Javascript: Parsing html table

Question

I am working on a html page. The page has a table with 3 columns and a button. One of the columns of the table is a check box (The number of rows are changed dynamically):

<table cellpadding="0" cellspacing="0" border="1" class="display" id="tag" width="100%">
    <tbody>
        <tr>
            <td align="center">
                <input type="checkbox" class="case" name="case" value="0"/>
            <td>fruit</td>
            <td>apple</td>
        </tr>
        <tr>
            <td align="center">
                <input type="checkbox" class="case" name="case" value="1"/>
            <td>fruit</td>
            <td>pear</td>
        </tr>
    </tbody>
</table>
<p><input type="button" value="Generate" onclick="generate()"></p>

When user click the "Generate" button, the generate() function will generate a special string base on the column of each row.

My question is that how can I check if the "checkbox" row is checked or not? I would like to filter those non-checked rows when I generate the string.

Thanks.

No idea whatsoever what you are asking. Please try again...

Scott Evernden
– Scott Evernden

2012-12-12 00:45:29 +00:00
Commented Dec 12, 2012 at 0:45 — Scott Evernden
– Scott Evernden, Commented Dec 12, 2012 at 0:45
Sorry. I have rewritten my question.

Kintarō
– Kintarō

2012-12-12 01:05:23 +00:00
Commented Dec 12, 2012 at 1:05 — Kintarō
– Kintarō, Commented Dec 12, 2012 at 1:05

Sampson · Accepted Answer · 2012-12-12 01:19:09Z

3

Filter the rows based upon which have a checked checkbox:

var rows = $(".case:checked").closest("tr");

This returns a jQuery object that contains all of your table rows housing checked checkboxes.

answered Dec 12, 2012 at 1:19

Sampson

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Comments

HMarioD · Accepted Answer · 2012-12-12 01:32:57Z

0

Call getElementsByName will give you an array of references of your checkboxes. Loop the array in order to get the values.

var arr = document.getElementsByName("case");
for(i = 0; i < arr.length; i++){
   if(arr[i].checked){
      doSomething;
   }
}

answered Dec 12, 2012 at 1:32

HMarioD

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