|
PHP - Trying To Display User Details After Login
after the user has logged in, I would like to display their details by barcode id
Login.php <?php $host=""; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name=""; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); session_start(); // username and password sent from form $barcodeID=$_POST['barcode']; // To protect MySQL injection (more detail about MySQL injection) $barcodeID = stripslashes($barcodeID); $barcodeID = mysql_real_escape_string($barcodeID); $sql="SELECT * FROM $tbl_name WHERE BarcodeID='$barcodeID'"; $result=mysql_query($sql); $count=mysql_num_rows($result); if($count > 0){ $data = mysql_fetch_array ($result); $_SESSION["user_id"] = $data["BarcodeID"]; $_SESSION["user_firstname"] = $data["Firstname"]; $_SESSION["user_surname"] = $data["Surname"]; $_SESSION["user_jobrole"] = $data["JobRole"]; $_SESSION["user_manager"] = $data["Manager"]; $_SESSION["user_priority"] = $data["Priority"]; $_SESSION["user_datejoined"] = $data["DateJoined"]; $_SESSION["user_times_loggged_in"] = $data["TimesLoggedOn"]; if ($_SESSION["user_priority"] == '1') { header("Location: AdminSection.php"); } else { header("Location:LoggedIn.php"); } if ($_SESSION["user_times_loggged_in"] == '0') { header("Location:UsingTheSystem.html"); } } ?> LoggedIn.php I keep getting the error undefined index "barcode"? <?php $barcodeID = $_POST["barcode"]; include 'dbcon.php'; $sql = "SELECT Firstname, Surname, JobRole, Manager" . " FROM users" . " WHERE BarcodeID = .'$barcodeID'" ; $rows = mysql_query($sql); echo $rows; ?> Any help will be greatly appreciated Thanks Similar Tutorialshi im new to php
im using a script that i found at the link below:
http://forums.devshe...sql-891201.html
It works fine but i have added a couple of fields to the database : telephone and mobile_telephone
Ive change the register.php to include these fields but im struggling with the edit_account
Could anyone help please
Hi all I need some help with displaying user account details i am currently able to show only the email address and i would like to show the name school name and yeargroup heres my code for myaccount.php <?php require_once('Connections/isn_1.php'); ?> <?php if (!isset($_SESSION)) { session_start(); } $MM_authorizedUsers = "1,2,3,4"; $MM_donotCheckaccess = "false"; // *** Restrict Access To Page: Grant or deny access to this page function isAuthorized($strUsers, $strGroups, $UserName, $UserGroup) { // For security, start by assuming the visitor is NOT authorized. $isValid = False; // When a visitor has logged into this site, the Session variable MM_Username set equal to their username. // Therefore, we know that a user is NOT logged in if that Session variable is blank. if (!empty($UserName)) { // Besides being logged in, you may restrict access to only certain users based on an ID established when they login. // Parse the strings into arrays. $arrUsers = Explode(",", $strUsers); $arrGroups = Explode(",", $strGroups); if (in_array($UserName, $arrUsers)) { $isValid = true; } // Or, you may restrict access to only certain users based on their username. if (in_array($UserGroup, $arrGroups)) { $isValid = true; } if (($strUsers == "") && false) { $isValid = true; } } return $isValid; } $MM_restrictGoTo = "login.php?login=false"; if (!((isset($_SESSION['MM_Username'])) && (isAuthorized("",$MM_authorizedUsers, $_SESSION['MM_Username'], $_SESSION['MM_UserGroup'])))) { $MM_qsChar = "?"; $MM_referrer = $_SERVER['PHP_SELF']; if (strpos($MM_restrictGoTo, "?")) $MM_qsChar = "&"; if (isset($_SERVER['QUERY_STRING']) && strlen($_SERVER['QUERY_STRING']) > 0) $MM_referrer .= "?" . $_SERVER['QUERY_STRING']; $MM_restrictGoTo = $MM_restrictGoTo. $MM_qsChar . "accesscheck=" . urlencode($MM_referrer); header("Location: ". $MM_restrictGoTo); exit; } ?