I find it misleading that if I create a form element using Javascript with
a particular value attribute, and then use Javascript to alter the value
attribute that Firebug shows only the original value, and not the new
value. For example imagine I have the following HTML:
<p id='row'>
<input type='text' id='Fred' name='Fred' value='Flintstone'>
</p>
And in Javascript I do:
var newinput = document.getElementById('Fred').clone(false);
newinput.id='Barney';
newinput.name='Barney';
newinput.value='Rubble';
document.getElementById('row').append(newinput);
Then Firebug displays the value attribute of the new input node as
"Flintstone", not "Rubble", while it shows the id and name attributes as
'Barney'.
This is the situation where element.value is not the same as
element.defaultValue. Because of the clone operation newinput.defaultValue
is "Flintstone" and newinput.value is "Rubble". To make the Firebug
display of the input element display the current value of the value
attribute, I must also change the value of the defaultValue attribute so
the two attributes agree. I feel that in this situation Firebug should
either display just the current value of the value attribute, which is what
the external user sees in the browser, or should display the values of both
the value attribute and the defaultValue attribute.
This is with Firebug 2.0.6 on Firefox 34.0 running on Ubuntu Linux.
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