php - Why aren't my dynamically-added form inputs posting? -

php - Why aren't my dynamically-added form inputs posting? -

September 15, 2013

I am working on a form where I dynamically input the user's click on the "more widget" There is a hidden div inside the form, and I connect the input to jQuery with it, such as:

  $ ('div # newwidgetinputs'). Show (). Append (newInputs);

They appear, have proper names, etc., but when I post a form, their content is not in the PHP $ _ POST array.

So I tried to include myself in the form:

  $ ('form # someform') Attachments (newInputs);

They can not be viewed on the page, but I default to them, and this time they do appear in '$ _POST'.

From this I think that div # newwidgetinputs is not considered to be part of the form, but why do not I see it;

Post text "itemprop =" text ">

If HTML is not well-formed Then it may consider moving dynamic input to form; For example:

  'Other HTML' can be understood out of the form by the browser, because the second  & lt; / div & gt;  closes the '# div_1' div form Has a container , So consider running the browser off the form after the second   .


















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