Change value of each td > input with JQuery

Question

I have a simple function to re-create id's in a table but it's not giving me the expected output. I am by no means an expert in jquery so it's probably something simple.

$("#myTable tr").each(function (index) {
    if (index > 0) {
        $(this).find(".tdWithId > input").val(index);
        console.log($(this).find(".tdWithId > input").val());
    }
});

In the console output, the values look correct (1, 2, 3, etc.), but when I inspect the html I still see the previous values (value="1", value="2", value="4", etc.).

What am I missing?

Update:

When I do this from console after the fact you can see what I mean.

$(".tdWithId>input").val();
output: "1"

$(".tdWithId>input")
output: [<input name="[0].Details[0].ID" type="text" value="2">]

What does your table html look like? Do you only had one td with an input per row? — mafafu
– mafafu, Commented Apr 10, 2014 at 19:38
@mafafu Well it's looking for td with a certain class > input but there's only one td with that class per row — aw04
– aw04, Commented Apr 10, 2014 at 19:40
Oh, I thought maybe there were more than one and they would all get the same value and was going to suggest moving the td > input selector into the main selector for your each. Still wouldn't hurt to have the html or a fiddle or something. — mafafu
– mafafu, Commented Apr 10, 2014 at 20:03
This looks like the case brought up in this question. There is a difference between an element's properties and its attributes. — ajp15243
– ajp15243, Commented Apr 10, 2014 at 20:16
@ajp15243 I think you're probably right, thanks. That makes alot of sense. Learn something every day :) — aw04
– aw04, Commented Apr 10, 2014 at 20:18

David Stetler · Accepted Answer · 2014-04-10 20:16:43Z

2

change this

$(this).find(".tdWithId > input").val(index);

to this:

$(this).find(".tdWithId > input").attr('value', index);

answered Apr 10, 2014 at 20:16

David Stetler

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4 Comments

ajp15243 Over a year ago

It's possible this won't update the property on the input that's used during form submission. I would have to test, but it's something to verify before going with this approach.

aw04 Over a year ago

@ajp15243 Good point. I will certainly test that, although I think it will work.

ajp15243 Over a year ago

@aw04 You could also just try setting both .val() and .attr(). It's also possible that setting .attr() will internally set the property as well, in which case this answer will work as is (but I'm fairly certain it doesn't work like that).

aw04 Over a year ago

@ajp15243 I need to do some further coding before I can test that, but I will update when I find out

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