I really can't figure out why this isn't working. I want to show the content stored in a php session variable as the value in an input field.
<?php
if($_SESSION['id'])
echo ' <input type="text" value="'.$_SESSION['friday'].'"></input>'; ?>
Thanks in advance!
Update:
At my login page, index.php, I'm starting the sessions and assigning them values.
$row = mysql_fetch_assoc(mysql_query("
SELECT *
FROM knine_settings_login, knine_school_db, knine_class_db
WHERE usr='{$_POST['username']}' AND pass='".md5($_POST['password'])."' AND knine_settings_login.School = knine_school_db.School AND knine_settings_login.ClassID = knine_class_db.ClassID AND knine_class_db.week = 0
"));
if($row['usr'])
{
// If everything is OK login
$_SESSION['usr']=$row['usr'];
$_SESSION['id'] = $row['id'];
$_SESSION['homework'] = $row['homework'];
$_SESSION['pe'] = $row['pe'];
$_SESSION['pm'] = $row['pm'];
$_SESSION['pc'] = $row['pc'];
$_SESSION['School'] = $row['School'];
$_SESSION['Class'] = $row['Class'];
$_SESSION['level'] = $row['level'];
$_SESSION['rememberMe'] = $_POST['rememberMe'];
$_SESSION['schoolurl'] = $row['schoolurl'];
$_SESSION['monday']=$row['monday'];
$_SESSION['tuesday'] = $row['tuesday'];
$_SESSION['wednesday'] = $row['wednesday'];
$_SESSION['thursday'] = $row['thursday'];
$_SESSION['friday'] = $row['friday'];
$_SESSION['peone'] = $row['peone'];
$_SESSION['petwo'] = $row['petwo'];
Session Needs to be start before using it.Try It.
<?php
session_start();
if(isset($_SESSION['id']))
echo '<input type="text" value="'.$_SESSION['friday'].'"></input>';
?>
Make sure you have called session_start() before echoing out any $_SESSION data.
<?php
session_start();
if ($_SESSION['id']) {
echo ' <input type="text" value="'.$_SESSION['friday'].'"></input>';
}
?>
<?php
// Start the session on every page at the top first
session_start();
?>
<!DOCTYPE html>
<html>
<body>
<?php
// Set session variables
$_SESSION["animals"] = "dog";
$_SESSION["cars"] = "ford";
echo ' <input type="text" value="'.$_SESSION['animals'].'"></input>';
?>
Related
I have been following a lesson on how to make an admin page. I got all the information out of my database to a table on the page. I have an update button and when I change the information and press the button I receive this error: Warning: undefined array key "WebID" in ..\Update.php on line 3
From my search online everyone is trying to change the code so that if array key does not exist: return null. I tried that and the error does not appear no more, but the table does not change.
Any thoughts?
This is the code:
<?php
require_once("DB/DB.php");
$SearchQueryParameter = $_GET["WebID"];
if (isset($_POST["Update"])) {
$Ename = $_POST["Ename"];
$Eid = $_POST["Eid"];
$Erank = $_POST["Erank"];
$Eemail = $_POST["Eemail"];
$Edate = $_POST["Edate"];
$Epassword = $_POST["Epassword"];
$Specialisms = $_POST["Specialisms"];
global $ConnectingDB;
$sql ="UPDATE emp_data SET Ename='$Ename', Eid='$Eid', Erank='$Erank', Eemail='$Eemail', Edate='$Edate', Epassword='$Epassword',
Specialisms='$Specialisms' WHERE WebID='$SearchQueryParameter'";
$Execute = $ConnectingDB->query($sql);
if ($Execute) {
echo '<script>window.open("adminpage.php?WebID=Recored Updated","_self")</script>';
}
}
?>
<?php
<?php
global $ConnectingDB;
$sql = "SELECT * FROM emp_data WHERE WebID='$SearchQueryParameter'";
$stmt = $ConnectingDB->query($sql);
while ($DataRows = $stmt->fetch()) {
$WebID = $DataRows["WebID"];
$Ename = $DataRows["Ename"];
$Eid = $DataRows["Eid"];
$Erank = $DataRows["Erank"];
$Eemail = $DataRows["Eemail"];
$Edate = $DataRows["Edate"];
$Epassword = $DataRows["Epassword"];
$Specialisms = $DataRows["Specialisms"];
}
?>
Html file used to update:
<form id="UpdateForm" method="post" action="Update.php?WebID<?php echo $SearchQueryParameter; ?>">
<div class="form-group">
<button type="submit" name="Update" class="form-control-submit-button">Update</button>
</div>
you have to write the form action like this.. you missed the = sign
action="Update.php?WebID=<?php echo $SearchQueryParameter; ?>"
<form id="UpdateForm" method="post" action="Update.php?WebID=<?php echo $SearchQueryParameter; ?>">
You missed the = sign, in the url
I am trying to edit form data by displaying the previous saved data on the form and then update it. It shows the data on the form which is saved in database but when i enter the new data it does not get the id of the row. I echo the update query, it shows the changed values but it shows id equals to empty. Here is my code for edit record and update; Edit record is working but update isn't:
