Data Management Issue - How to manage retrieved data from mysql? - php

I am developing a game(c#) with database(mysql) and web service(php) to retrieving the data. The issue is the data management. There is a table on database with the name of items and it has some columns like id, item_name, item_description, item_prop, update_date, ownerId. I added 70 items to this table manually. The users can also add some items to this table or they can update the items they have already added in the game. My purpose is retrieving the whole affected rows of the table when the user is first logged in and save it as a json file in the game folder. After, read that file to fill the game environment with those items.
I try a way to achieve this. Firstly, i hold an updateDate variable in the game which is past like "1990-05-10 21:15:43". Second, i send this variable to the webservice as '$lastUpdateDate'; and make a query according to that variable at the database. select * from channels where update_date >= '$lastUpdateDate'; and write these rows in a json file as items.json. after that make a second query to retrieve the time and refresh the updateDate variable in the game. select select now() as 'Result';. In this way user would not have to get the whole table and write in json file every login process. So, it would be good for the performance and the internet usage. The problem occurs when the users update an item which is already added before. I can see the updated item, too with the first query, but I wrote it in json file twice in this way.
php code part of the getItems of my loginuser.php :
<?php
include './function.php';
// CONNECTIONS =========================================================
$host = "localhost"; //put your host here
$user = "root"; //in general is root
$password = ""; //use your password here
$dbname = "yourDataBaseName"; //your database
$phpHash = "yourHashCode"; // same code in here as in your Unity game
mysql_connect($host, $user, $password) or die("Cant connect into database");
mysql_select_db($dbname)or die("Cant connect into database");
$op = anti_injection_register($_REQUEST["op"]);
$unityHash = anti_injection_register($_REQUEST["myform_hash"]);
if ($unityHash == $phpHash)
{
if($op == "getItems")
{
$lastUpdateDate = anti_injection_login($_POST["updateDate"]);
if(!$lastUpdateDate)
{
echo "Empty";
}
else
{
$q = "select * from items where update_date >= '$lastUpdateDate';";
$rs = mysql_query($q);
if (!$rs)
{
echo "Could not successfully run query ($q) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($rs) == 0)
{
echo "No rows found, nothing to print so am exiting";
exit;
}
$arr = array();
while ($row = mysql_fetch_row($rs))
{
$arr = $row;
}
echo json_encode($arr);
}
mysql_close();
}
}
?>
So, how can i solve this problem? Or do you have any better idea for this approach. Help would be much appreciated. Thank you for your time.

Related

mysql problem using php when update table

This system is based on invitation codes, if u have a code that is present in the database you can submit the input therefore change a value in a row. There are 2 inputs, 1) Invitation Code (key), if exist in the database the user can submit the value 2)Name (user). I done the following code but it doesn't work, any suggestions?
<?php
//get value pass from form in login.php
$username = $POST['user'];
$password = $POST['key'];
//connect to the server and select database
mysql_connect("localhost", "...","...");
mysql_select_db("...");
// Query the database for user
$result = mysql_query("UPDATE invitation_keys SET name ='$username' WHERE key = '$password'";)
or die("Failed to query database".mysql_error());
$row = mysql_fetch_array($result);
if ($row['key'] == $password) {
echo "Login success!!!".$row['key'];
} else {
echo "Failed to login";
}
?>
When you are coding in PHP, var_dump($var) is your best friend.
So the first thing to do here, is to print the query.
You will see, that your $username and $password vars are NULL, because you missed the syntax of $_POST[].
After, you can put in var_dump what you want, and that's why its interesting, because you will debug faster with this.

PHP login script using bind_result in subsequent query mysqli

I'm trying to build a relatively simple PHP login script to connect to MySQL database running on my home server. I know the connection works as I've gotten some data returned as I would expect. However, I am having trouble getting the full script to work.
Essentially, I'm taking in a username/password from the user, and I first do a lookup to get the user_id from the users table. I then want to use that user_id value to do a comparison from user_pswd table (i'm storing usernames and passwords in separate database tables). At one point, I was able to echo the correct user_id based on the username input. But I haven't been able to get all the issues worked out, as I'm pretty new to PHP and don't really know where to see errors since I load this onto my server from a remote desktop. Can anyone offer some advice/corrections to my code?
The end result is I want to send the user to another page, but the echo "test" is just to see if I can get this much working. Thanks so much for the help!
<?php
ob_start();
$con = new mysqli("localhost","username","password","database");
// check connection
if (mysqli_connect_errno()) {
trigger_error('Database connection failed: ' . $con->connect_error, E_USER_ERROR);
}
$users_name = $_POST['user'];
$users_pass = $_POST['pass'];
$user_esc = $con->real_escape_string($users_name);
$pass_esc = $con->real_escape_string($users_pass);
$query1 = "SELECT user_id FROM users WHERE username = ?;";
if ($result1 = $con->prepare($query1)) {
$result1->bind_param("s",$user_esc);
$result1->execute();
$result1->bind_result($userid);
$result1->fetch();
$query2 = "SELECT user_pswd_id FROM user_pswd WHERE active = 1 AND user_id = ? AND user_pswd = ?;";
if ($result2 = $con->prepare($query2)) {
$result2->bind_param("is",$userid,$pass_esc);
$result2->execute();
$result2->bind_result($userpswd);
$result2->fetch();
echo "test", $userpswd;
$result2->free_result();
$result2->close();
} else {
echo "failed password";
}
$result1->free_result();
$result1->close();
}
$con->close();
ob_end_clean();
?>

