Dynamically Add Rows in Select Tag Using PHP

byMohd Farooq January 26, 2016
Today I am going to explain you how we can add rows in Select Tag Using PHP code. You have to read the example below.

Step1: Create Database

           Create database test_db;
           Create Table location;
           Create table location(country char(50),state char(50));
           Insert some data into the table location for testing purpose
           insert into location values('India','Uttar Pardesh');
           insert into location values('India','New Delhi');
           insert into location values('India','MP');
           insert into location values('Saudi Arabia','Jeddah');
           insert into location values('Saudi Arabia','Riyad');
           insert into location values('Saudi Arabia','Jizan');

 Step2: create a file index.php

 <html>
<head>
<script>
function showState(str) {
    if (str == "") {
        document.getElementById("txtHint").innerHTML = "";
        return;
    } else {
        if (window.XMLHttpRequest) {
            // code for IE7+, Firefox, Chrome, Opera, Safari
            xmlhttp = new XMLHttpRequest();
        } else {
            // code for IE6, IE5
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange = function() {
            if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                document.getElementById("state").innerHTML = xmlhttp.responseText;
            }
        };
        xmlhttp.open("GET","getState.php?q="+str,true);
        xmlhttp.send();
    }
}
</script>
</head>
<body>
    <?php

    $con = mysqli_connect('localhost','root','root','test_db');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

    $sql = "SELECT distinct country FROM location";

    $result = mysqli_query($con,$sql);
  
           echo "<select name='country' onchange='showState(this.value)'>";
 echo "<option value=''>Select Country:</option>";

       while($row = mysqli_fetch_array($result)){
 echo "<option value='".$row['country']."'>".$row['country']."</option>";
 
        }
        echo  "</select>";
       
        ?>
   <select name="state" id="state">
       <option value="">Select State:</option>
    </select>
</body>
</html>
 
Related Post:
Display data in Select Tag from Database Table and Insert the Selected Value in Different Table
Step3: create a file getState.php

<?php
//for integer value
//$q = intval($_GET['q']);
//for string value
$q = ($_GET['q']);
echo $q;
$con = mysqli_connect('localhost','root','root','test_db');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"ajax_demo");
$sql="SELECT state FROM location WHERE country = '".$q."'";
$result = mysqli_query($con,$sql);
if($result>0){


 echo "<option value=''>Select State:</option>";

       while($row = mysqli_fetch_array($result)){
 echo "<option value='".$row['state']."'>".$row['state']."</option>";
 
        }   
}
mysqli_close($con);
?>

Output:






Download Project




Note: If you like this code or having any question then reply me in comment i will help you.
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3 Comments

  1. Anonymous28 January 2016 at 03:44

    Nice code, good job

    ReplyDelete
  2. Unknown3 February 2016 at 04:04

    Nice :) thanks :D

    ReplyDelete
  3. Mueeid Khan10 January 2019 at 07:22

    The PHP programmers can easily enhance the web application's performance by not executing database queries in loop.plakatų spausdinimas

    ReplyDelete