CODEWITHSUNDEEP
phpjqueryhtmlmysqldrop-down-menu
user3315848
user3315848

Reputation: 77

Populate input fields with values when option is selected in php mysql

I have a form that i would like to submit details when an option is selected from the database.

The mysql database table :-

USERS contains fields like [email],[age],[name].

I want to be able to populate the other input fields values when one field is selected from the menu.

        <form>
        User
        <select name="user" id="user">
          <option>-- Select User --</option>
          <option value="Mark">Mark</option>
          <option value="Paul">Paul</option>
          <option value="Hannah">Hannah</option>
        </select>

        <p>

          Age 
          <input type="text" name="age" id="age">

        </p>
        <p>
          Email 
          <input type="text" name="email" id="email">
        </p>
        </form>

How do i acheive this using jquery or javascript.

Upvotes: 0

Views: 6023

Answers (1)

Mr. J
Mr. J

Reputation: 320

Please Try this,

on your HTML page:

write this to your html page,

<script>
$(document).ready(function(){
    $('#user').on('change',function(){
        var user = $(this).val();
        $.ajax({
            url : "getUser.php",
            dataType: 'json',
            type: 'POST',
            async : false,
            data : { user : user},
            success : function(data) {
                userData = json.parse(data);
                $('#age').val(userData.age);
                $('#email').val(userData.email);
            }
        }); 
    });
});
</script>

in getUser.php:

getUser.php

<?php

$link = mysqli_connect("localhost", "user", "pass","mydb");

$user = $_REQUEST['user'];    
$sql = mysqli_query($link, "SELECT age,email FROM userstable WHERE name = '".$user."' ");
$row = mysqli_fetch_array($sql);

json_encode($row);die;

Upvotes: 2

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