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how to retain drop down values as the value selected after submit

My drop down is showing blank then when i select the value of dropdown the same value is showing, but i have to show dropdown value as select first then when I click on button the respective value should show
I am doing a Php program
<form class="form-horizontal" name="form" method="post" action="<?php $_PHP_SELF?>">
<label for="courseDisp" class="col-sm-2" style="margin-top:10px;">Course : </label>
<?php
$course="SELECT * from course";
$res= $conn->query($course);
if($res->num_rows>0)
{
echo '<select name="courseDisp" id="courseDisp" class="form-control col-sm-3" style="margin-top:8px;display:inline;padding:10px;">';
echo '<option value="0" selected> -- SELECT --</option>';
while($row=$res->fetch_assoc())
{
echo '<option value='.$row["course_id"].'>'.$row['shortname'].'</option>';
}
echo '</select>';
} else {
echo "0 result";
}
?>
<label for="yearDisp" class="col-sm-2" style="margin-top:10px;">Year : </label>
<?php
$year="SELECT distinct(year) from syllabus";
$res= $conn->query($year);
if($res->num_rows>0)
{
echo '<select name="yearDisp" id="yearDisp" class="form-control col-sm-3" style="margin-top:8px;display:inline;padding:10px;">';
echo '<option value="0">-- SELECT --</option>';
while($row=$res->fetch_assoc())
{
echo '<option value='.$row["year"].'>'.$row['year'].'</option>';
}
echo '</select>';
} else {
echo "0 result";
}
?>
<script type="text/javascript">
document.getElementById('courseDisp').value = "<?php echo $_POST['courseDisp'];?>";
document.getElementById('yearDisp').value = "<?php echo $_POST['yearDisp'];?>";
<input type="submit" class="btn col-sm-2" style="margin-left:15px;margin-top:10px;width:60px;font-weight:bold;font-size:15px;" value="GO" name="btnGo" id="btnGo" />
</form>

I think you are doing it in a wrong way:
your code should look like this
<script type="text/JavaScript">
var valueSelected=document.getElementById('course').value;
alert(valueSelected);// do here according to the need
</script>

This is because there is no $_POST variables present before you submit a form.
$_POST variables can only be 'accessed' whenever a POST form is submitted, so when the form is not submitted, $_POST['course'] will be undefined. If you want to use persistant, but also relative variables, use $_GET.
This can be done the following way:
<script type="text/javascript">
document.getElementById('course').value =<?php echo $_GET['course'];?>";
</script>
(this will cause an error if value is not set, make sure to make exceptions for that, using if statements in PHP)
but the value also needs to be fetched from the URL.
so your url needs to have ?course=<course_value> in it, for example:
https://example.com/index.php?course=Course%201
Click here for more about POST vs GET requests

Instead of setting the value with javascript, you should directly write the selected attribute.
<select name="course">
<?php foreach ($options as $key => $value): ?>
<option value="<?= $key ?>"<?php if ($key == $_POST['course']) echo " selected" ?>>
<?= $value ?>
</option>
<?php endforeach; ?>
</select>
If you have to do this in javascript, keep sure, you use the correct syntax. Your example has a wrong " at the end of the line. Also you should use json_encode, if you want to output vars into javascript. And a last thing - if you don't put this inside the document ready event, the script has to be placed after the select element, which you wan't to manipulate
<select name="course">...</select>
...
<script type="text/javascript">
document.getElementById('course').value = <?= echo json_encode($_POST['course']) ?>;
</script>

Needed to keep the <option value="">-Select-</option>

select not sending $_POST data

I am confused as to why my select is not passing any post data to the results page. If I run var_dump($_POST); the select (#boxdest or #boxdest2) is not being displayed. Is there some special way to pass select to php results page. What I normally do is as an example: $var=$_POST['boxdest'];. Problem is the select is not being sent from original code. I have posted my code and would be grateful if someone could show me where I have gone wromg. Many thanks.
<?php
$conn = mysql_connect("localhost","root","");
mysql_select_db("sample",$conn);
$result = mysql_query("SELECT * FROM boxes where department = '{$_GET['dept']}'");
?>
<select id="boxdest" name="boxdest[]"size="7">
<?php
$i=0;
while($row = mysql_fetch_array($result)) {
?>
<option value="<?php echo $row["custref"];?>"><?php echo $row["custref"];?></option>
<?php
$i++;
}
?>
</select>
<input type="button" id="submit2" name="submit2" value=">" />
<input type="button" id="submit3" name="submit3" value="<" />
<select id="boxdest2" name="boxdest2[]" size="7"></select>
<script type="text/javascript">
$("#submit2").click( function()
{
//alert('button clicked');
$box1_value=$("#boxdest").val();
$box1_text=$("#boxdest option:selected").text();
$("#boxdest2").append('<option value="'+$box1_value+'">'+$box1_text+'</option>');
$("#boxdest option:selected").remove();
});
$("#submit3").click( function()
{
//alert('button3 clicked');
$box2_value=$("#boxdest2").val();
$box2_text=$("#boxdest2 option:selected").text();
$("#boxdest").append('<option value="'+$box2_value+'">'+$box2_text+'</option>');
$("#boxdest2 option:selected").remove();
}
);
</script>

How to make HTML form fields automatically change when an option selected on a select field

Let's say I have a form like this in my CodeIgniter project.
<?php echo form_open(); ?>
<select>
<?php foreach($status_list as $status): ?>
<option value="<?php echo $status->id; ?>"><?php echo $status->name; ?></option>
<?php endforeach; ?>
</select>
<!-- Show this only if status type is, let's say "E" -->
<input type="text" name="E_happened">
<!-- Show this only if status type is, let's say "S" -->
<input type="text" name="S_happened">
<?php echo form_close(); ?>
What I want to do is if an user select one status, according to the it's type, show a text field to get an input.
I've made a way to get a type of the status like this: http://localhost/myapp/index.php/status/type/{status_id} where users can pass status ID and it will "echo" the type of the status.
I want to receive it back to the HTML page via a JavaScript method and show those text input fields. How do I do that?
Thank you. :)