> <!DOCTYPE HTML> <html> <head> <title>My Account - <?php echo($_SESSION['MM_Username']); ?></title> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <style type="text/css"> @import url("style.css"); </style> </head> <body class="about"> <!-- Start NavBar --> <div id="topnavbar"> <dl> <dt id="home"><a href="index.php">Home</a></dt> <dt id="about"><a href="about.php">About</a></dt> <dt id="account"><a href="myaccount.php">Account</a></dt> <dt id="login"><a href="login.php">Login</a></dt> </dl> <dl id="rightnavbar"> <dt id="ISN"><a href="index.php">ISN</a></dt> </dl> </div> <!-- End NavBar --> <div id="page-container"> <div id="header"> </div> <div id="sidebar-a"></div> <div id="content"> <div class="padding"> <center> <table width="631" border="0"> <tr> <td colspan="2">Personal Details</td> </tr> <tr> <td width="229"> </td> <td width="648"></td> </tr> <tr> <td>Name</td> <td></td> </tr> <tr> <td>Email</td> <td><?php echo($_SESSION['MM_Username']); ?></td> </tr> <tr> <td>School Name</td> <td></td> </tr> <tr> <td>Year Group</td> <td></td> </tr> <tr> <td>DOB</td> <td></td> </tr> <tr> <td> </td> <td><a href="updateprofile.php">Modify my details</a></td> </tr> </table> <a href="logout.php">Logout?</a> </center> </div> </div> <div id="footer"> <div id="altnav"> <a href="index.php">Home</a> - <a href="login.php">Login</a> - <a href="register.php">Register</a> - <a href="about.php">About</a> - <a href="terms.php">Terms & Conditions</a> </div> <div id="copyright">© 2011 InterSchoolsNetwork, All Rights Reserved - A <a href="http://jordansmithsolutions.co.uk">Jordan Smith Solutions</a> & <a href="http://www.joecocorp.webs.com/">JoeCo Corp Production</a><br /> </div> </div> </div> </body> </html> <?php mysql_free_result($rsUpdateUser); ?> If you need any other code to help answer it for me then let me no please Actually, what i want to do is to use the email to fetch the $email,$password and $randomnumber from database after i am working on a project in which i am facing a problem. my question is that is it possible to get information/details of a user who is not logged in or who is not registered. if a user logged in then we can find get details easily. but how can i do this. is it possible. Obviously when connecting to php Im not going to show all of my login details; mysql_connect("details","details","password") or die(mysql_error()); mysql_select_db("details") or die(mysql_error()); whats the best way to hide them? Ive seen some people using an include file with their login details on but say for eg. <?php include('con.php'); ?> Whats to stop somone looking at www.myweb/con.php and obtaining my details there instead? I have created a button which when pressed should present the user with their details (whoever is logged in), here is the form code: <form id="form1" name="form1" method="post" action="getdetails.php"> <input type="submit" name="Get Details" value="Get Details" /> </label> </p> </form> Here is the getdetails.php file <?php mysql_connect("localhost","root",""); mysql_select_db("test"); $username = $_POST['textfield']; echo '</br>'; $query = mysql_query("SELECT * FROM membersdetails WHERE name=`$username` "); while($result = mysql_fetch_array($query)) { //display echo $result['firstname']; echo $result['surname']; } ?> Its not workin at all I have attacthed the error i am getting Any help please? Hello everyone.
I have a fully working form that gets data from a user (with $_post array) , and stores it in a database (mysql).
After successfulIy filling the form, I refer him to the "dashboard" page. In this page, i am having trouble to get his details from the database. How should I recognize him as the user that just registered?
should I use a $_post? or maybe a session? could you please give me a clue how to solve this?
