Updating something in a database with a php file - php

Hello iam working on a game on unity3d and i have a login/register system that have a database so i have these level thing i want to save in to the users database, just for testing i have created this php file:
<?PHP
$username = $_POST['username'];
$pass = $_POST['password'];
$con = mysql_connect("host","user","password") or ("Cannot connect!" . mysql_error());
if (!$con)
die('Could not connect: ' . mysql_error());
mysql_select_db("database" , $con) or die ("could not load the database" . mysql_error());
$check = mysql_query("SELECT * FROM users WHERE `username`='".$username."'");
$numrows = mysql_num_rows($check);
if($username == null)
{
die ("You are not allowed to access this site! \n");
}
else
{
if ($numrows == 0)
{
die ("Username does not exist \n");
}
else
{
$pass = md5($pass);
while($row = mysql_fetch_assoc($check))
{
if ($pass == $row['pass'])
{
die("saved-SUCCESS");
$sqlUpdate = "UPDATE users SET level='2' WHERE id='1'";
//$SQL = "UPDATE users SET level = '2' WHERE id = 1";
}
else
die("Password is wrong \n");
}
}
}
?>
it does says "saved-SUCCESS" in unity3d so that means this:
$sqlUpdate = "UPDATE users SET level='2' WHERE id='1'";
should work but it doesent
also i have tryed this:
$SQL = "UPDATE users SET level = '2' WHERE id = 1";
but it doesent work either
here is a picture of the user with id 1:
http://i.imgur.com/NZslFBN.png
and here is a picture of the structur tab:
http://i.imgur.com/ZN6PyW1.png
and yes my table is named users :)

You are not executing your second query but only printing saved-SUCCES
$sqlUpdate = "UPDATE users SET level='Rune' WHERE id='1'";
mysql_query($sqlUpdate);
die("saved-SUCCESS"); //then exit.
As side note i would advise thta your code is highly vulnerable mo mysql injection, you should use prepared statments either with PDO or mysqli, also mysql_* api are deprecated and soon will be no longer mantained.

Related

MySQL query no results

I want echo my DB results from Session but i get no results or errors:
$_SESSION['username'];
$link = mysqli_connect("$myHost", "$myUser", "$myPass", "$myDB");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$username = mysqli_real_escape_string($link, $_SESSION['username']);
$sql = "SELECT * FROM users where username = $username";
$result = mysqli_query($link, $sql);
echo $result;
Anyone know why not? Session works.
Thanks
You should change your query, like this:
$sql = "SELECT user FROM yourtablename WHERE username = $username"
Where "user" is what you want to SELECT if you want to select all data, you can use "*", yourtablename is table name of table you want to select.
After your edits, your code should look like
$_SESSION['username'];
$link = mysqli_connect("$myHost", "$myUser", "$myPass", "$myDB");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$username = mysqli_real_escape_string($link, $_SESSION['username']);
$sql = "SELECT * FROM users where username = $username";
if ($result = $link->query($sql)) {
while ($row = $result->fetch_row()) {
var_dump($row);
}
$result->close();
}
More info here
Notice: mysqli_real_escape_string it's not very security. A better option to protect against SQL injections is using prepared statements, more info here

Delete From a Database Using Session ($_SESSION)

I have a record i want to remove from the database. I have so far been able save to the mysql database now i have several information in several rows, now suppose the information isnt the needed one, i want to delete it from the database. Thats what i am trying to achieve here
I tried this
<?php
session_start();
require_once('inc/config.php');
if(!isset($_SESSION['username'])){
header('Location: signon.php');
}
?>
<?php
require_once('inc/config.php');
$con = mysqli_connect($host, $user, $pass, $db) or die ('Cannot connect: '.mysqli_error());
$sql = "SELECT * FROM education_info WHERE username = '" . $_SESSION['username'] . "'";
$result = mysqli_query($con,$sql);
$row = mysqli_fetch_array($result);
$id = $row['id'];
$username = $row['username'];
$sql2 = "DELETE FROM education_info WHERE id = $id" ;
$result = mysqli_query($con,$sql);
mysqli_close($con);
header("Refresh:0; url=EDWE.php");
?>
Only that the information still remains present in the database, How do i go about deleting it completely, if not needed?
You are passing wrong variable while executing delete query:
$sql2 = "DELETE FROM education_info WHERE id = $id" ;
$result = mysqli_query($con,$sql2); //<---pass $sql2

