I was trying to make a page where you can log in and then change your nickname or/and password. Everything in mySQL database, but when I try to save the id to session variable, it doesn't work. Any suggestions?
I am using XAMPP, users is my table in database users, I'm not posting login form code, because it's very simple.
Everything is connected, code doesn't give any warnings or errors.
login.php (fragment):
$sql = "SELECT * FROM users WHERE nickname = '$myusername' and pass = '$mypassword' and confirmed = 1";
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$count = mysqli_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count == 1) {
$logged = true;
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["id"];
$_SESSION['currentId'] = $row["id"];
echo 'Id: ' . $_SESSION['currentId'];
}
}else {
$error = "Your Login Name or Password is invalid";
}
}
change.php (whole):
<?php
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Users";
$currentId = $_SESSION['currentId'];
if($currentId<1){echo 'No Id.';}
else {echo 'CurrentId: ';
echo $currentId;}
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully <br>";
if($_SERVER["REQUEST_METHOD"] == "POST") {
// username and password sent from form
$aCUname = mysqli_real_escape_string($conn,$_POST['CUname']);
$aCUpass = mysqli_real_escape_string($conn,$_POST['CUpass']);
$sql = "UPDATE users SET nickname = '$aCUname', pass = '$aCUpass' WHERE id = '$currentId';";
$result = mysqli_query($conn,$sql);
echo 'Updated successfully.';
}
?>
Thanks for help.
I got a solution. I just had to delete
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
from login.php. Thanks to #Shashikumar Misal !
Related
I have looked at similar questions on here and have found none that help me with this issue.
I have a database which contains a user table for name, email and password.
My HTML is a very basic form:
<form action = "" method = "post">
<label>Email :</label><input type = "text" name = "email" >
<label>Password :</label><input type = "password" name = "password" />
<input type = "submit" value = " Submit "/>
</form>
My php is:
$servername = "localhost";
$username = "student_user";
$password = "";
$db_name = "Student_db";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $db_name);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
if($_SERVER["REQUEST_METHOD"] == "POST") {
$email - mysqli_real_escape_string($conn,$_POST['email']);
$password = mysqli_real_escape_string($conn,$_POST['password']);
$sql = ("SELECT * FROM user WHERE u_email = '$email' and u_password = '$password'");
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$active = $row['active'];
$count = mysqli_num_rows($result);
// If result matched $email and $mypassword, table row must be 1 row
if($count == 1) {
echo "Login Successful";
}else {
echo "Login Unsuccessful";
}
}
When I enter the correct generic email and password I have in the database, it still echos that the login is unsuccessful. I'm not sure why.
$email - replace with $email =
$servername = "localhost";
$username = "student_user";
$password = "";
$db_name = "Student_db";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $db_name);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
if($_SERVER["REQUEST_METHOD"] == "POST") {
$email = mysqli_real_escape_string($conn,$_POST['email']);
$password = mysqli_real_escape_string($conn,$_POST['password']);
$sql = ("SELECT * FROM user WHERE u_email = '$email' and u_password = '$password'");
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$active = $row['active'];
$count = mysqli_num_rows($result);
// If result matched $email and $mypassword, table row must be 1 row
if($count == 1) {
echo "Login Successful";
}else {
echo "Login Unsuccessful";
}
}
There is issue with your assignment operator in your php code.
You are using
$email - mysqli_real_escape_string($conn,$_POST['email']);
Which is wrong you have to use
$email = mysqli_real_escape_string($conn,$_POST['email']);
you are using "-" insted of "="
here am trying to get username and password from the database and if their result is found then redirect to some page but mysqli_num_rows returns 0 always i dunno why `
if (isset($_POST['login'])) {
$username = $_POST['username'];
$password = $_POST['password'];
$query = "SELECT * FROM `login` WHERE username = '$username' AND password = '$password'";
$run = mysqli_query($con,$query);
if (mysqli_num_rows($run) == 1) {
header("location: login.php");
}}?>
`
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "ornament"; // My local DB Name
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$username = "test"; // $_POST["username"] value
$password = "test"; // $_POST["password"] value
$sql = "SELECT * FROM `ornament` WHERE username = '$username' AND password = '$password'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) == 1) {
echo "correct";
} else {
echo "wrong";
}
mysqli_close($conn);
?>
</body>
</html>
OUTPUT:
correct
I have a database name and table name are same, But for me it's working fine.
NOTE: Please check the whether you are using correct password while getting from POST.
Any how you got a solution, Have a great day
Thanks
Muthu
I have a database which stores data. How can I view data in my database with the same username as my session? What I have tried is below. There is a session and the username is uploading in each row in the database.
