None of the other methods work so could you please help? Here is my code:
<?php
include("db.php");
If(isset($_SESSION['name']))
{
$main = "<h2>hey</h2>";
}
else
{
$main = "<h2>welcome</h2>"
}
?>
<html>
<body>
<?php echo $main;?>
</body>
</html>
//db.php consists of
<?php
mysql_connect("localhost", "root", "");
mysql_select_db("notes");
echo mysql_error();
session_start();
?>
Whenever I run the code it always shows: welcome. No matter if I'm logged in or logged out of my website.
Your html is looks good but
It seems that you did not set session_start() at the top of your php file or you did not set session variable.
try like this:
<?php
session_start();
if(isset($_SESSION['name']))
{
$main = "<h2>hey</h2>";
}
else
{
$main = "<h2>welcome</h2>"
}
?>
<html>
<body>
<?php echo $main;?>
</body>
</html>
Related
UPDATED: I have a variable in PHP mailuid that I want to show in my HTML. It displays the error the value of mailuid is undefined on the webpage. How can I show the value of height to html page?
index.php
<?php
require "header.php";
?>
<main>
<link rel="stylesheet" type="text/css" href="styl.css">
<div class="wrapper-main">
<section class="section-default">
<h2><?php echo "$mailuid" ?></h2>
<?php
?>
</section>
</div>
</main>
<?php
require "footer.php";
?>
loginbackend.php
<?php
if(isset($_POST['login-submit'])) {
require 'db.php';
$mailuid = $_POST['mailuid'];
$password = $_POST['pwd'];
if (empty($mailuid) || empty($password)) {
header("Location: ./index.php?error=emptyfields");
exit();
} else {
$sql = "SELECT * FROM users WHERE uidUsers=? OR emailUsers=?;";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
header("Location: ./index.php?error=sqlerror");
exit();
} else {
mysqli_stmt_bind_param($stmt, "ss", $mailuid, $mailuid);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
if ($row = mysqli_fetch_assoc($result)) {
$pwdCheck = password_verify($password, $row['pwdUsers']);
if($pwdCheck == false) {
header("Location: ./index.php?error=wrongpwd");
exit();
} else if ($pwdCheck == true) {
session_start();
$_SESSION['userId'] = $row['idUsers'];
$_SESSION['userUid'] = $row['uidUsers'];
$username = substr($mailuid, 0, strpos($mailuid, "#"));
header("Location: ./index.php?login=success".$username);
exit();
} else {
header("Location: ./index.php?error=wrongpwd");
exit();
}
} else {
header("Location: ./index.php?error=nouser");
exit();
}
}
}
} else {
header("Location: ./signup.php");
exit();
}
As per your latest comment:
To get the mailuid from the URL (GET parameters) add the following code to your index.php
<?PHP
require "header.php";
$mailuid = !empty($_GET['mailuid']) ? $_GET['mailuid'] : null;
// You can also specify the default value to be used instead of `null` if the `mailuid` is not specified in the URL.
?>
<main>
<link rel="stylesheet" type="text/css" href="styl.css">
<div class="wrapper-main">
<section class="section-default">
<h2><?php echo "$mailuid"?></h2>
</section>
</div>
</main>
<?php
require "footer.php";
?>
From PHP7 you can use
$mailuid = $_GET['mailuid'] ?? null;
instead of
$mailuid = !empty($_GET['mailuid']) ? $_GET['mailuid'] : null;
The mistake you've made:
I think you're confusing forms and file including with how post works.
Let me explain:
A form sends data to the server, which is then pushed into the $_POST global variable. You can read this data and use this data easily by echoing or dumping it.
This is what you should do:
In this case, your data value will be empty as you're not passing anything to it.
You can solve this by creating a form and passing it to your PHP file.
You can also just require your php script.
Normally you would put data.php in your action, but since you wish to use the variable before you entered the form, you have to include it first.
index.html
<?php require 'data.php'; ?>
<form method="POST" action="">
<h1>Height: <?=$height?></h1>
<input type="text" placeholder="Enter the height..." name="height">
<input type="submit" name="submit" value="Submit">
</form>
data.php
<?php
if (!empty($_POST)) {
$height = $_POST['height'];
} else {
$height = 0; //Default height
}
My apologies if i didn't get your question properly.
===========================================
Option B, if this is what you mean, is just doing this:
index.html
<body>
<div class="container">
<?php
require 'data.php'; //Get the data.php file so we can use the contents
?>
<h1><?php echo $height; ?></h1>
</div>
</body>
data.php
<?php
$height = 100; //Height variable
At that time I need two view pages in one page. So that in the session i do something like this:
<?php
session_start();
if(isset($_SESSION['login_id']) && !empty($_SESSION['login_id'])){
?>
///My HTML CODE///
<?
}
else {
?>
//Diffetent View HeRe//
<?
}
require_once('../libraries/config/configPDO.php');
?>
But now I just need to make it become one page and redirect it if they not logged in.
<?php
session_start();
if(isset($_SESSION['login_id']) && !empty($_SESSION['login_id'])){
?>
///My HTML CODE///
<?
}
else {header("Location: login.php");
?>
<?
}
require_once('../libraries/config/configPDO.php');
?>
But as you see, I put the header redirect at the bottom of the page.
How to make it simple, so that I set header location in the top of the page.
Try with -
session_start();
if(empty($_SESSION['login_id'])){
header("Location: login.php");
exit;
} else {
//HTML
}
try this,
<?php
ob_start();
session_start();
if(isset($_SESSION['login_id']) && !empty($_SESSION['login_id'])){
?>
///My HTML CODE///
<?
require_once('../libraries/config/configPDO.php');
}
else {
header("Location: login.php");
}
?>
reverse the if to
if(!isset($_SESSION['login_id']) || empty($_SESSION['login_id'])){
and swap the cases
page show2.php:
<?php
session_start();
$_SESSION["abc"]=2;
print $_SESSION["abc"];
include 'import/function.php';
?>
<html>
.............
