CODEWITHSUNDEEP
phpcheckbox
user12358408
user12358408

Reputation:

How to show echo "checked" edit in the checkbox?

I'm writing a system and have a checkbox in a form.

My problem is, when i press the edit button, the checkbox cannot show me the tick in the checkbox, I have tried to change checkbox type to text, it can show me the value. I have set the value if value = 1 is tick, if value = 0 is no tick. How can shows up tick in the form? Anyone can guide me to solve it?

Below is my coding:

Checkbox

<div class="form-group col-lg-6">
<label class="control-label col-lg-4">Pricing<span style="color:red;">&nbsp;</span></label>
<div class="col-lg-8">
<input type="text" name="rm_option" id="rm_option" value="1"  <?php if(_POST[$value]=='1'){echo "checked='checked'";} ?> ><strong> RM </strong></input>&nbsp;&nbsp;&nbsp;

<input type="text" name="point_option" id="point_option" value="1"    <?php if(_POST[$value]=='1'){echo "checked='checked'";} ?>><strong> Full Point </strong></input>&nbsp;&nbsp;&nbsp;
<input type="text" name="partial_option" id="partial_option" value="1"   <?php if(_POST[$value]=='1'){echo "checked='checked'";} ?>><strong> Partial Point + RM </strong></input>

</div>
</div>

Checkbox function

<?php
$sql = "select * from promotion_list where id=" . $_GET['id'];
$arr_sql = db_conn_select($sql);
foreach ($arr_sql as $rs_sql) {             
foreach ($rs_sql as $key => $value) {
        ?>
                                $("#<?php echo $key ?>").val("<?php echo $value?>");

        <?php
    }
        ?>

When I press the edit button, other column can show up in the form, only the checkbox cannot show me the tick. Below is the output picture:

Output

If I change Checkbox type to text, below is the output (Prove inside the checkbox got value)

Output 2

I have stuck in this problem already 1 week, hope someone coding hero can guide me to solve this problem. Thanks a lot.

Upvotes: 0

Views: 134

Answers (1)

Akam
Akam

Reputation: 1052

Add below code exactly after this line:

$("#<?php echo $key ?>").val("<?php echo $value?>");

it became:

$("#<?php echo $key ?>").val("<?php echo $value?>");
<?php if($value == 1){ ?>
$("#<?php echo $key ?>").attr("checked", true).prop("checked", true);
<?php } ?>

its clear that you are using jquery to work on the checkbox.

Upvotes: 0

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