PHP echo select menu value but delete duplicate - php

Question

I'm working on a profile update page and i have a form field where users can select enabled or disabled. When i echo the value, it has a duplicate and can be confusing for the user. How can i get rid of the duplicate?
<div class="form-group">
<label for="status" class="col-sm-3 control-label">Change Status</label>
<div class="col-sm-9">
<select class="form-control btn-danger" id="status" name="status">
<option value="<?php echo $status ?>"><?php echo $status ?></option>
<option value="Disabled">Disabled</option>
<option value="Enabled">Enabled</option>
</select>
</div>
</div>
As you can see, if the status is enabled and i hit the select menu its gonna say Enabled twice, same thing for disabled.
The only thing i thought of was to have another field labeled current status and have the select menu to change status.
I really dont want to do that.
<div class="form-group">
<label for="currentstatus" class="col-sm-3 control-label">Current Status</label>
<div class="col-sm-9">
<input type="text" class="form-control" id="currentstatus" name="currentstatus" value="<?php echo $status ?>">
</div>
</div>
<div class="form-group">
<label for="status" class="col-sm-3 control-label">Change Status</label>
<div class="col-sm-9">
<select class="form-control btn-danger" id="status" name="status">
<option value="<?php echo $status ?>"><?php echo $status ?></option>
<option value="Disabled">Disabled</option>
<option value="Enabled">Enabled</option>
</select>
</div>
</div>
This worked for me:
<option <?php if ($status=="Disabled") echo " selected"; ?>>Disabled</option>
<option <?php if ($status=="Enabled") echo " selected"; ?>>Enabled</option>

Answer 1

Instead of outputting the selected value as another option, output it as the selected attribute of an existing option. Something like this:
<select class="form-control btn-danger" id="status" name="status">
<option value="Disabled" <?php echo ($status == "Disabled") ? "selected" : ""; ?>>Disabled</option>
<option value="Enabled" <?php echo ($status == "Enabled") ? "selected" : ""; ?>>Enabled</option>
</select>
There are likely a variety of ways to achieve the same end result, but the point is to not add options based on the data you have but instead use that data to determine which of the existing, finite, hard-coded options you already have should be selected.

Answer 2

Here is some jQuery that will let you change the value of the select box.
function setStatusSelect(val) {
$("select#status").val(val);
}
setStatusSelect('Enabled');
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="form-group">
<label for="status" class="col-sm-3 control-label">Change Status</label>
<div class="col-sm-9">
<select class="form-control btn-danger" id="status" name="status">
<option value="Disabled">Disabled</option>
<option value="Enabled">Enabled</option>
</select>
</div>
</div>
If you include the Jquery function at the top of the page (so that it loads quickly) then you can do something like this:
<?php
print("<script type=\"text/javascript\">");
print("setStatusSelect(".$status.")");
print("</script>");
?>
to set the value.
I haven't tested the PHP portion, but I know the jQuery portion works.