PHP Insert values into Database at certain position

So I'm currently creating a Online Game and already got the login and registering working but now I'm working on submitting Stats such as health and a player's level and retrieving that.
So in this PHP file I'm able to retrieve the info of a certain player but I don't know how to submit new stats into a certain player's row. I'm still learning PHP so please help me out.
<?php
// Database Things =========================================================
$host = "localhost";
$user = "lyth_com_Spillnk";
$password = "INTERESTED?";
$dbname = "lythumn_com_Spillnk";
mysql_connect($host, $user, $password) or die("Can't connect into database");
mysql_select_db($dbname)or die("Can't connect into database");
// =============================================================================
$Act = $_GET["Act"];// what is action, Submit or Retrieve?
$nick = $_GET["User"];
$health = $_GET["Health"];
$level = $_GET["Level"];
$xcood = $_GET["X"];
$ycood = $_GET["Y"];
if($Act == "Retrieve"){
$SQL = "SELECT * FROM Stats WHERE Username = '" . $nick . "'";
$result_id = #mysql_query($SQL) or die("DB ERROR");
$total = mysql_num_rows($result_id);
if($total) {
$datas = #mysql_fetch_array($result_id);
echo ($datas["Health"], $datas["Level"], $datas["X"], $datas["Y"]);
}
}
if($Act == "Submit"){
$SQL = "SELECT * FROM Stats WHERE Username = '" . $nick . "'";
$result_id = #mysql_query($SQL) or die("DB ERROR");
$query = "INSERT INTO Stats (Username, Health, Level, X, Y) VALUES('$nick', '$health', '$level', $'xcood', $'ycood')";
mysql_query($query) or die("ERROR");
mysql_close();
echo "Submitted";
}
// Close mySQL Connection
mysql_close();
?>
It's mostly concercing this piece of code:
if($Act == "Submit"){
$SQL = "SELECT * FROM Stats WHERE Username = '" . $nick . "'";
$result_id = #mysql_query($SQL) or die("DB ERROR");
$query = "INSERT INTO Stats (Username, Health, Level, X, Y) VALUES('$nick', '$health', '$level', $'xcood', $'ycood')";
mysql_query($query) or die("ERROR");
mysql_close();
echo "Submitted";
}
As you are able to see I already retrieve the Index of where the player's stats are located so how do I insert values there?
Thanks in advance!

