I downloaded a php login source and now it works and it even logs in. when it logs in it shows your name with
<?php echo $_SESSION['username']; ?>
but i also want to display there balance by putting
<?php echo $_SESSION['balance']; ?>
But it doesnt display the balance?
https://gyazo.com/7cd0a7888ac976590391888925d0c18f
I added the balance table to the existing tables.
I really dont know what to do! :(
Please Insert;
<?php session_start(); ?>
In every page where you want to access $_SESSION global array.
in your case;
<?php session_start(); ?>
$_SESSION['balance'] = 1000;
echo $_SESSION['balance']; // will output 1000.
suppose this page was set-session-value.php and you want to get this $_SESSION['balance'] value in get-session-value.php do as follow in get-session-value.php file;
<?php session_start(); ?>
echo $_SESSION['balance'];
In your login.php add this line after setting session for username
$_SESSION["balance"]=$balance;
And in your home, it is always advised to check if session exists and then print it. So,
if(isset($_SESSION["balance"])){echo $_SESSION["balance"]; }
start session before use
<?php
session_start();//start session
$_SESSION['username']='abc';//Set value session
$_SESSION['balance']=10;
echo $_SESSION['username']; //Use value of session
echo $_SESSION['balance'];
?>
please use this process for session use
When you logged into your application.After that you should want to add following code into top of the page.
session_start();
$_SESSION['username'] = $userNameValue;//set user name
echo $_SESSION['username'];
And then you should want to fetch data related to who logged user from your user table by using query.
Example
$query = "SELECT * YOURTABLENAME WHERE username='$_SESSION['username']'";
mysql_query($query, $connection) or die(mysql_error());
$data = mysql_fetch_array($query);
Then you can assign to $_SESSION['balance'] for value like this;
$_SESSION['balance'] = $data['balance'];
echo $_SESSION['balance'];
Related
I want to try to make a session but always session Empty.Use this code rate this product this link send an email open email and click link Active than code working session empty please help me...
<?php
session_start();
include("myhomeportal/setting/config.php");
$conform = $_GET['conform'];
$query = mysqli_query($conn, "SELECT * FROM item_users where com_code='$conform'");
$row = mysqli_fetch_array($query);
if ($row) {
// now update `com_code`
$sql = "UPDATE item_users SET com_code='active', user_type='user' WHERE com_code='$conform'";
$result = mysqli_query($conn, $sql) or die(mysqli_error());
$inventory_id = $row['inventory_id'];
$active = $row['com_code'];
$_SESSION['sess_active'] = $active;
header("Location: category.php?inventory_id=$inventory_id");
} else {
// confirm code not found, show error
}
?>
Try to debug first.
Echo this $row['com_code']; then $_SESSION['sess_active'].
If they print something then go ahead.
There ir an error in your test page.
"sir other page is not working session just simple test code – Pankaj"
Instead you are only testing, you must start the session always with session_start() in every .php you want to use session.
Solving testing page.
"<h1>welcome <?php session_start(); echo $_SESSION['sess_active'];?> </h1> – Panka"
Note from w3schools:
The session_start() function must be the very first thing in your document. Before any HTML tags.
you are writting some html code before start the session, you should do that way:
<? php session_start(); ?>
<h1>welcome <?= $_SESSION['sess_active']; ?> </h1>
You can read more about this https://www.w3schools.com/php/php_sessions.asp
I am using a simple PHP/MYSQL login system using sessions and I need to print the username for the person currently logged in to the session.
The current code I am using to do this is:
<?php
session_start();
include_once 'dbconnect.php';
$res=mysql_query("SELECT username FROM users WHERE user_id=".$_SESSION['user']);
$userRow=mysql_fetch_row($res);
print_r($userRow)
?>
But the output of this is:
Array ( [0] => USERNAME )
While I would like it to be:
USERNAME
How would I achieve this, or is this possible?
Thank you in advance!
The value is in array. you have to access the index to get the value like this,
echo $userRow['username'];
I would suggest to store the username in session and access it directly instead of accessing from database everytime.
$_SESSION['username'] = "name here";
echo $_SESSION['username'];
EDIT
echo $userRow[0];
this should help you.
echo $userRow['username'];
Try this out.
I know I can't use two session start codes in a same php page but for the sake of updating user account, I need the below code and I need to use session_start twice. One, to check if the user is not logged in, then redirect them and banned them from seeing the update info page and also the other session start has to be there so that my session variables could be set automatically in the update info page if the user is logged in.
anyways, I am getting this error can you guys please show me a work around way? if there's any?
thanks.