<?php
include('connection.php');
$id = '';
if( isset( $_GET['id'])) {
$id = $_GET['id'];
}
$udfname = mysql_real_escape_string($_POST["udfname"]);
$udlname = mysql_real_escape_string($_POST["udlname"]);
$udpwd = mysql_real_escape_string($_POST["udpwd"]);
$udeml = mysql_real_escape_string($_POST["udeml"]);
$udnum = mysql_real_escape_string($_POST["udnum"]);
$query="UPDATE form
SET fname = '$udfname', lname = '$udlname', pwd = '$udpwd', eml = '$udeml', num = '$udnum'
WHERE id='$id'";
$res= mysql_query($query);
if($res){
echo "<p> Record Updated<p>";
}else{
echo "Problem updating record. MY SQL Error: " . mysql_error();
}
?>
Form for editing record:
<?php
include('connection.php');
$id = (int)$_GET['id'];
$query = mysql_query("SELECT * FROM form WHERE id = '$id'") or die(mysql_error());
while($row = mysql_fetch_array($query)) {
echo "";
$fname = $row['fname'];
$lname = $row['lname'];
$pwd = $row['pwd'];
$eml = $row['eml'];
$num = $row['num'];
}
?>
<html>
<head>
<title>Edit</title>
<script>
'
'
Jquery code here
'
'
</script>
</head>
<body>
<form action="update.php" method="post">
<input type="hidden" name="ID" value="<?=$id;?>">
First Name: <input type="text" name="udfname" value="<?=$fname;?>"><br>
Last Name: <input type="text" name="udlname" value="<?=$lname?>"><br>
Password: <input type="text" name="udpwd" value="<?=$pwd?>"><br>
Email: <input type="text" name="udeml" value="<?=$eml?>"><br>
Contact Number: <input type="text" name="udnum" value="<?=$num?>"><br>
<input type="Submit">
</form>
</body>
</html>
At update time your form is submitted using POST request. So you need to get ID using POST method. So to get ID of hidden field change your code as below:
$id = '';
if( isset( $_POST['ID'])) {
$id = $_POST['ID'];
}
Please try below code
if( isset( $_POST['id']) && $_POST['id']!=null) {
$id = $_POST['id'];
}
Dear i think the problem with your method you are sending the data using post method and its very simple instead of this code
if( isset( $_GET['id'])) {
$id = $_GET['id'];
}
write
if( isset( $_POST['id'])) {
$id = $_POST['id'];
}
and one more thing that is you are using the mysql deprecated function for database kindly use the pdo for this or new mysqli functions.
I am having a bit of an issue passing array variables that are fetched from the database and stored in variables and passing them into session variables to use all over the site.
Code:
$rid=$_SESSION['SESS_MEMBER_ID'] ;
$recquery = 'SELECT * FROM clinic_receptionist,clinic_table where clinic_receptionist.recep_clinic_id = clinic_table.ID AND recep_id ='.$rid;
$recresult = mysqli_query($conn,$recquery);
if(mysqli_num_rows($recresult)>0)
{
while($row = mysqli_fetch_assoc($recresult))
{
$cid = $row['clinic_id'];
$cname = $row['clinic_name'];
$name = $row['recep_full_name'];
$phone = $row['recep_mobile'];
$email = $row['recep_email'];
$address = $row['recep_address'];
$gender = explode(",",$row['recep_gender']);
$dob = $row['recep_dob'];
$doj = $row['recep_doj'];
}
}
?>
The above code is where the select query is happening. I wanted to know how do I assign $cid and $cname to a $_SESSION variable in different pages. Any help would be much appreciated. Thanks
CODE:
<?php session_start();
include('header.php');
include('menu.php');
include('config.php');
$appquery = 'SELECT * FROM appointment_table,clinic_table where clinic_table.ID = appointment_table.clinic_id';
$appresult = mysqli_query($conn,$appquery);
if(mysqli_num_rows($appresult)>0)
{
while($row = mysqli_fetch_array($appresult))
{
$_SESSION['cid'] = $row['ID'];
$_SESSION['cname'] = $row['clinic_name'];
$ptdate = $row['date'];
$pttime = $row['time'];
$ptname = $row['pt_name'];
$ptphone = $row['pt_ph_num'];
$ptemail = $row['pt_email_id'];
}
}
?>
The above code is the place from where I want to receive the id and clinic name.
<?php
// Start the session
session_start();
//Assigning the values to $_SESSION
$_SESSION['cid'] = $row['clinic_id'];
$_SESSION['cname'] = $row['clinic_name'];;
?>
session_start
You need to call session_start() on each page to initialize the session, of course; after that you can just
$_SESSION['cid'] = $cid;
$_SESSION['cname'] = $cname;
to store those values in $_SESSION.
By the way, someone's going to tell you to use a prepared statement for that query, so it might as well be me.
