Asked  7 Months ago    Answers:  5   Viewed   24 times

Is it valid html to have the following:

<form action="a">
    <input.../>
    <form action="b">
        <input.../>
        <input.../>
        <input.../>
    </form>
    <input.../>
</form>

So when you submit "b" you only get the fields within the inner form. When you submit "a" you get all fields minus those within "b".

If it isn't possible, what workarounds for this situation are available?

 Answers

84

A. It is not valid HTML nor XHTML

In the official W3C XHTML specification, Section B. "Element Prohibitions", states that:

"form must not contain other form elements."

http://www.w3.org/TR/xhtml1/#prohibitions

As for the older HTML 3.2 spec, the section on the FORMS element states that:

"Every form must be enclosed within a FORM element. There can be several forms in a single document, but the FORM element can't be nested."

B. The Workaround

There are workarounds using JavaScript without needing to nest form tags.

"How to create a nested form." (despite title this is not nested form tags, but a JavaScript workaround).

Answers to this StackOverflow question

Note: Although one can trick the W3C Validators to pass a page by manipulating the DOM via scripting, it's still not legal HTML. The problem with using such approaches is that the behavior of your code is now not guaranteed across browsers. (since it's not standard)

Tuesday, June 1, 2021
 
Kenny
answered 7 Months ago
55

You can't. Your only option is to divide this into multiple tables and put the form tag outside of it. You could end up nesting your tables, but this is not recommended:

<table>
  <tr><td><form>
    <table><tr><td>id</td><td>name</td>...</tr></table>
  </form></td></tr>
</table>

I would remove the tables entirely and replace it with styled html elements like divs and spans.

Tuesday, June 8, 2021
 
Asher
answered 6 Months ago
37

<form> is valid inside a <td> element. You can check this sort of thing at http://validator.w3.org. Choose "Validate by direct input", then paste the following HTML:

<table>
  <tbody>
    <tr>
      <td><form action="test"><div><input type="text"></div></form></td>
    </tr>
  </tbody>
</table>

Under "More options", select "Validate document fragment". This allows you to check a HTML snippet without writing an entire page for it. I use it for checking HTML fragments all the time.

References:

  • The TD element
  • The FORM element
Saturday, July 31, 2021
 
Mirko
answered 5 Months ago
52

You must listen for the onopen websocket event before sending your first message.

socket.onopen = function(){
    // send some message   
};
Tuesday, August 3, 2021
 
Kevin
answered 5 Months ago
73

From 1.9 "Program execution:

conforming implementations are required to emulate (only) the observable behavior of the abstract machine

and in an informational footnote:

This provision is sometimes called the “as-if” rule, because an implementation is free to disregard any requirement of this International Standard as long as the result is as if the requirement had been obeyed, as far as can be determined from the observable behavior of the program. For instance, an actual implementation need not evaluate part of an expression if it can deduce that its value is not used and that no side effects affecting the observable behavior of the program are produced.

The standard does specifically note that the "as-if" requirement is binding on a replacement version of the nothrow version of operator new(). However, as I read it, that requirement would fall to the programmer overriding operator new() not the compiler. The flip side of this responsibility is that I think the standard pretty much requires the default implementation of the nothrow operator new() provided by the library must do something along the lines of calling the throwing new in a try/catch and return 0 if std::bad_alloc is caught.

Where the "as if rule" could come in to play here is if the compiler/linker/whatever were smart enough to figure out that when the default throwing new() was in being used, the default non-throwing new() could take the shortcut, but if the default throwing new() was overridden, the default non-throwing new() would have to act differently. I'm sure this is technically possible for an implementation (even if you probably can't express it in standard C++). I'd be surprised if there was ever an implementation that did this.

I might be reading too much into the requirement, but I think that's what can be inferred.

Thursday, August 5, 2021
 
lena
answered 4 Months ago
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