Notice: A session had already been started - ignoring session_start() in ....
<?php session_start();
if(isset($_SESSION['userid'])) {
} else {
header('Location: login.php');
}
?>
<?php
$user = $_SESSION['userid'];
$myquery = "SELECT * FROM our_users WHERE `userid`='$user'";
$result = mysqli_query($conn, $thequery);
$row = mysqli_fetch_array($result, MYSQLI_BOTH);
session_start(); /* Basically this right here gets ignored. */
$_SESSION["user_first_name"] = $row['fn'];
$_SESSION["user_last_name"] = $row['ln'];
$_SESSION["user_email"] = $row['em'];
$_SESSION["user_password"] = $row['pw'];
?>
i am making my own php game. So far i have made almost everything. Now to finish it, i need to get id from user who is logged in. I'm not so familiar with the functions and sessions. Please help.
This is what i made so far:
In my index page people login. then they are redirected to this.
So $_POST['username'] is where user type his user name in index.
<?php
$username = $_POST['username'];
include("Files/config.php");
$connect = #mysql_connect(DB_SERVER, DB_USER, DB_PASSWORD);
if($connect) {
if(mysql_select_db(DB_NAME)) {
$sql = mysql_query("SELECT * FROM users WHERE `username`='$username'") or die(mysql_error());
$gatherinfo = mysql_fetch_array($sql);
global $getid;
$getid = $gatherinfo['id'];
echo $getid;
function getuid() {
$_SESSION['getuid'] = $getid;
echo $getid;
}
}
}
else{ echo "Can not connect";}
?>
I searched other scripts for this, i found on one it says just $session->uid and it shows his id from mysql.
In mysql database i have table users with info about them
Id, username, password (password is hashed), email,...
Please help me if you can :D
At the beginning of index file (where your user logging in) start named session (be careful to avoid echo or print any values before session_start:
<?php //index.php
session_name('SAMPLESESSION');
session_start();
then when you will get the logged User ID, write this value to the session variable, like this:
.....
$_SESSION['uid'] = $getid;
.....
in the script you was redirected by your index file start session with the same name and get your user ID:
<?php //redirectedfromindex.php
session_name('SAMPLESESSION');
session_start();
echo $_SESSION['uid'];
....
If I right understand you, these that you need.
I need the login details in another page for retrieving the data from the database. Basically, I need to display the editable form with the details of the user logged in. I tried session_register() for storing the username in login.php page. But for some reason I am not able to display the username using $_SESSION[] in my edit.php page. I am doing this after the function session_start() as well.
I am new to php, so don't know whether I misunderstood session! Or is there any other way to pass the login details?
Thanks in advance
My code:
**Login.php**
<?php
$userName = $_POST['username'];
$password = $_POST['password'];
//Connect to the database
//query the database
if($rows==1)
{
session_start();
$_SESSION['user']=$userName;
header("location:edit_user.php");
}
else
{
echo 'Data Does Not Match <br /> Re-Enter UserName and Password';
}
?>
**In edit.php**
<?php
session_start();
if(!isset($_SESSION['user']))
{
header("location:login_form.php");
}
else
{
echo $_SESSION['user'];
}
?>
First of all make sure that you place session_start() at the very beginning of any script you use it in. There can be no output to the browser before you call session_start() and that includes spaces or new-lines before the opening <?php tag.
So:
<?php
session_start();
...
Second, make sure you terminate your script after a redirect, for example:
header("location:edit_user.php");
exit();
That makes sure that no code after the redirect gets executed, so sessions won't get unset or session variables changed by accident.
session_register() is a deprecated function. Just use $_SESSION["bar"] = "foo" to store something.
for future references, please post parts of your code when you are asking questions. It helps everyone to give you an answer in more specific cases.
<?php
session_start();
if(!isset($_SESSION['Foo']))
{
$_SESSION['Foo'] = "Bar";
}
?>
Source : http://php.net/manual/en/features.sessions.php
you can retrive data from the database like this
//start connection
$connect = mysql_connect(DB_SERVER,DB_USER,DB_PASSWORD);
if(!$connect){
die("Database connection Error".mysql_error());
}
//select database
$db = mysql_select_db(DB_NAME);
if(!$db){
die("Database selection Error".mysql_error());
}
//get data
$login = mysql_query("SELECT * FROM TABLENAME where user_id={$_SESSION['user_id']}");
$login_data = mysql_fetch_array($login);
now $login_data array has the user details which you can point to form text field values..
the $_session['user_id']=$login_data['user_id'] value has to be assigned earlier which stays in the $_SESSION global variable through out the session