Multiple selection from Drop down List

Question

Hie,I had asked this question already but I have make it briefly now please share some logic. 1st dropdown List- enter image description here Contain Class-1, class-2, class-3..etc. 2nd dropdown List- Depends on class.If we chose class-1 and class-2 then respective Student Id will show on 2nd dropdown list. I want selected Classes in respective variables and selected turbines in respective variable. Then I will get this selected variables and use in another page.

I want multiple selection in each dropdown list. suppose I have select one Class-1 then, On the classId second dropdown list will fill with respective studentID,again I am select another class-2 then again Add respective studentID list in 2nd dropdown list.

In my databse script ClassId and StudentId this attribute present in same table only (not diffrent tables).

Table view somewhat like this->

 ClassID    StudentID   StudentNm 
      1        101         abc 
      1        102         xyz 
      1        103         jkl 
      2        201         uio
      2        202         tyu 
      3        302         qwe 
      3        305         zxc

@bablu I want multiple selection from both dropdown list and 2nd dropdown dependent on 1st dropdown. — Pratiksha
– Pratiksha, Commented Feb 28, 2015 at 10:47
As you have already asked the same question ( see stackoverflow.com/questions/28758879/… ) you should clarify the original question rather than creating a new question. — AdrianHHH
– AdrianHHH, Commented Feb 28, 2015 at 10:58
You are populating second dropdown from classID, It means you need to bind ajax function on change event of first dropdown. — Mihir Bhatt
– Mihir Bhatt, Commented Feb 28, 2015 at 11:15

user3386779 · Accepted Answer · 2015-02-28 11:43:57Z

try this sample

your table shoule be like this

   CREATE TABLE 'data'
   (
   'id' int primary key auto_increment,
    'data' varchar(50),
     'weight' int(2),
   );

  CREATE TABLE `data_parent` 
   (
  `pid` int(11) primary key auto_increment,
  `did` int(11) unique,
  `parent` int(11),
  Foreign key(did) references data(id)
  )

<script type="text/javascript" src="http://ajax.googleapis.com/
ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function()
{
$(".country").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;

$.ajax
({
type: "POST",
url: "ajax_city.php",
data: dataString,
cache: false,
success: function(html)
{
$(".city").html(html);
} 
});

});

});
</script>

Country :
<select name="country" class="country">
<option selected="selected">--Select Country--</option>
<?php
include('db.php');
$sql=mysql_query("select id,data from data where weight='1'");
while($row=mysql_fetch_array($sql))
{
$id=$row['id'];
$data=$row['data'];
echo '<option value="'.$id.'">'.$data.'</option>';
} ?>
</select> <br/><br/>

City :
<select name="city" class="city">
<option selected="selected">--Select City--</option>
</select>

Anto S · Accepted Answer · 2015-02-28 12:10:07Z

0

Try this

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<select multiple="multiple" name="classId" class="select-box">
    <option value="1">Option 1</option>
    <option value="2">Option 2</option>
    <option value="3">Option 3</option>
    <option value="4">Option 4</option>
    <option value="5">Option 5</option>
</select>

 <script src="https://code.jquery.com/jquery-2.1.3.js"></script>
 <script>
    $('.select-box').change(function(){
        var classId = $(this).val();

        $.ajax({
            url : 'getSub.php',
            type: 'POST',
            dataType : 'JSON',
            data : {
                'classId' : classId,
            },
            success : function(data){
                console.log(data);
            }

        });
    });
 </script>
</body>
</html>

When you print selectedValues values will be printed in array format. in getSub.php you can get values and send it as json

getSub.php

<?php

$classIds = implode(',', $_POST['classId']);

$stmet = "SELECT StudentID from TBL where ClassID IN ('$classIds')";

$query = mysql_query($stmet);

$result = mysql_fetch_array($query);

echo json_encode($result);
?>

edited Feb 28, 2015 at 12:10

answered Feb 28, 2015 at 11:30

Anto S

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13 Comments

Mihir Bhatt Over a year ago

You need to pass selectedValues as parameter. Otherwise dropdown will not be based on class value, also event will be "change" not "click".

Pratiksha Over a year ago

@bablu You have to mean, the selected Class Id get on getSub .php page. and on that page select student value and show on second dropdown.

Pratiksha Over a year ago

@bablu It work on button event or on selection of 1st dropdown? because, In my html view 2nd drop down exactly bottom of the first dropdow so,

Pratiksha Over a year ago

I want it on change on select-box and also multiple selection on both dropdown

Pratiksha Over a year ago

@bablu Yes your code is working like charm! Thank you! Your approch is good to question. but,I have one query that, Here we have use list box, Can we use dropdownList only..meance I have to say, If I select one class then dropdown list is close again, then again select another class and so on...Here you have using dropdown list box, but I want in dropdown list. acn we approch ths?

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