Using HTML5 forms with oninput inside a table

Question

Can anyone explain why i can not use oninput inside a table and how i can get it to work?

For example this works:

<form onsubmit="return false" oninput="o.value = parseInt(a.value) + parseInt(b.value)">
  <input name="a" type="number" step="any"> +
  <input name="b" type="number" step="any"> =
  <output name="o"></output>
</form>

And this does not:

<table>
<form onsubmit="return false" oninput="o.value = parseInt(a.value) + parseInt(b.value)">
  <input name="a" type="number" step="any"> +
  <input name="b" type="number" step="any"> =
  <output name="o"></output>
</form>
</table>

Possible duplicate of [Form inside a table](http://stackoverflow.com/questions/5967564/form-inside-a-table) — Peyman Mohamadpour, Feb 09 '16 at 05:39

Ankit Deshpande · Accepted Answer · 2016-02-08T09:40:27.093

0

<form onsubmit="return false" oninput="o.value = parseInt(a.value) + parseInt(b.value)">
<table>

  <input name="a" type="number" step="any"> +
 <input name="b" type="number" step="any"> =
  <output name="o"></output>

</table>
</form>

edited Feb 08 '16 at 09:40

answered Feb 08 '16 at 09:18

Ankit Deshpande

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It has nothing to do with those missing tags, but your answer lead me to the real cause. You started the form outside the table, then it works, when you start it after it doesn't!
– witipedia Feb 08 '16 at 09:35

Using HTML5 forms with oninput inside a table

1 Answers1