
validate that there was a selection made before it is submitted to the
database. But I am not doing something correctly.
select name=market class=ajax
onchange=javascript:get(this.parentNode);
option value=Choose.../option
?php
foreach($market_prefix as $key = $value)
{
$selected = '';
if($value == $market)
{
$selected = 'selected';
}
echo 'option value=', htmlspecialchars($value), ' ', $selected,
'', htmlspecialchars($market_name[$key]), '/option';
}
?
/select
I am using the folling on the posted page.
if (! array_key_exists($_POST['market'], $market_name))
{
echo You did not select a market.;
}
Thank you,
Chris
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Re: [PHP] form validation
On Fri, Aug 12, 2011 at 11:42, Chris Stinemetz chrisstinem...@gmail.com wrote:
I have a select menu created by a foreach loop. I am trying to
validate that there was a selection made before it is submitted to the
database. But I am not doing something correctly.
Try using a combination of isset, empty, and is_null() instead:
?php
if (!isset($_POST['market']) || empty($_POST['market']) ||
is_null($_POST['market'])) {
// Wasn't set
}
?
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Re: [PHP] form validation
On 12 August 2011 16:42, Chris Stinemetz chrisstinem...@gmail.com wrote:
I have a select menu created by a foreach loop. I am trying to
validate that there was a selection made before it is submitted to the
database. But I am not doing something correctly.
select name=market class=ajax
onchange=javascript:get(this.parentNode);
option value=Choose.../option
?php
foreach($market_prefix as $key = $value)
{
$selected = '';
if($value == $market)
{
$selected = 'selected';
}
echo 'option value=', htmlspecialchars($value), ' ', $selected,
'', htmlspecialchars($market_name[$key]), '/option';
}
?
/select
I am using the folling on the posted page.
if (! array_key_exists($_POST['market'], $market_name))
{
echo You did not select a market.;
}
Thank you,
Chris
$_POST['market'] won't exist if you haven't chosen one.
Turn on your error reporting and you should see something appropriate.
At a bare minimum, adding ...
isset($_POST['market'])
as the first thing to test (before seeing if the value is in
$market_name (though I would have thought $market_names would have
been a better name for the variable - implies more than 1 market).
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Twitter : EE : Zend : PHPDoc
@RQuadling : e-e.com/M_248814.html : bit.ly/9O8vFY : bit.ly/lFnVea
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