How to write PHP code inside jquery for retrieving values from mysql database?

Question

I am trying to get the values for the next drop-down from a database, which will be dependent on two previous drop-downs. The values for first two drop-downs are listed in the file itself. I want the second drop-down to be enable after selecting values from first, and similarly for third after second. Kindly help. HTML code below:

<form>
<select id="branch" name="branch" class="form-control" onchange="showUser(this.value)">
 <option value="">Select</option>
 <option value="1">Civil</option>
 <option value="2">Computer</option>
 <option value="3">Mechanical</option>
 <option value="4">Electronics and Telecommunications</option>
 <option value="5">Electrical and Electronics</option>
 <option value="6">Information Technology</option>
</select>
<select id="semester" name="semester" class="form-control" disabled>
 <option value="1">I</option>
 <option value="2">II</option>
 <option value="3">III</option>
 <option value="4">IV</option>
 <option value="5">V</option>
 <option value="6">VI</option>
 <option value="7">VII</option>
 <option value="8">VII</option>
</select>
</form>

jquery is:

<script>
 $(document).ready(function(){
 document.cookie = "s =hello" ;
 console.log('hello');
 $("#semester").attr("disabled", true);
 $("#branch").change(function(){
 $("#semester").attr("disabled", false); 
 $b = $('#branch').val();
 $("#semester").change(function(){
 $s = $('#semester').val();
 $("#sub_code").attr("disabled", false);
 console.log($s);
 if($s!=1||$s!=2)
 $s = $b+$s;
 <?php 
 $s= $_COOKIE['s'];
 $sql = "SELECT * FROM subjects WHERE sem_code=`$s`";
 ?>
 }); 
 });
 });
 </script>

I did not run the query since it is not assigned properly yet.

• dreamweiver: post relevant code here, what u have tried
• user5296864: Post your code i will modify whatever require
• asprin: I reckon you might have mistaken SO for a freelancing site. Btw, a hint for you - Lookup these following terms in google: Ajax+jquery+php and I'm sure you'll find a dozen tutorials
• Raj Sf: I think you need this - thesoftwareguy.in/multiple-dropdown-with-jquery-ajax-and-php
• Ru11: @RajSf It helped me.. I needed almost same thing. Thank you

Answer 1

You can't include php code in javascript , the first is executed on the server side, the second is executed on the client side which means that you can't re-execute only if you resend a request to the server, obviously this is usually done by submitting forms, BUT sometimes -like in your case- we don't want to reload the whole page for each request ! and for this purpose we use AJAX ajax sends post/get request to a specified php page which does the desired server-side tasks (updating data in the database, selecting some results from database, dealing with sessions maybe, etc...) try something like this:

var pathurl='/path/to/your/php/file';
var params={}; //the parameters you want to send
params['semester']=$s;
var requestData= $.ajax({
 type: 'POST',
 url: pathurl,
 cache: 'false',
 data: params,
 dataType: "json",
 beforeSend: function () {
 //here you can begin an animation or anything...
 },
 complete: function () {
 //here you can stop animations ...
 },
 success: function (response) {
 console.log(response); //check your console to verify the response
 //loop other elements inside response
 $.each(response, function (index, resultArray) {
 //do something : in your case append to dropdown a new option
 });
 }
 });
 requestData.error(function () {
 alert("error");
});

you should create a php file with the path specified in the above code, and there you can extract what you need from the database table, store values in an array and finally use:

echo json_encode($resultArray);

hope this was helpful