This is what I'm trying to do: say I logged in as jack I typed data in and sent it to the database. It saves the name as jack and then only views the results with jack. But it is saying 0 results. Why?
<?php
session_start();
if (isset($_SESSION['username'])) {
$username = $_SESSION['username'];
echo "$username";
}
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "score";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, name, description FROM all_scores WHERE username = '".$username."' ORDER BY id DESC LIMIT 5";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<p></p>";
echo "". $row["name"]. "";
echo "<p>". $row["description"]. "</p>";
}
} else {
echo "0 results";
}
$conn->close();
?>
you have two mistakes
1- SQL syntax error, correct syntax is
$sql = "SELECT id, name, description FROM all_scores WHERE username = '".$username."'";
2- the variable $username is overwritten by the username of the database
try this:
$sql = "SELECT id, name, description FROM all_scores WHERE username = '".$_SESSION['username']."'";
I believe my PHP to be functioning perfectly, therefore I think it's a query error. When I proceed, with form details stored in the session... it happily returns my Posted information but doesn't seem to be pulling anything from my database - there is a row in my database containing the email address I am using. Does anybody see anything blatantly wrong with this PHP?
Thanks for your help.
<?php
session_start();
$servername = "localhost";
$username = "privatedbroot";
$password = "not4ulol";
$dbname = "pdb_inventory";
$status = $_GET["action"];
$_SESSION["Cemail"] = $_POST["CEMAIL"];
$_SESSION["Access"] = md5($_POST["ACCESS"]);
$conn = new mysqli($servername, $username, $password, $dbname);
$sql = "SELECT CEMAIL, ACCESS FROM POPU WHERE `CEMAIL`= ".$_SESSION['Cemail'];
echo $sql;
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
if ($_SESSION["Access"] == $row["ACCESS"]){
echo "password correct!";
} else {
echo "password wrong!";
}
}
}else{
echo "ur email is wrong m8.";
}
?>
Try this:
$cemail = $_SESSION['Cemail'];
$sql = "SELECT CEMAIL, ACCESS FROM POPU WHERE `CEMAIL`= '$cemail'";
The website has a login system, however when a user logs into the website I simply want their details to appear on the next page. This is my code I so far. Problem is, I only want to display the logged in users details, not all the databases details.
<?php $servername = "localhost"; $username = "root"; $password = ""; $dbname = "loginsystem";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, firstname, lastname FROM members";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<table><tr><th>ID</th><th>Name</th></tr>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["id"]. "</td><td>" . $row["firstname"]. " " . $row["lastname"]. "</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
?>
LOG IN SYSTEM
<?php
session_start();
if (isset($_POST['username'])) {
include_once("dbConnect.php");
// Set the posted data from the form into local variables
$usname = strip_tags($_POST['username']);
$paswd = strip_tags($_POST['password']);
$usname = mysqli_real_escape_string($dbCon, $usname);
$paswd = mysqli_real_escape_string($dbCon, $paswd);
$sql = "SELECT id, username, password FROM members WHERE username = '$usname' AND activated = '1' LIMIT 1";
$query = mysqli_query($dbCon, $sql);
$row = mysqli_fetch_row($query);
$uid = $row[0];
$dbUsname = $row[1];
$dbPassword = $row[2];
// Check if the username and the password they entered was correct
if ($usname == $dbUsname && password_verify($paswd,$dbPassword)) {
// Set session
$_SESSION['username'] = $usname;
$_SESSION['id'] = $uid;
// Now direct to users feed
header("Location: MemberDetails.php");
} else {
echo "Oops that username or password combination was incorrect.
<br /> Please try again.";
}
}
?>
Add
session_start();
to the top of the page and then on the next page as well and then you will be able to carry over those variables once they are set.
For example:
$_SESSION['user'] = $_POST['user'];
Then on the next page call:
echo $_SESSION['user'];
You first have to implement the user login part. and after that, get the specified user id or login credentials and use that in your query.
In your LOG IN SYSTEM file, put session_start(); before including the db connection.
Then in the member details page do this:
session_start(); //put this on the first line.
Then your query will now look like below:
<?php
$servername = "localhost"; $username = "root"; $password = ""; $dbname = "loginsystem";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$user_id = $_SESSION['id'];
$sql = "SELECT user_id, firstname, lastname FROM members WHERE user_id = ".$user_id;
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<table><tr><th>ID</th><th>Name</th></tr>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["id"]. "</td><td>" . $row["firstname"]. " " . $row["lastname"]. "</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
?>
Database structure