PHP, JS, Ajax දන්න අයගෙන් උදව්වක් ඕනේ

Sonique · Aug 17, 2021

geeko said:
ppodi example ekak danna puluwanda

davas ganak google kala eka eka widihata solution ekak na bro. kiyala thiyena eka widihakwath hariyanne na
------ Post added on Aug 17, 2021 at 12:13 PM

Example ekak danna nan amarui machan man karanne django. File thogayak methanta atha arinna wenawa ehema danna. Mama web developer kenek newe.

karanne mehemai
Page ekak thigannawa queries walata. Eka pennana ekak newe user ta. Ajax walin karanne js walin fetch karana data ara hidden page ekata yawana eka (select options data wage). Cookies use karala POST karanawa incoming data backend ekata. back end eke quaries duwala return karanawa json ekak hidden page ekata. Eka fetch karanawa original page eken ajax walin.
Meka safe da nadda man danne na. Habai mata wadak na ewwa man web developer kenek newe hinda

wade wenawa

Lakshan-Seram · Aug 17, 2021

onna weda karana code eka:

HTML:

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>testphpajax</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script>
</head>

<body>
<select name="brand" id="brand" onChange="make( this )" required>
    <option value="0" disabled>Select...</option>
    <option value="1">Elakiri</option>
    <option value="2">Google</option>
    <option value="3">Something</option>
</select>
<div id="response">
   
</div>
<script>
function make( el ) {
    var x = el.value;
    $.ajax( {
        url: 'php.php',
        type: 'POST',
        data: { x:x },
        dataType : 'json',
        success: function( response ){
            var id = response.id;
            var name = response.name;
            var url = response.url;
            var message = response.message;
            $( '#response' ).html( id + ', ' + name + ', ' + url + ', ' + message );
        },
        error: function (xhr, ajaxOptions, thrownError) {
            alert( thrownError );
        }
    } );
}
</script>
</body>
</html>

PHP:

<?php

$x = @$_REQUEST['x'];

if( $x ) {
   
    $message = '';
    $data = array();
   
    $servername = 'localhost';
    $username = 'root';
    $password = '';
    $database = 'testphpajax';

    $conn = new mysqli( $servername, $username, $password, $database );

    if ( $conn->connect_error ) {
     
        die( $conn->connect_error );
     
    }
   
    $sql = 'SELECT * FROM testphpajax WHERE id = ' . $x;
    $result = $conn->query( $sql );

    if ( $result->num_rows > 0 ) {
       
        $row = $result->fetch_assoc();
       
        $data = array (
            'id' => $row['id'],
            'name' => $row['name'],
            'url' => $row['url']
        );
       
        $message = 'Success!';
        http_response_code( 200 );
       
    } else {
       
        http_response_code( 404 );
        $message = 'No results found!';
       
    }
   
} else {
   
    http_response_code( 400 );
    $message = 'x is empty!';
   
}

$data['message'] = $message;

die( json_encode( $data ) );

?>

SQL:

DROP TABLE IF EXISTS `testphpajax`;
CREATE TABLE IF NOT EXISTS `testphpajax` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(250) NOT NULL,
  `url` varchar(250) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;

INSERT INTO `testphpajax` (`id`, `name`, `url`) VALUES
(1, 'elakiri', 'http://elakiri.com'),
(2, 'google', 'https://google.com');
COMMIT;

geeko · Aug 17, 2021

lkzombie said:
data php file ekata post wenawada?

na

MachoMan said:
On change,
Grab the parameters and make the ajax call
Sent it to differnt page on ajax call, run the mysql query and get the data on JSON
Pass it to main page.
Grab date from JSON
Dispaly hidden div with query data...

Try this, hope this work..

any example code?
------ Post added on Aug 17, 2021 at 1:29 PM

Lakshan-Seram said:

onna weda karana code eka:

HTML:

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>testphpajax</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.6.0/jquery.min.js"></script>
</head>

<body>
<select name="brand" id="brand" onChange="make( this )" required>
    <option value="0" disabled>Select...</option>
    <option value="1">Elakiri</option>
    <option value="2">Google</option>
    <option value="3">Something</option>
</select>
<div id="response">
  
</div>
<script>
function make( el ) {
    var x = el.value;
    $.ajax( {
        url: 'php.php',
        type: 'POST',
        data: { x:x },
        dataType : 'json',
        success: function( response ){
            var id = response.id;
            var name = response.name;
            var url = response.url;
            var message = response.message;
            $( '#response' ).html( id + ', ' + name + ', ' + url + ', ' + message );
        },
        error: function (xhr, ajaxOptions, thrownError) {
            alert( thrownError );
        }
    } );
}
</script>
</body>
</html>

PHP:

<?php

$x = @$_REQUEST['x'];

if( $x ) {
  
    $message = '';
    $data = array();
  
    $servername = 'localhost';
    $username = 'root';
    $password = '';
    $database = 'testphpajax';

    $conn = new mysqli( $servername, $username, $password, $database );

    if ( $conn->connect_error ) {
    
        die( $conn->connect_error );
    
    }
  
    $sql = 'SELECT * FROM testphpajax WHERE id = ' . $x;
    $result = $conn->query( $sql );

    if ( $result->num_rows > 0 ) {
      
        $row = $result->fetch_assoc();
      
        $data = array (
            'id' => $row['id'],
            'name' => $row['name'],
            'url' => $row['url']
        );
      
        $message = 'Success!';
        http_response_code( 200 );
      
    } else {
      
        http_response_code( 404 );
        $message = 'No results found!';
      
    }
  
} else {
  
    http_response_code( 400 );
    $message = 'x is empty!';
  
}

$data['message'] = $message;

die( json_encode( $data ) );

?>

SQL:

DROP TABLE IF EXISTS `testphpajax`;
CREATE TABLE IF NOT EXISTS `testphpajax` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(250) NOT NULL,
  `url` varchar(250) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;

INSERT INTO `testphpajax` (`id`, `name`, `url`) VALUES
(1, 'elakiri', 'http://elakiri.com'),
(2, 'google', 'https://google.com');
COMMIT;

Thanks machan man meka try karala balannam
------ Post added on Aug 17, 2021 at 1:42 PM

shenic88 · Aug 17, 2021

AJAX
people have helped you already..

geeko · Aug 17, 2021

shenic88 said:
AJAX
people have helped you already..

yes i got the solution. Thanks for all

උදව් කරපු හැමෝටම ගොඩාක් ස්තූතියි. @Lakshan-Seram ස්තූතියි මම වෙනුවෙන් වෙලාවක් වෙන් කරලා මේ වැඩේ කරලා දුන්නාට.

hodakolla · Aug 18, 2021

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PHP, JS, Ajax දන්න අයගෙන් උදව්වක් ඕනේ

Sonique

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