Hi, I got this code which is meant to display the login details of the person that is logged in, however it just displays the details of the last person in the mysql table. I have set up some test logins, so if I login as paul1 the details for paul3 are displayed...confused Anyway, here is the page which displays the details Code: [Select] <?php session_start(); // This checks to make sure the session variable is registered // WARNING: DO NOT DELETE THE FOLLOWING LINE OF TEXT if( isset($_SESSION['username']) && isset($_SESSION['sid'])) { // You are free to edit the following code to suit your requirements include_once("../../data/server.php"); include_once("../../lib/userdata.php"); // THIS BIT WORKS AND DISPLAYS THE USERNAME $data = mysql_query("SELECT * FROM members") or die(mysql_error()); while($info = mysql_fetch_array( $data )) { include("../../lib/userinfo.php"); //////////////////////////////////////////// WARNING: YOU SHOULD NOT EDIT ANYTHING ABOVE THIS LINE //////////////////////////////////////////////////////// ECHO <<<PAGE <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>$siteName</title> <link rel="stylesheet" href="../../userstylesheet.css" type="text/css" /> </head> <div id="page"> <img alt="" src="../../images/leftCurve" height="6" width="6" id="left" /> <img alt="" src="../../images/rightCurve.gif" height="6" width="6" id="right" /> <div id="pageName"> <h1>$siteName</h1> </div> <div id="pageNav"> <div id="sectionLinks"> <a href="profile.php?username=$username">My Profile</a> <a href="modify.php?username=$username">Personal Details</a> <a href="message.php?username=$username">Messages</a> <a href="../../logout.php?username=$username">Logout</a></div> </div> <div id="content"> <div class="feature"> <h2>Welcome $username </h2> <p>This is the demonstration home.html template. You are free to edit this or any of the other templates to suit your own needs. </p> <p>This is the first page your member will see once they have logged in. </p> <p>If you look at the code for this page, you will see that all HTML code is placed between the ***PAGE and PAGE; tags. Please note that the three * should be replaced with the < character. This format must be kept to ensure that the user variables work. Changing this format may result in errors being returned.</p> <p>You may call member information using the $ tag and the variable name eg $ firstname without the space, will show the members first name, such as $firstname</p> <p>For any information please visit our site http://www.membersitemaker.co.uk. User guides will be added shortly and the forum will soon be full of help. </p> </div> </div> <div id="information"> <a href="#">About Us</a> | <a href="#">Site Map</a> | <a href="#">Privacy Policy</a> | <a href="#">Contact Us</a> | ©2011 $siteName </div> </div> </body> </html> PAGE; } //////////////////////////////////////// WARNING: DO NOT DELETE ANYTHING BELOW THIS LINE ////////////////////////////////////////////////////////// } else { // This will redirect the user to the login page if the session variables do not exist header( "Location: ../../../login.html" ); } ?> And here is the code for userdata.php Code: [Select] <?php // Decode sitename function decode_variable(&$siteName) { $siteName = urldecode($siteName); $siteName = str_replace('%20',' ',$siteName); return $siteName; } decode_variable($siteName); // Connnect to MySQL database include_once("../../data/mysql.php"); $mysqlPassword = (base64_decode($mysqlpword)); $db = mysql_connect("$localhost", "$mysqlusername", "$mysqlPassword") or die ("Error connecting to database"); mysql_select_db("$dbname", $db) or die ("An error occured when connecting to database"); // Carry out MySQL query ?> and userinfo.php Code: [Select] <?php $username = $info['username']; $firstname = $info['firstname']; $lastname = $info['lastname']; $address = $info['address']; $town = $info['town']; $county = $info['county']; $postcode = $info['postcode']; $email = $info['email']; $birth_year = $info['birth_year']; $country = $info['country']; $telephone_number = $info['telephone_number']; $mobile_number = $info['mobile_number']; $nickname = $info['nickname']; As always, your help is much appreciated Paul Incorrect login attempt 1 \/ Incorrect login attempt 2 \/ Incorrect login attempt 3 -->> ?forgot your login details? What's the most effecient way of achieving this? Is it to: 1. create a session for the user who hasn't logged in 2. the user login fails once, session['fail']=1 3. the user login fails twice, session['fail']=2 