PHP login system help needed for deletion

Help with user deletion:
Hello I am creating a user creation system for a project of mine, I am still very new to PHP, my issue is getting the user from the MySQL database and then deleting it, I will show you my code below:
<?php
require_once("config/db.php");
if ($login->isUserLoggedIn() == true) {
if ($_SESSION['user_perm'] == 1) {
//Create Connection
$db_connection = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);
// Check connection
if ($db_connection->connect_error) {
die("Connection failed: " . $db_connection->connect_error);
}
$sql = "SELECT user_name FROM users";
$result = $db_connection->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo $row["user_name"];
$user_name_delete = $row["user_name"];
$_SESSION['user_name_delete'] = $user_name_delete;
echo 'Delete User<br>';
}
}
}
$db_connection->close(); ?>
On the deleteuser.php page my code is, note this was just a test:
<?php
echo $_SESSION['user_name_delete'];
?>
My issue with this is grabbing the user who you selected to delete as at the moment it only outputs the last user grabbed from the database.
Any help is much appreciated.
This value:
$_SESSION['user_name_delete']
Is going to contain only the last user in your data. Because you keep overwriting it in your loop:
$_SESSION['user_name_delete'] = $user_name_delete;
The short answer is... Don't use session state for this. (Really, you shouldn't use session state for much of anything unless you absolutely have to.) The identity of the user to be deleted should be included in the request to delete the user. In this case, you can add it to the link. Something like this:
echo 'Delete User<br>';
(Or whatever you use to identify the user in the data row.)
Then in deleteuser.php you can get that value from:
$_GET['id']
Validate the inputs, validate that the user is authorized to perform the delete, and then use that value in the WHERE clause of your DELETE query.
Get the users id and insert it into a delete query.
$id = $db->real_escape_string($_GET['id']);
$sql = "delete from users where id = " . $id; and then run the query to delete the user from the database.

Echo out user information in the same table to their page base on their store information without echoing out the same information to another user

