Select options taking same value every time - php

Question

Looping from database duplicating the option values in options tag.
Please help me out.
My code is below.
<input type="hidden" name="beautician<?php echo $data[0]; ?>[]" id="beautician">
<select name="beautician[<?php echo $data[0]; ?>]" id="beauty" onchange="pvalue()">
<option value="null">Select</option>
<?php
$query = "select id,title from beautician order by id";
$run = mysql_query($query);
while ($row = mysql_fetch_array($run)) {
$id = $row[0];
$name = $row[1];
?>
<option value="<?php echo $id; ?>"> <?php echo $name; ?> </option>
<?php } ?>
</select>
whenever the user is selecting beautician it is taking id of the first option.
If id of the first selected beautician is 6 then every other beautician will have id 6

Answer 1

Because you forgot to print $id
Correct
<option value="<?php $id; ?>"> <?php echo $name; ?> </option>
To
<option value="<?php echo $id; ?>"> <?php echo $name; ?> </option>

Answer 2

Instead this:
<input type="hidden" name="beautician<?php echo $data[0]; ?>[]" id="beautician">
<select name="beautician[<?php echo $data[0]; ?>]" id="beauty" onchange="pvalue()">
<option value="null">Select</option>
<?php
$query = "select id,title from beautician order by id";
$run = mysql_query($query);
while ($row = mysql_fetch_array($run)) {
$id = $row[0];
$name = $row[1];
?>
<option value="<?php $id; ?>"> <?php echo $name; ?> </option>
<?php } ?>
</select>
Try this:
<input type="hidden" name="beautician <?php echo $data[0]; ?>[]" id="beautician">
<select name="beautician[<?php echo $data[0]; ?>]" id="beauty" onchange="pvalue()">
<option value="null">Select</option>
<?php
$query = "select id,title from beautician order by id";
$run = mysql_query($query);
while ($row = mysql_fetch_array($run)) {
$id = $row['id'];
$name = $row['title'];
?>
<option value="<?php echo $id; ?>"> <?php echo $name; ?> </option>
<?php } ?>
</select>

Answer 3

In the below statement you are not echoing $id in value. To get the value you have to <?php echo $id;?> OR <?=$id?>. Using your existing statement you have not printed the $id so the select option will take the <option> tag's text value, In your case value will be from $name variable.
Your statement:
<option value="<?php $id; ?>"> <?php echo $name; ?> </option>
Correct Statement:
<option value="<?php echo $id; ?>"> <?php echo $name; ?> </option>
As per my thinking you are getting value from the pvalue() on the change event of <select> tag and in this function you might be getting value based on the #beauty selector, because of this you are getting first dropdown value this happens because of duplicate id selector #beauty
Checkout below fiddle. There are two dropdowns and those have onchange="pvalue(this)" function on change event. Instead of no parameters in pvalue() function I have passed this in pvalue(this) function. Based on this parameter it will give you current <select> option.
<select name="beautician[0]" id="beauty" onchange="pvalue(this)">
<option value="null">Select</option>
<option value="one - 1">One - 1</option>
<option value="Two - 1">Two - 1</option>
</select>
<select name="beautician[1]" id="beauty" onchange="pvalue(this)">
<option value="null">Select</option>
<option value="one - 2">One - 2</option>
<option value="Two - 2">Two - 2</option>
</select>
<script>
function pvalue(obj, rowid){
var x = obj.value;
var y = document.getElementById('beautician'+ ).value = x;
alert(x);
}
</script>
<input type="hidden" name="beautician<?php echo $data[0]; ?>[]" id="beautician-<?php echo $data[0]; ?>">
<select name="beautician[<?php echo $data[0]; ?>]" id="beauty" onchange="pvalue(this, '<?php echo $data[0]; ?>')">
<option value="null">Select</option>
</select>
<script>
function pvalue(obj, rowid) {
var x = obj.value;
var y = document.getElementById('beautician-' + rowid).value = x;
}
</script>

Answer 4

This line:
<option value="<?php $id; ?>"> <?php echo $name; ?> </option>
Should be:
<option value="<?php echo $id; ?>"> <?php echo $name; ?> </option>