4. the user login fails for a third time pushing the session['fail'] count to three: this triggers an 'if' on the index.php prompting the user to retrieve their details through the "forgot login details system" However if the session['fail'] count never reaches 3 then this temp session is destroyed and the proper one created allowing the user into the site?? As usual any pointers into the correct direction here would be very much appreciated (and i try to repay by answering other peoples questions [where i can ]) hi i had database with field of name,title,post,content i want to fetch the post and content for a specific user from giving name of that user by form help me to get that ps just give me idea to how to do that/ Code: [Select] <form id="form1" name="form1" method="post" action="view.php"> <label>Name <input type="text" name="textfield" /> </label> <p> <label> <input type="submit" name="Submit" value="Submit" /> </label> </p> </form> I am ok with codeing but NOT half-as good as some of your GUYS here. I'm trying to write a simple sql query based on returning certain values from a table for that user. I would like to just beable to include this as a page user.info.php and render it on any page as include to return need values . See My CODE. PLESE HELP ME. Code: [Select] <?php @session_start(); $user = $_SESSION['username']; // Set cookie $userid = JRequest::getVar('userid'); $data = new stdClass(); $model =& $this->getModel('profile'); $my = CFactory::getUser(); // Test if userid is 0, check if the user is viewing its own profile. $db =& JFactory::getDBO(); $user =& JFactory::getUser(); $userId = $user->id; // Return with empty data if($userId == null || $userId == '') { //return false; } $user =& JFactory::getUser($userId); if($user->id == null){ //return false; } $id = & JFactory::getUser($userId); $query = 'SELECT user_id, id, format_id, year, name FROM #__muscol_albums WHERE user_id = ' . $id; //$query = 'SELECT user_id FROM #__muscol_albums WHERE id = ' . $album_id ; $result = mysql_query($query) or die('Error, No Album Search failed'); list($name, $user_id, $id, $year) = mysql_fetch_array($result); echo $id; echo $user_id; echo $year; // preform id return check and redirecto to correct url if ($user->get('id') == 0 || $userid == 0 || $userid <> $user->get('id')){ $url=JURI::root().'index.php?'.$component.'&id='.$id.'&tmpl=component&print=1'; } else { $url=JURI::root().'index.php?option=com_community&view=profile&id=1&tmpl=component&print=1'; //redirect is a function } ?> I would appreciate your assistance, there are tons of login scripts and they work just fine. However I need my operators to login and then list their activities for the other operators who are logged in to see and if desired send their clients on the desired activity. I have the login working like a charm and the activities are listed just beautifully. How do I combine the two tables in the MySQL with PHP so the operator Logged in can only make changes to his listing but see the others. FIRST THE ONE script the member logges in here to the one table in MSQL: <?php session_start(); require_once('config.php'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } $login = clean($_POST['login']); $password = clean($_POST['password']); if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'"; $result=mysql_query($qry); if($result) { if(mysql_num_rows($result) == 1) { session_regenerate_id(); $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['member_id']; $_SESSION['SESS_FIRST_NAME'] = $member['firstname']; $_SESSION['SESS_LAST_NAME'] = $member['lastname']; session_write_close(); header("location: member-index.php"); exit(); }else { header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?> ................................................. ................................ Now I need the person who logged in to the table above to be able to make multiple entries to the table below <? $ID=$_POST['ID']; $title=$_POST['title']; $cost=$_POST['cost']; $activity=$_POST['activity']; $ayear=$_POST['aday']; $aday=$_POST['ayear']; $seats=$_POST['special']; $special=$_POST['seats']; mysql_connect("xxxxxx", "xxx350234427", "========") or die(mysql_error()); mysql_select_db("xxxx") or die(mysql_error()); mysql_query("INSERT INTO `activity` VALUES ('ID','$title', '$cost','$activity', '$aday', '$ayear', '$special', '$seats')"); Print "Your information has been successfully added to the database!" ?> Click <a href="member-profile.php">HERE</a> to return to the main menu <?php ?