I have created the following sql statement in order to check if a user already exists in the database and if so to prevent his registration. However, I have done something wrong because now every registration even from not duplicate users is denied.
$res = mysqli_query($mysqli, "SELECT * FROM Users WHERE Email = '{$_POST['email']}' OR Fullname = '{$_POST['fullname']}'");
if (count($res) > 0) {
echo "<script type=\"text/javascript\">window.alert('User already exists!');window.location.href = '/Register.php';</script>";
exit;
}
If you want to prevent duplicate users, then use constraints or unique indexes (really the same thing):
create unique index users_email on users(email);
create unique index users_fullname on users(fullname);
This will prevent duplicates from going into these fields at the database level. Much safer than trying to do this at the application level.
As you can read in the manual, mysqli_query()
will return a mysqli_result object.
You can use mysqli_num_rows() to properly do this:
$result = mysqli_query($mysqli, "SELECT * FROM Users WHERE Email = '{$_POST['email']}' OR Fullname = '{$_POST['fullname']}'");
$row_count = mysqli_num_rows($result);
if ($row_count > 0) {
//etc.
}
Related
Here is my login process, I want a same dashboard but data will be different for each user. But I am stuck with creating uid variables to get data for each login user.
if(isset($_POST['login_btn']))
{
$email_login=$_POST['email'];
$password_login=$_POST['password'];
$admin="admin";
$co_admin="co_admin";
$query = "SELECT * FROM registered_users WHERE email='$email_login' AND password='$password_login' AND usertype='$admin' ";
$query_run = mysqli_query($connection, $query);
$query_co = "SELECT * FROM registered_users WHERE email='$email_login' AND password='$password_login' AND usertype='$co_admin' ";
$query_run_co = mysqli_query($connection, $query_co);
if(mysqli_fetch_array($query_run))
{
$_SESSION['username'] = $email_login;
$_SESSION['usertype'] = $admin;
header('Location: index.php');
}
else if(mysqli_fetch_array($query_run_co))
{
$_SESSION['username'] = $email_login;
$_SESSION['usertype'] = $co_admin;
header('Location: company_view.php');
}
else
{
$_SESSION['status'] = 'Email ID / Password / User Type is Invalid';
header('Location: login.php');
}
}
Above source code is for separating Co-admin and Admin. Now Any Co-Admin login to the portal he should get his own details, I would like to know which function I have to call or how should I declare a uid variable to fetch data tables for each current logged in user. I found some other source codes but which is not related to me so i am confused with how I fix it with those code. Can anyone do it in my codes.
I think you are asking how to get data for the current user from mysql tables. Yes, the standard way of doing this is via a unique ID for each user that is pulled from the registered_users table, storing this in the session, and then referencing this in the other tables and filtering by this ID. I would not suggest storing anything else from this table in the session as the ID is likely to have a stronger guarantee of imutibility.
For example if you have a table of recently visited pages per user, you would get this via:
$query = 'SELECT * from recently_visited WHERE user_id = ?';
$stmt = mysqli_prepare($query);
$stmt->bind_param("i", $_SESSION['user_id']);
$stmt->execute();
You can check the mysqli documentation for how to then extract what you need from the executed statement. I've shown this example of a prepared statement so you can see how to avoid SQL injection as well.
You may want to look into using foreign keys to enforce this connection.
I have developed a game with Javascript and when the user finishes it, I must save his record in a database. Here you see the code:
$temp = $_POST['playername']; //username
$text = file_get_contents('names.txt'); //list with all usernames
//this text file contains the names of the players that sent a record.
$con=mysqli_connect("localhost","username","pass","my_mk7vrlist");
if (stripos(strtolower($text), strtolower($temp)) !== false) {
//if the username is in the list, don't create a new record but edit the correct one
mysqli_query($con, "UPDATE `my_mk7vrlist`.`mk7game` SET `record` = '".$_POST['dadate']."' WHERE `mk7game`.`playername` = ".$temp." LIMIT 1 ");
} else {
//The username is not in the list, so this is a new user --> add him in the database
mysqli_query($con, "INSERT INTO `mk7game` (`playername`,`record`,`country`,`timen`) VALUES ('".$_POST['playername']."', '".$_POST['dadate']."', '".$_POST['country']."', '".$_POST['time_e']."')");
file_put_contents("names.txt",$text."\n".$temp);
//update the list with this new name
}
//Close connection
mysqli_close($con);
When I have a new user (the part inside my "else") the code works correctly because I have a new row in my database.