Perhaps an illustration is in order.
page1.php:
<?php
session_start(); // find session, or create new one
include('header.php');
include('menu.php');
include('config.php');
$appquery = 'SELECT * FROM appointment_table,clinic_table where clinic_table.ID = appointment_table.clinic_id';
$appresult = $conn->query($appquery);
if(($rowcount = $appresult->num_rows()) > 0)
{
// create arrays to pass data
$cids = array($rowcount);
$cnames = array($rowcount);
$i = 0;
while($row = $appresult->fetch_array())
{
// add values to arrays
$cids[$i] = $row['ID'];
$cnames[$i] = $row['clinic_name'];
$i++;
}
// add arrays to session
$_SESSION['cids'] = $cids;
$_SESSION['cnames'] = $cnames;
}
page2.php:
<?php
session_start(); // find session, or create new one
include('header.php');
include('menu.php');
include('config.php');
// retrieve arrays from session
$cnames = $_SESSION['cnames'];
$cids = $_SESSION['cids'];
// ... and print them
$count = count($cids);
for ($i = 0; $i < $count; $i++)
{
echo "cid = " . $cid[$i] . "; cname = " . $cname[$i] . "<br>";
}
page1.php performs the query and stores cid and cname values in arrays which are then stored in $_SESSION. page2.php retrieves the arrays from $_SESSION and echoes them out. In production, of course, it would be considered more efficient to run the query on the page where it is displayed, and only pass key values between pages.
I've also used object syntax instead of procedural syntax for mysqli, just to show what it looks like. As usual, I leave error handling as an exercise for the reader.
I have two sessions here and use the $POST ['username'] in both related queries, but on the $_SESSION['address'] it still is taking my old $POST ['username'] so therefore it is not refreshed.
<?php
include_once'config/connect.php'; //database connection
if($_POST['submit']) {
$retrieve = mysql_query("SELECT username,password FROM iangadot_user
WHERE username='".$_POST['username']."' AND password='".md5($_POST['password'])."' ");
$address = mysql_query("SELECT location FROM venue
WHERE vid in (select vid from user_venue where id in (select id from iangadot_user where username='".$_POST['username']."' )) ");
if(mysql_num_rows($retrieve)) {
//redirect to your client page...
//SET YOUR session
$_SESSION['username_profile'] = $_POST['username'];
header("Location: clientpag.php");
} else {
echo "<script language=javascript>alert('Wrong username or password')</script>";
//header("Location: ./?errorlogged=err");
}
if(mysql_num_rows($address)) {
$row = mysql_fetch_assoc($address);
$_SESSION['address'] = $row['location'];
header("Location: clientpag.php");
}
}
?>
Here is my Html code:
<?php session_start(); ?>
<!DOCTYPE html>
<html>
<body>
<h1 id = "title" name= venue> Welcome <?php echo $_SESSION['username_profile'] ?> </h1>
<input class = "field" placeholder= "<?php echo $_SESSION['address'] ?>" type="text" >
<input class = "button" type = "submit" Value = "Save"><br>
......... code continued......
You need to start the session in every file where you set or use $_SESSION
at the beginning of the script: session_start();
In your php file add session start:
<?php
include_once'config/connect.php'; //database connection
session_start();
if it executes
$_SESSION['address'] = $row['location'];
line correctly then you will have $_SESSION['address'] it in your next page.
REMEMBER:
You need start session by session_start() in every files where you want to use session.
See details: http://www.w3schools.com/php/php_sessions.asp
I'm trying to create a system that when i submit the form, after the page refresh it should show the new values that i get from the database. The values work well when they go into the databse but they dont show after submited, only when i refresh again. Thanks for helping
<?php
include("connect.php");
$query = "SELECT * FROM `laliga`";
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result)){
$id = $row['id'];
$home = $row['home'];
$away = $row['away'];
$win = $row['win'];
$draw = $row['draw'];
$lose = $row['lose'];
}
echo "<h2>La Liga</h2>";
echo $home, " - ", $away;
if (isset($_POST) && $_POST['laliga'] == 1){
$select = mysql_real_escape_string($_POST['laliga']);
$select = $win + $select;
mysql_query("UPDATE laliga SET win='$select'");
}else if (isset($_POST) && $_POST['laliga'] == 'X'){
$select = mysql_real_escape_string($_POST['laliga']);
$select = '1';
$select = $draw + $select;
mysql_query("UPDATE laliga SET draw='$select'");
}else if (isset($_POST) && $_POST['laliga'] == 2){
$select = mysql_real_escape_string($_POST['laliga']);
$select = '1';
$select = $lose + $select;
mysql_query("UPDATE laliga SET lose='$select'");
}
header('Location: ../laliga.php');
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="radio" name="laliga" value="1">1
<input type="radio" name="laliga" value="X">X
<input type="radio" name="laliga" value="2">2
<input type="submit" name="submit" value="Submit"/>
</form>
<br/>
<?php
echo $home, " -> ", $win;
echo "<br/>Barazim -> ", $draw,"<br/>";
echo $away, " -> ", $lose;
?>
You should handle all of the post data at the top of the PHP file, whilst the header function will solve your problem it's a silly and inefficient way of approaching it. by handling the post data and updating the database first, by the time you query the database the data is there! at the moment you are trying to find the data and then adding it. does this make sense?
Good luck!
Add:
header('Location: <mypage.php>');
After mysql_query.