> I'm trying to display relevant user details when click after user details button.it is css pop up window.i want to know how to catch relevant user when click the button.my primary key is email. hi, i'm new in php/mysql. i'm stored student marks values in following format in mysql db table. id student_code Tamil English Maths Science Social 1 1 100 75 78 88 95 2 2 85 90 88 80 100 But i want to search and display the specific student marks in following format. id:1 student_code:1 Tamil:100 English:75 Maths:78 Science:88 Social:92 Total:? Avg:? please give correct code for this format. friends my url is like this Category=Love&ID=1 now in url if i do something like this Category4637=Hate19203&ID=1 the details are still showing, how can i fix this? if someone type Capegory instead of Category and a wrong category name which do not exist with that id in db send them to index.php? Hi guys, Can anyone assist me. I am trying to create a login for admin and user (if user not a member click register link) below is my code: But whenever I enter the value as: Username: admin Password:123 - I got an error message "That user does not exist!" Any suggestion and help would be appreciated. Thanks. login.php <?php //Assigned varibale $error_msg as empty //$error_msg = ""; session_start(); $error_msg = ""; if (isset($_POST['submit'])) { if ($a_username = "admin" && $a_password = "123") { //Define $_POST from form text feilds $username = $_POST['username']; $password = $_POST['password']; //Add some stripslashes $username = stripslashes($username); $password = stripslashes($password); //Check if usernmae and password is good, if it is it will start session if ($username == $a_username && $password == $a_password) { session_start(); $_SESSION['session_logged'] = 'true'; $_SESSION['session_username'] = $username; //Redirect to admin page header("Location: admin_area.php"); } } $username = (isset($_POST['username'])) ? $_POST['username'] : ''; $password = (isset($_POST['password'])) ? $_POST['password'] : ''; if($username && $password) { $connect = mysql_connect("localhost", "root", "") or die ("Couldn't connect!"); mysql_select_db("friendsdb") or die ("Couldn't find the DB"); $query = mysql_query ("SELECT * FROM `user` WHERE username = '$username'"); $numrows = mysql_num_rows($query); if ($numrows != 0){ while ($row = mysql_fetch_array($query)) { $dbusername = $row['username']; $dbpassword = $row['password']; } //Check to see if they are match! if ($username == $dbusername && md5($password) == $dbpassword) { header ("Location: user_area.php"); $_SESSION['username'] = $username; } else $error_msg = "Incorrect password!"; //code of login }else $error_msg = "That user does not exist!"; //echo $numrows; } else $error_msg = "Please enter a username and password!"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Login Page</title> </head> <body> <br /> <?php require "header.php"; ?><br /> <div align="center"> <table width="200" border="1"> <?php // If $error_msg not equal to emtpy then display error message if($error_msg!="") echo "<div id=\"error_message\"style=\"color:red; \">$error_msg</div><br />";?> <form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post"> <!--form action="login_a.php" method="post"--> Username: <input type="text" name="username" /><br /><br /> Password: <input type="password" name="password" /><br /><br /> <input type="submit" name = "submit" value="Log in" /> </form> <p> </p> Register a <a href="register.php">New User</a> </table> </div> </body> </html> hi, i have made a website where people resgister their details of them and products. they have to enter the following details in form Name of company name of the product company address email id password mobile number contact and brief details about their company
user can then login with email id and pwd. now after login ..user will get a page where he can upload the photos of products images and their price, so now my question is that when he finishes uploading (|by clicking on upload button) the product images and price text box ..then on final uploaded webspage it should show all other things which he registerd before (company name , mobile number etc) along with images and price...hence the main question that user does not need to enter mobile and address while uploading images and filling proce ..but on the final page it should show mobile and address along with price and images..as user is not going to enter mobile and address again and again as he will have multiple products to upload.