When the username already exists in the list, it means that this player has already sent his record and so I must update the table. By the way I cannot edit the record on the player that has alredy sent the record.
mysqli_query($con, "UPDATE `my_mk7vrlist`.`mk7game` SET `record` = '".$_POST['dadate']."' WHERE `mk7game`.`playername` = ".$temp." LIMIT 1 ");
It looks like this is wrong, and I can't get why. I am pretty new with PHP and MySQL.
Do you have any suggestion?
You're missing quotes around $temp in the UPDATE statement:
mysqli_query($con, "UPDATE `my_mk7vrlist`.`mk7game`
SET `record` = '".$_POST['dadate']."'
WHERE `mk7game`.`playername` = '".$temp."'
^ ^
LIMIT 1 ") or die(mysqli_error($con));
However, it would be better to make use of prepared statements with parameters, rather than inserting strings into the query.
Escape your user input!
$temp = mysqli_real_escape_string($con, $_POST['playername']);
Make sure to stick your mysqli_connect() above that
$select = mysqli_query($con, "SELECT `id` FROM `mk7game` WHERE `playername` = '".$temp."'");
if(mysqli_num_rows($select))
exit("A player with that name already exists");
Whack that in before the UPDATE query, and you should be good to go - obviously, you'll need to edit it to match your table setup
I have written a code to check whether the username exists in the database or not. It seems to return that there is no such username exists even if there's a same username existing.
$conu=mysqli_connect("localhost","db_user","db_pass","db_name");
$result = mysql_query("SELECT 1 FROM member WHERE username = $username");
if ($result && mysql_num_rows($result) > 0) {
$user_err = "<i><span class='error'>Usernme already exists</span></i>";
$errflag = true;
}
elseif(preg_match("/^[0-9a-zA-Z_]{5,}$/", $username) === 0) {
$user_err = "<i><span class='error'>Usernme must be bigger than 5 chararacters and can contain only digits, letters and underscore</span></i>";
$errflag = true;
}
Try
mysql_query("SELECT username FROM member WHERE username = '$username' LIMIT 1;");
SELECT 1 is not actually using the database; it's always returning 1, hence why your result is always the same regardless of the contents of the member table.
Usernames I take it are some sort of varchar? If that is the case, you might want to put its value in quotes:
$result = mysql_query("SELECT `username` FROM `member` WHERE `username` = '".$username."' LIMIT 1;");
your query is subject to sql injections btw.
At first, you are trying to return a column that probably doesn't exist: "1"
Second, I hope that you are cleaning the $username or else you are allowing anyone to inject your database.
The correct query is
mysql_query("SELECT * FROM `member` WHERE `username`='$username'");
You are using mysqli to connect, but mysql to perform your query.
When you SELECT 1 FROM member WHERE username = $username, the result will always be 1.
You need to put $username in the query in quotes. Something like SELECT username FROM member WHERE username = '$username'.
You forgot to include the part of the code for when there is no such username in your posting.
I have a PHP login script. This is the part where the person can create a new user. My issue is I want to check if the user exists, and if the username does not exist the the table, than create the new user. However, if the user does exist, I want it to return an error in a session variable. Here is the code I have right now. This doesn't include my DB connections, but I know they do work. Its num_rows() that is being written as an error in the error_log file. Here is the code:
$username = mysql_real_escape_string($username);
$query = "SELECT * FROM users WHERE username = '$username';";
$result = mysql_query($query,$conn);
if(mysql_num_rows($result)>0) //user exists
{
header('Location: index.php');
$_SESSION['reg_error']='User already exists';
die();
}
else
{
$query = "INSERT INTO users ( username, password, salt )
VALUES ( '$username' , '$hash' , '$salt' );";
mysql_query($query);
mysql_close();
header('Location: index.php');
The error it is giving me is
mysql_num_rows(): supplied argument is not a valid MySQL result resource in [dirctory name]
mysql_num_rows()
Retrieves the number of rows from a result set. This command is only valid for statements like SELECT or SHOW that return an actual result set. To retrieve the number of rows affected by a INSERT, UPDATE, REPLACE or DELETE query, use mysql_affected_rows().