As you have jQuery tag, so i suggest you this:
$('select').change(function(){
var inp = this.value.trim();
$(this).parent().find('input[type="text"][name^="'+inp+'"]').show().siblings(':text').hide();
});

I have posted the sample flow as per you want to do. hope this will helpful for you.
PHP:
<?php echo form_open(); ?>
<select>
<?php foreach($status_list as $status): ?>
<option value="<?php echo $status->id; ?>"><?php echo $status->name; ?></option>
<?php endforeach; ?>
</select>
<!-- Show this only if status type is, let's say "E" -->
<input type="text" id="E_happened" name="E_happened">
<!-- Show this only if status type is, let's say "S" -->
<input type="text" id="S_happened" name="S_happened">
<?php echo form_close(); ?>
JAVASCRIPT :
$('select').change(function(){
var statusId = $(this).val();
var statusType = $.get("http://localhost/myapp/index.php/status/type/"+statusId);
if(statusType == 'E')
{
$('#E_happened').value("what do you want here");
}
if(statusType == 'S')
{
$('#S_happened').value("what do you want here");
}
});

PHP Form fetching MySQL using AJAX [duplicate]

This question already has answers here:
jQuery Ajax POST example with PHP
(17 answers)
Closed 8 years ago.
I developed a php form which initiates a database request to fetch data depending on a drop down choice.
PHP Form:
<form method="get" action="<?php echo $url = basename($_SERVER['PHP_SELF']); ?>">
<select name="town" onchange='this.form.submit()'>
<?php $result= mysql_query('Query'); ?>
<option value="x" selected>Select Choice</option>
<?php while($row= mysql_fetch_assoc($result)) { ?>
<option value="<?php echo htmlspecialchars($row['town']);?>" >
<?php echo htmlspecialchars($row['town']); ?>
</option>
<?php } ?>
<input type="hidden" name="action" value="submit" /><br>
</select>
</form>
Form action:
<?php
if(isset($_GET["action"])) {
$var1= $wpdb->get_results("Query");
$var2= $wpdb->get_results("Query");
Content to show once executed }
?>
How can I make the form fetch the Data using AJAX not to stay refreshing the whole page continuously but only the form part?

<!DOCTYPE html>
<html>
<head>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
</head>
<body>
<form id="form_id" action="<?php echo $url = basename($_SERVER['PHP_SELF']); ?>" method="post">
<select id="town" name="town" onchange="send_to_server()">
<?php $result= mysql_query("Query"); ?>
<option value="x" selected>Select Choice</option>
<?php while($row= mysql_fetch_assoc($result)){ ?>
<option value="<?php echo htmlspecialchars($row['town']); ?>">
<?php echo htmlspecialchars($row['town']); ?>
</option>
<?php } ?>
<input type="hidden" name="action" value="submit" /><br>
</select>
</form>
<script type='text/javascript'>
/* attach a submit handler to the form */
function send_to_server(){
var value = $("#town").val();
/* get some values from elements on the page: */
var $form = $("#form_id"); var url = $form.attr('action');
/* Send the data using post */
var posting = $.post( url, { option_value: $("#town").val() } );
posting.done(function( data ) {
alert('success');
});
}
</script>
</body>
</html>
The above does exactly what you want. Check it in localhost

multiple JS form submit on change inside a PHP while

I have a while in PHP that is building a table. Inside my while, I have a dropdown menu that I want to execute code on change.... The form submit on change doesn'T work...
This is my code:
<?
while ($data2 = mysql_fetch_array( $data))
{ ?>
<form name="FormSize" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" >
<td>
<select name="size[]" class="form-field3" OnChange="document.FormSize.submit();" >
<? foreach ($sizearray as $value) { ?>
<option value="$value" <? if($data2['size'] == $value) { echo "selected";} ?> ><? echo $value; ?></option>
<? } ?>
</select>
<input type="hidden" name="product_id[]" value="<? echo $data2['product_id']; ?>">
</td>
</form>
<? } ?>
and on top of my page, I have this to execute the code..
<?
//submit size form
if (isset($_POST['size']))
{
foreach($_POST['product_id'] as $key => $id)
{
$product_id = $id;
$newsize = $_POST['size'][$key];
$sql3 = mysql_query("update cart SET size = '".$newsize."' where product_id = '".$product_id."' ");
}
?>
any idea why it's not executing ?

You have multiple forms with the same name="FormSize". So document.FormSize will be an array-like object, and you need to access the appropriate one. You can do this by using this.form to refer to the <form> that contains the <select>.
You also need to fix the element nesting -- the <form> has to be inside the <td>, not the other way around.
<?
while ($data2 = mysql_fetch_array( $data))
{ ?>
<td>
<form name="FormSize" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" >
<select name="size[]" class="form-field3" OnChange="this.form.submit();" >
<? foreach ($sizearray as $value) { ?>
<option value="$value" <? if($data2['size'] == $value) { echo "selected";} ?> ><? echo $value; ?></option>
<? } ?>
</select>
<input type="hidden" name="product_id[]" value="<? echo $data2['product_id']; ?>">
</form>
/td>
<? } ?>

Could be because the html is not valid, you cannot put your form tag as a direct child of a tr.
Apart from that, naming all your forms the same will make it pretty hard to submit a specific one by its name.
As you are using arrays for your input names, I assume that you mean all of them to be in the same form, so you should just have your form wrap the table and take it out of the while loop.