使用jquery将数据传递到新页面$ _POST

Question

I have two tables and an input inside a form that more or less goes like this:

the table code (the two tables are the same, except for a slight difference in what table it calls):

<table cellpadding="0" cellspacing="0" class="paperTable searchResultContainer">
    <thead>
        <tr>
            <th>Name</th>
            <th>ID</th>
            <th>Company Reg.</th>
            <th>Postcode</th>
        </tr>
    </thead>

    <tbody>
        <?php while ($row = $items->fetch_assoc()) : ?>
            <tr class="row" data-address1="<?php echo $row['address1']; ?>"
                data-address2="<?php echo $row['address2']; ?>" data-address3="<?php echo $row['address3']; ?>"
                data-county="<?php echo $row['address4']; ?>" data-postcode="<?php echo $row['postcode']; ?>"
            >
                <td><?php echo (strlen($row['name'])           > 0 ? $row['name']           : '-'); ?></td>
                <td><?php echo (strlen($row['id'])             > 0 ? $row['id']             : '-'); ?></td>
                <td><?php echo (strlen($row['company_reg_no']) > 0 ? $row['company_reg_no'] : '-'); ?></td>
                <td><?php echo (strlen($row['postcode'])       > 0 ? $row['postcode']       : '-'); ?></td>
            </tr>
        <?php endwhile; ?>
    </tbody>
</table>

and my input is a simple one-liner checkbox that asks if none of the above matches the search..

what I'm trying to do, if the user clicks a table row, it posts the .data() attributes of the <tr> tag to the next step in the form, which is another page.

I've tried the hidden input idea, but because it's in a while loop where the inputs share the same name, the value of the input will be the last in the loop because it's last set. I've also tried ajax:

$.ajax({
    data:    $(this).data(),
    type:    'post',
    url:     'stepDos.php',
    success: function(data)
             {
                 window.location.replace('/path/to/stepDos.php');
             }
});

which didn't work when I var_dumped $_POST it returned empty.

I also tried a $.post function:

$.post('/path/to/stepDos.php', {data: data}, function(ev)
{
    window.location.replace('/path/to/stepDos.php');
});

again no avail - anything I'm missing?

Thanks

Answer 1

As I mentioned in the comment, you don't have to make one single form with all the data inside it.. Instead you can go for 1 form per row in a loop.

This will give you the intended result!