Selecting certain row in mysql

I am completely new to MYSQL and PHP, so i just need to do something very basic.
I need to select a password from accounts where username = $_POST['username']... i couldn't figure this one out, i keep getting resource id(2) instead of the desired password for the entered account. I need to pass that mysql through a mysql query function and save the returned value in the variable $realpassword. Thanks!
EDIT:
this code returned Resource id (2) instead of the real password
CODE:
<?php
$con = mysql_connect('server', 'user', 'pass');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
echo '<br/> ';
// Create table
mysql_select_db("dbname", $con);
//Variables
//save the entered values
$enteredusername = $_POST['username'];
$hashedpassword = sha1($_POST['password']);
$sql = "SELECT password from accounts where username = '$enteredusername'";
$new = mysql_query($sql,$con);
echo "$new";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
mysql_close($con);
?>
It will be a lot better if you use PDO together with prepared statements.
This is how you connect to a MySQL server:
$db = new PDO('mysql:host=example.com;port=3306;dbname=your_database', $mysql_user, $mysql_pass);
And this is how you select rows properly (using bindParam):
$stmt = $db->prepare('SELECT password FROM accounts WHERE username = ?;');
$stmt->bindParam(1, $enteredusername);
$stmt->execute();
$result = $stmt->fetch(PDO::FETCH_ASSOC);
$password = $result['password'];
Also, binding parameters, instead of putting them immediately into query string, protects you from SQL injection (which in your case would be very likely as you do not filter input in any way).
I think your code looks something like this
$realpassword = mysql_query("SELECT password
from accounts where username = '$_POST[username]'");
echo $realpassword;
This will return a Resource which is used to point to the records in the database. What you then need to do is fetch the row where the resource is pointing. So, you do this (Note that I am going to use structural MySQLi instead of MySQL, because MySQL is deprecated now.)
$connection = mysqli_connect("localhost", "your_mysql_username",
"your_mysql_password", "your_mysql_database")
or die("There was an error");
foreach($_POST as $key=>$val) //this code will sanitize your inputs.
$_POST[$key] = mysqli_real_escape_string($connection, $val);
$result = mysqli_query($connection, "what_ever_my_query_is")
or die("There was an error");
//since you should only get one row here, I'm not going to loop over the result.
//However, if you are getting more than one rows, you might have to loop.
$dataRow = mysqli_fetch_array($result);
$realpassword = $dataRow['password'];
echo $realpassword;
So, this will take care of retrieving the password. But then you have more inherent problems. You are not sanitizing your inputs, and probably not even storing the hashed password in the database. If you are starting out in PHP and MySQL, you should really look into these things.
Edit : If you are only looking to create a login system, then you don't need to retrieve the password from the database. The query is pretty simple in that case.
$pass = sha1($_POST['Password']);
$selQ = "select * from accounts
where username = '$_POST[Username]'
and password = '$pass'";
$result = mysqli_query($connection, $selQ);
if(mysqli_num_rows($result) == 1) {
//log the user in
}
else {
//authentication failed
}
Logically speaking, the only way the user can log in is if the username and password both match. So, there will only be exactly 1 row for the username and password. That's exactly what we are checking here.
By seeing this question we can understand you are very very new to programming.So i requesting you to go thru this link http://php.net/manual/en/function.mysql-fetch-assoc.php
I am adding comment to each line below
$sql = "SELECT id as userid, fullname, userstatus
FROM sometable
WHERE userstatus = 1"; // This is query
$result = mysql_query($sql); // This is how to execute query
if (!$result) { //if the query is not successfully executed
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) { // if the query is successfully executed, check how many rows it returned
echo "No rows found, nothing to print so am exiting";
exit;
}
while ($row = mysql_fetch_assoc($result)) { //fetch the data from table as rows
echo $row["userid"]; //echoing each column
echo $row["fullname"];
echo $row["userstatus"];
}
hope it helps
try this
<?php
$con = mysql_connect('server', 'user', 'pass');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
echo '<br/> ';
// Create table
mysql_select_db("dbname", $con);
//Variables
//save the entered values
$enteredusername = mysql_real_escape_string($_POST['username']);
$hashedpassword = sha1($_POST['password']);
$sql = "SELECT password from accounts where username = '$enteredusername'";
$new = mysql_query($sql,$con);
$row = mysql_fetch_array($new) ;
echo $row['password'];
if (!$new)
{
die('Error: ' . mysql_error());
}
mysql_close($con);
?>
<?php
$query = "SELECT password_field_name FROM UsersTableName WHERE username_field_name =".$_POST['username'];
$result = mysql_query($query);
$row = mysql_fetch_array($result);
echo $row['password_field_name'];
?>
$username = $_POST['username'];
$login_query = "SELECT password FROM users_info WHERE users_info.username ='$username'";
$password = mysql_result($result,0,'password');

undefined variable php updating mysql data [duplicate]

This question already has answers here:
Reference - What does this error mean in PHP?
(38 answers)
Closed 9 years ago.
This is the code for attempting to do a update on mysql data errors stating undefined variable
mysql_connect ("localhost", "root", "");
mysql_select_db("supplierdetails");
$con = mysql_connect("localhost", "root", "");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
//Run a query
$result = mysql_query ("SELECT * FROM users WHERE id= '$id'");
while ($row = mysql_fetch_array($result))
{
$username=$row['username'];
$password=$row['password'];
}
$query = "UPDATE users SET username = '$username', password = '$password' WHERE id = '$id'";
$result = #mysql_query($query);
//Check whether the query was successful or not
if($result) {
header("message= Users Updated");
}else {
die("Query failed");
}
?>
You miss the $id value?
And can use echo to debug or check script result, not header
http://php.net/manual/en/function.header.php
Please be more specific with regards to which variable is undefined.
In the code you've posted $username and $password are only set if $result returns a result, if it doesn't then your while loop will not run and therefore $username and $password will never be set.
Also $id doesn't look as if that has been set either, unless this has been set outside of the code which you have included in your question.
Hope this helps :)
you used 2 connect no need to do while and you forgot $id
$con = mysql_connect("localhost", "root", "");
mysql_select_db("supplierdetails");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
$id = $_POST['id'];
$username=$_POST['username'];
$password=$_POST['password'];
$query = "UPDATE users SET username = '".$username."', password = '".$password."' WHERE id = '".$id."'";
$result = mysql_query($query);
//Check whether the query was successful or not
if($result) {
echo "message= Users Updated";
}else {
die("Query failed");
}
?>

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