Adding multiple check boxes from the db in php does not show up

Question

I have the following code that I would like to use to add check boxes to the interface depending on the selected drop down value

    <!--drop down list for the floors--> 
<div class="form-group">
    <label for="ddlFloorNo"><?php echo $_data['add_new_form_field_text_10']; ?> :</label>
    <select onchange="getUnitReport(this.value)" name="ddlFloorNo" id="ddlFloorNo" class="form-control">
        <option value="">--<?php echo $_data['add_new_form_field_text_11']; ?>--</option>
        <?php
        $result_floor = mysqli_query($link, "SELECT * FROM tbl_add_floor order by fid ASC");
        while ($row_floor = mysqli_fetch_array($result_floor)) {
            ?>
            <option <?php
            if ($floor_id == $row_floor['fid']) {
                echo 'selected';
            }
            ?> value="<?php echo $row_floor['fid']; ?>">
                <?php echo $row_floor['floor_no']; ?></option>
        <?php } ?>
    </select>
</div>
 <!--Check boxes to be displayed based on the selected floor inn the drop down list above--> 
    <div class="form-group">
        <label for="ChkOwnerUnit"><?php echo $_data['add_new_form_field_text_8']; ?> : </label>
        <?php
        $result_unit = mysqli_query($link, "SELECT * FROM tbl_add_unit where floor_no ='" . (int) $row_floor['fid'] . "' order by uid ASC");
        while ($row_unit = mysqli_fetch_array($result_unit)) {
            ?>
        <input type="checkbox" class="form-control" name="ChkOwnerUnit[]" value="<?php echo $row_unit['uid']; ?>"/>
            <?php } ?>
    </div>

My problem is that the checkboxes do not show up. What could I be doing wrong

Are you sure there is data in that table with a floor_no matching the fid you are passing — RiggsFolly
– RiggsFolly, Commented Sep 12, 2018 at 9:38
How exactly did you determine that they don't show up? What have you tried to debug the problem? — Nico Haase
– Nico Haase, Commented Sep 12, 2018 at 9:38
Did you check the page source using your browser to see if there is anything there, but it is just not showing up due to some HTML issue — RiggsFolly
– RiggsFolly, Commented Sep 12, 2018 at 9:40

Jigar Shah · Accepted Answer · 2018-09-12 09:53:13Z

3

Its because you are not getting data in your line

$result_unit = mysqli_query($link, "SELECT * FROM tbl_add_unit where floor_no ='" . (int) $row_floor['fid'] . "' order by uid ASC");

$row_floor is not accessible at there. You need to make changes in either html or create array to store ids.

answered Sep 12, 2018 at 9:53

Jigar Shah

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6 Comments

The Only Smart Boy Over a year ago

I did the same thing with option select but it worked would you give me a clue on how to go about making it work? I am new with php

Jigar Shah Over a year ago

you are trying to use that $row_floor variable out of the while loop so, you do not have data in (int) $row_floor['fid'], thus query will not return any thing and checkboxes are not showing

The Only Smart Boy Over a year ago

the $row_floor is in the dropdown menu

Jigar Shah Over a year ago

yes and you are using the ids (int) $row_floor['fid'] from that variable for getting checkbox values, but these ids are not accessible there in another query. Please create an array to store the ids and use that array. if that is dependent data then you should go for AJAX call

The Only Smart Boy Over a year ago

Would you proovide me with a link to some tutorial on the ajax?

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