Instead of doing SELECT * and then mysql_num_rows(), you can do a SELECT COUNT(*) and then retrieve the number of rows, by fetching the field (that should be 0 or 1). SELECT COUNT will always return a result (provided that the query syntax is correct of course).
Also, change the query:
$query = "SELECT * FROM users WHERE username = '$username';";
into
$query = "SELECT * FROM users WHERE username = '"
. mysql_real_escape_string($username) . "';";
Just out of curiosity, have you ever heard of upserts? I.E., "insert on duplicate key". They'd be useful to you in this situation, at least if your username column is a unique key.
http://dev.mysql.com/doc/refman/5.0/en/insert-on-duplicate.html
$username = mysql_real_escape_string($username);
i think you have to replace the above to
$username = mysql_real_escape_string($_POST[$username]);
I want have an insert query, but before inserting I check whether the username and email are used by someone else. If used, I want to cancel insert query and echo a message to say whether username or email is in use.
Here my code:
$sql = "SELECT 1 FROM user WHERE username='".$_POST['username']."'";
if(!$result = mysql_query($sql))
die(mysql_error());
while($row = mysql_fetch_array($result))
die('This username is already exists');
$sql = "SELECT 2 FROM user WHERE email='".$_POST['email']."'";
if(!$result = mysql_query($sql))
die(mysql_error());
while($row = mysql_fetch_array($result))
die('This email address is already exists');
$sql = "insert into user (username,email,password,tel,type) values ('".$_POST['username']."','".$_POST['email']."','".$_POST['password']."','".$_POST['telnumber']."','member')";
if(!mysql_query($sql))
die(mysql_error());
I want these three sql statements in one. It can be either using cases or something else that you suggest. So,
Is it possible to zip this code into one sql query?
As a result what I need is
sql = "sql_query"
if(!$result = mysql_query($sql))
die(mysql_error());
while($row = mysql_fetch_array($result)){
if($row['result']==1)
die('This username is already exists');
else if($row['result']==2)
die('This email is already exists');
}
die('you have succesfully registered');
thanks for any advice.
While I suggest you follow #cularis' answer, you may be interested in the following alternative:
Give email and username the UNIQUE constraint, by creating a unique index for both of these.
run your INSERT query, and if this fails... (due to duplicate keys)
run the suggested combined SELECT, to determine which field existed (username or email)
You can combine the first two queries like this:
$sql = "SELECT * FROM user WHERE username='".$_POST['username']."' OR email='".$_POST['email']."'";
Have look at mysql_real_escape string to sanatize your input.
Assuming you don't care about a more specific error case you could probably just do the following:
$sql = "SELECT * FROM user WHERE username='".$_POST['username']."' OR email='".$_POST['email']."'";
if(!$result = mysql_query($sql))
die(mysql_error());
while($row = mysql_fetch_array($result))
die('The username or email address is already being used');
$sql = "insert into user (username,email,password,tel,type) values ('".$_POST['username']."','".$_POST['email']."','".$_POST['password']."','".$_POST['telnumber']."','member')";
if(!mysql_query($sql))
die(mysql_error());
This isn't the best of designs if you're looking for, as I said, specific error cases. So if you are okay with just telling the person there is an error that one or both are in use then that should work.
I am not sure as I am very rusty in PHP/MySQL but I assume that if such cases of both exist then multiple rows may be returned and I forget exactly how mysql_fetch_array works but I assume it's an array of all results valid for the query so you should be set. As long as the array exists, you know there was a hit in the db.