Form fields updating values from mysql on selecting a value in drop down
Question
I am developing a page in PHP. It has a select drop down(user names) and a form below containing user details. On selecting a particular user from drop down, details of selected user should be populated in form below.
Am beginner to PHP.
How to do this? Any simple solution
2 answers
solution1
0 2014-04-18 07:20:40
You need to put dropdown in a form. On change event of dropdown just post form on same URL with get method. Check the paramenter in url and fetch data and display.
<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
$('#userid').change(function(){
$('#getUserData').submit();
});
});
</script>
</head>
<body>
<form name='getUserData' id='getUserData' action='#' method='GET'>
Select User : <select id='userid' name='userid'>
<option value='1'>Lokendra</option>
<option value='2'>Amit</option>
<option value='3'>Nitin</option>
<option value='4'>Rishabh</option>
</select>
</form>
<?php
$userArray=array(
1 => 'Lokendra',
2 => 'Amit',
3 => 'Nitin',
4 => 'Rishabh',
);
$postedData=$_REQUEST;
// Fire your select query here and diplay data
if(isset($postedData['userid'])){
echo "Selected User name =>".$userArray[$postedData['userid']];
}
?>
</body>
</html>
Dont forget to accept this answer if helps :)
solution2
0 2014-04-18 07:34:44
If you want to make it an AJAX request.
Follow the steps
Lets assume you make your post requests to user.php
<form id='getDetails' type='post' action='user.php'>
<select id='users'>
<option>AMAN</option>
<option>ABHAY</option>
</select>
</form>
<script>
$(document).ready(function(){
$('users').on('change', function(){
var userVal = $('users option:selected').text();
$.post('user.php',{user:userVal},function(data){
console.log(data);
//Populate the form by using data variable
//which contains the data you need
});
});
});
</script>
Make sure to include jQuery
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