I am new to JQUERY and I am trying to search for the something and based on the searched text I am doing an ajax call which will call php function and the PHP is returning me with JSON data. I want to display the returned data in the Datatable form. I have my PHP file table.php and JavaScript file jss.js and my main.php. The PHP file is returning the JSON data and I able to use alert to display it.
I want to know how can I display it in datatable.
<div>
<input type="text" name="search_query" id="search_query" placeholder="Search Client" size="50" autocomplete="off"/>
<button id="search" name="submit">Search</button>
</div>
my ajax/jss.js file
$(document).ready(function(){
$('#search').click(function(){
var search_query = $('#search_query').val();
if(search_query !='')
{
$.ajax({
url:"table.php",
method:"POST",
data:{search_query:search_query},
success: function(data)
{
alert("HEKKI "+data);
}
});
}
else
{
alert("Please Search again");
}
});
});
my table.php file
<?php
$data=array();
$dbc = mysqli_connect('localhost','root','','acdc') OR die('Could not connect because: '.mysqli_connect_error());
if (isset($_REQUEST['search_query']))
{
$name = $_REQUEST['search_query'];
}
if($dbc)
{
if (!empty($name))
{
$sql = "select c.res1 res1,
cc.res2 res2,
cc.res3 res3,
cc.res4 res4,
cc.res5 res5
from table1 c
inner join table2 cc
on c.id = cc.id
where c.name like '".$name."%'
and cc.ENABLED = 1";
$res = mysqli_query($dbc,$sql);
if(!(mysqli_num_rows($res)==0))
{
while($row=mysqli_fetch_array($res))
{
$data['RES1'] = $row['res1'];
$data['RES2'] = $row['res2'];
$data['RES3'] = $row['res3'];
$data['RES4'] = $row['res4'];
$data['RES5'] = $row['res5'];
}
}
else
{
echo "<div style='display: block; color:red; text-align:center'><br/> Not Found,Please try again!!!</div>";
}
}
}
echo json_encode($data);
/*
*/
?>
Can you please guide me how to display the result in main page.