Hallo everybody,
i have the following code.
but i get allways this error while the user exist in the database.
User not found!
what do i do wrong?
thank you very much for your help
Rafal
<html>
<head>
<?php
$connection = mysql_connect("db.xyz.com", "username", "password")
or die ("connection fehler");
mysql_select_db("db0123456789")
or die ("database fehler");
$email = $_POST["inp_email"];
$pwd = $_POST["inp_pwd"];
if($email && $pwd)
{
$chkuser = mysql_query("SELECT email FROM gbook WHERE email = '($email)' ");
$chkuserare = mysql_num_rows($chkuser);
echo $email;
echo $pwd;
if ($chkuserare !=0)
{
$chkpwd = mysql_query("SELECT pwd FROM gbook WHERE email = '($email)' ");
$pwddb = mysql_fetch_assoc($chkpwd);
if ($pwd != $pwddb["pwd"])
{
echo "password is wrong!";
}
else
{
echo "login successed";
}
}
else
{
echo "User not found!";
}
}
else
{
echo "Pleas enter your email and password!";
}
mysql_close($connection);
?>
</head>
<body>
<form action="login.php" method="post">
Email <input type="text" name="inp_email"><br>
Password <input type="text" name="inp_pwd"><br>
<input type="submit" name="submit" value="login">
</form>
</body>
</html>
Edited by rafal, 21 September 2014 - 04:33 PM. Hallo everybody,
the user is in the table, but i get error (user not found!).
thank you very much for your help
Rafal
<!DOCTYPE html>
<html>
<head>
<title>index</title>
<meta http-EQUIV="CONTENT-LANGUAGE" content="en">
<?php
SESSION_START();
include("abc.php");
$link2 = mysqli_connect("$hoster", "$nameuser", "$password", "$basedata")
or die ("connection error" . mysqli_error($link2));
$email = $_POST["inp_email"];
$pwd = $_POST["inp_pwd"];
if($email && $pwd)
{
$chkuser = mysqli_query("SELECT email FROM $table2 WHERE email = '$email' ");
$chkuserare = mysqli_num_rows($chkuser);
if ($chkuserare !=0)
{
$chkpwd = mysqli_query("SELECT pwd FROM $table2 WHERE email = '$email'");
$pwddb = mysqli_fetch_assoc($chkpwd);
if (md5($pwd) != $pwddb["pwd"])
{
echo "Password is wrong!";
}
else
{
$_SESSION['username'] = $email;
header ('Location:list.php');
}
}
else
{
echo "user not found!";
}
}
else
{
echo "enter your Email and Password!";
}
mysqli_close($link2);
?>
</head>
<body style="font-family: arial;margin: 10; padding: 0" bgcolor="silver">
<font color="black">
<br>
<form action="index.php" method="post">
<b>Login</b><br><br>
<table width="100%">
<tr><td>
Email:<br><input type="text" name="inp_email" style="width:98%; padding: 4px;"><br>
Password:<br><input type="password" name="inp_pwd" style="width:98%; padding: 4px;"><br>
<br>
<input type="submit" name="submit" value="Login" style="width:100%; padding: 4px;">
</td></tr>
</table>
</form>
</font>
</body>
</html>
Hi, I have successfully implemented a master details page with the results aligned in columns linking to a details page. I wish to maintain the recordID passed from the master details page and make the dynamic text, which reads Shade A tree that is capable of..... in the attached screen shot a link to another details page referencing the same recordID. The detailspage2.php would look the same as the screenshot except the Shade text and description below will be highlighted, which I can do, there will be a new image and a new image description. All other dynmaic elements on the page will remain the same. I tried to simply save as my detailspage.php to detailspage2.php and create a link to detailspage2.php. It linked to detailspage2.php but none of the record info showed up in their respective table cells. I have all the names desc's, images, etc setup in a table in my database. Please let me know what code and other info you need to help me out with this procedure. Thanks. |
|||||||