</html>
page show1.php:
<?php
session_start();
$_SESSION["abc"]=1;
print $_SESSION["abc"];
include 'import/function.php';
?>
<html>
.............
</html>
I've bind an ajax_call to a btn in html that print $_SESSION["abc"]
session_start();
include 'import/function.php';
if(isset($_POST["data"]))
{
print $_SESSION["abc"];
}
When I go in show1.php the page print 1 and ajax_call print 1.
When I go in show2.php the page print 2 but ajax_call print 1! why?
I am new to php. Here is a code to check if the user is logged in using session and then allowing the user.
VALIDATION
<?php
session_start();
$uname = $_POST['uname'];
$pass = $_POST['pass'];
if($uname == "admin" && $pass == "admin")
{
$_SESSION['uname'] = $uname;
$_SESSION['auth'] = 1;
echo "Welcome Mr ".$uname.". You are now logged in ";
echo "<br>";
echo "<a href='TakeMeHome.html'>Click here to access the application </a>";
}
else
{
echo "Invalid username or password";
}
?>
Page
<?php
session_start();
if($_SESSION['auth'] != 1)
{
echo "You are not logged in! ";
echo "<a href = \"TakeMeHome.html\">";
echo "Access Application";
echo "</a>";
exit();
}
?>
<html>
You are now logged in
</html>
But the link tag is displaying
"; echo "Access Application"; echo ""; exit(); } ?>
along with the html data. No verification is done. I know there are many better ways to validate user is logged in or not. But i am learning sessions and hence i am using sessions.
Can you please tell me where i am going wrong?
regards.
use single quote in your echo codes like this:
<html>
<head>
</head>
<body>
<?php
echo "<a href='pageToGoTo.html' title='Page to go to' class='whatEver'>Anchor text</a>";
?>
</body>
</html>
What is told already, html should be put in the body...
I have no idea why #Andy has suggested you put your PHP in your head tags - it isn't javascript. You have 2 ways you can format your PHP and HTML, the first is to put all your PHP above your opening html tag, like so
<?php
session_start();
if($_SESSION['auth'] != 1) {
$message = 'You are not logged in! Access Application';
} else {
$message = 'You are logged in!';
}
?>
<html>
<head>
</head>
<body>
<?php echo $message; ?>
</body>
</html>
Or, place it in the body of your page, like so:
<?php
session_start();
?>
<html>
<head>
</head>
<body>
<?php
if($_SESSION['auth'] != 1) {
echo 'You are not logged in! Access Application';
} else {
echo 'You are logged in!';
}
?>
</body>
</html>
If you are still not getting the desired results then use var_dump($_SESSION); to print out your session array and make sure it holds the correct information.
My code works to a point. What I want is that when this if statement is false, the <div> doesn't show
<?php
$query3 = mysql_query($query3);
$numrows = mysql_num_rows($query3);
if ($numrows > 0) {
$fvisit = mysql_fetch_array($result3);
}
else {
}
?>
You can use css or js for hiding a div. In else statement you can write it as:
else{
?>
<style type="text/css">#divId{
display:none;
}</style>
<?php
}
Or in jQuery
else{
?>
<script type="text/javascript">$('#divId').hide()</script>
<?php
}
Or in javascript
else{
?>
<script type="text/javascript">document.getElementById('divId').style.display = 'none';</script>
<?php
}
This does not need jquery, you could set a variable inside the if and use it in html or pass it thru your template system if any
<?php
$showDivFlag=false
$query3 = mysql_query($query3);
$numrows = mysql_num_rows($query3);
if ($numrows > 0){
$fvisit = mysql_fetch_array($result3);
$showDivFlag=true;
}else {
}
?>
later in html
<div id="results" <?php if ($showDivFlag===false){?>style="display:none"<?php } ?>>
A fresh look at this(possibly)
in your php:
else{
$hidemydiv = "hide";
}
And then later in your html code:
<div class='<?php echo $hidemydiv ?>' > maybe show or hide this</div>
in this way your php remains quite clean
Use show/hide method as below
$("div").show();//To Show
$("div").hide();//To Hide
<?php
$divStyle=''; // show div
// add condition
if($variable == '1'){
$divStyle='style="display:none;"'; //hide div
}
print'<div '.$divStyle.'>Div to hide</div>';
?>
Probably the easiest to hide a div and show a div in PHP based on a variables and the operator.
<?php
$query3 = mysql_query($query3);
$numrows = mysql_num_rows($query3);
?>
<html>
<?php if($numrows > null){ ?>
no meow :-(
<?php } ?>
<?php if($numrows < null){ ?>
lots of meow
<?php } ?>
</html>
Here is my original code before adding your requirements:
<?php
$address = 'meow';
?>
<?php if($address == null){ ?>
no meow :-(
<?php } ?>
<?php if($address != null){ ?>
lots of meow
<?php } ?>
from php you can invoke jquery like this but my 2nd method is much cleaner and better for php
if($switchView) :?>
<script>$('.container').hide();</script>
<script>$('.confirm').show();</script>
<?php endif;
another way is to initiate your class and dynamically invoke the condition like this
$registerForm ='block';
then in your html use this
<div class="col" style="display: <?= $registerForm?>">
now you can play with the view with if and else easily without having a messed code example
if($condition) registerForm = 'none';
Make sure you use 'block' to show and 'none' to hide. This is far the easiest way with php
If you wrap the whole block within the <?php ?> tags in the initial 'if' statement and don't specify an 'else', this condition is not satisfied, returns false so therefore won't display the <div>