First of all I stored users in the same table and I created a page called welcome.php, where I want it to be echoing out user info from MySQL based on their entry.
Now when I created first user and echo it out to this welcome.php, it comes out from the table, and if I create another user info in the same table for it to echo out at the same welcome.php based on the user login info such as, if I create a user called John Fred etc and a user called Michael Kenneth etc.
So user John Fred comes out to the welcome.php with its information from the same table, and then user Michael Kenneth doesn't come to welcome.php when i sign with user Michael Kenneth instead it shows only user John Fred. I don't know where this error comes from; maybe from the login.php, or from welcome.php.
Here is my code echoing in welcome.php
<?php
$tnumber2 = "{$_SESSION['tnumber2']}";
// Connect to the database
$db = mysql_connect("$Sname","$Uname","$Pname") or die("Could not connect to the Database.");
$select = mysql_select_db("$Dname") or die("Could not select the Database.");
$sql="SELECT * FROM `$Tname` LIMIT 0, 25 ;";
$result=mysql_query($sql);
$rows=mysql_fetch_array($result);
?>
<? echo $rows['tnumber2']; ?>
Another script for other user info which I store for another table:
<?php
// Connect to the database
$tnumber2 = "{$_SESSION['tnumber2']}";
$db = mysql_connect("$Sname","$Uname","$Pname") or die("Could not connect to the Database.");
$select = mysql_select_db("$Dname") or die("Could not select the Database.");
$sql="SELECT * FROM `$UPname` LIMIT 0, 25 ;";
$result=mysql_query($sql);
?>
<?php
while($rows=mysql_fetch_array($result)){ // Start looping table row
?>
<? echo $rows['pdate']; ?>
<?php
// Exit looping and close connection
}
mysql_close();
?>
And here is my login.php in this case am using one input form:
<?php
session_start();
ob_start();
?>
<?php
if ($_POST['submit']) {
$tnumber2 = $_POST['user'];
if ($tnumber2) {
require("connect.php");
$query = mysql_query("SELECT * FROM users WHERE tnumber2='$tnumber2'");
$numrows = mysql_num_rows($query);
if($numrows == 1) {
$row = mysql_fetch_assoc($query);
$id = $row['id'];
$tnumber2 = $row['tnumber2'];
if ($tnumber2 == $tnumber2) {
$_SESSION['id'] = $id;
$_SESSION['tnumber2'] = $tnumber2;
header("Location: welcome.php");
}
}
else
include "error.php";
}
}
?>
I have tried all I can on this, maybe I might be a fool to think that such thing is possible but I am not a PHP professional, just a learner, please any help will be gladly appreciated.
Assuming the session has indeed stored the data of the logged-in user, you need to change "welcome.php" so it reads the correct user with a WHERE clause:
<?php
// Retrieve the ID of the user (and untaint it too)
$id = (int) $_SESSION['id'];
// Connect to the database (I've removed the unnecessary quotes)
$db = mysql_connect($Sname, $Uname, $Pname) or die("Could not connect to the Database.");
$select = mysql_select_db($Dname) or die("Could not select the Database.");
// Here is the query from the users table, we're selecting one user here
$sql="SELECT * FROM `users` WHERE `id` = $id;";
$result = mysql_query($sql);
$rows = mysql_fetch_array($result);
?>
<!-- Let's see what is in rows now, should be just one record -->
<?php print_r($rows) ?>
I would advise that you try to understand each part of the code above, and indeed the same for the code you have - don't just copy-and-paste without knowing what each bit does. If you get stuck on something, don't be afraid to look it up in the manual!
I've used print_r to just dump the row result - you can use the contents of that to determine what columns and other data you wish to extract out of it. After you have done that, the print_r can be removed.
Bear in mind that your login is not testing for password correctness - it only checks that someone has entered a particular username in login.php. If you want users to log on with a username and password, that needs to be designed and implemented as well. There are many questions on this site with best-practice techniques on how to do that, if that's of interest to you.
It has, incidentally, been rather difficult to understand what you are doing. I don't think this is a problem with your English, which seems fine to me. Rather, it's worth remembering to write in short sentences (no more than 20 words, say) and short paragraphs (no more than 4 or 5 sentences). And keep your descriptions as short as you can - it makes the difference between people helping you and their deciding they don't understand what you are trying to do. I expect this advice would be just as relevant in your native language as well!
Also, remember to add as much useful information to a question as you can, and if people ask for clarification, make sure you answer all their questions. Remember that people here are volunteers, and you need to make their job as easy as possible.

Determining Account Type from table and Registering Session

Basically what i am trying to do here is to read from the table in my database using the customers login details, then retrieve the record that matches this information. In this table there is a column called "AccountType", this differentiates the average user from a manager, if this column is 1, they are a average user. If this column is 2, they are a manager.
Now im having issues implementing this in my code, below is the snippet of my process script for the login:
<?php
***session_start()
$query = mysql_query("SELECT * FROM accounts WHERE username='$username' and password='$password'", $db) or die ("Query failed with error: ".mysql_error());
$count=#mysql_num_rows($query);
if(***$count == 1)
{
***$user_row = mysql_fetch_array($result)
$userid = $user_row["userid"];
$_SESSION['userid'] = $userid;
$customername = $user_row["customername"];
$_SESSION['customername'] = $customername;
$AccountType = $user_row["accounttype"];
if ($AccountType == 2)
{
$_SESSION['manager'] = $AccountType;
}
Depending on this, when my check login script which every page includes, it will display specific links on the navigation depending what there account type is, if they are user they will have access to normal links, but if they are a manager they have access to admin functions, below is the code snippet for this also:
***session_start();
if (***isset($_SESSION['userid']))
{
$employeeid = $_SESSION['userid'];
$firstname = $_SESSION['customername'];
if (***isset($_SESSION['manager']))
{
$User_Options .='Manager links go here';
}
else
{
$Links .='Normal Links go here';
}
}
Thats just a basic truncated version, but that gives the basis of what im trying to accomplish. I am guessing down to using the while loop its overwriting the session, which i understand, however there will only be one record for the information i am searching. It works to some extent, however even if the AccountType is 1, it displays the options for 2.
Can anyone assist me further in solving this issue? Thankyou!
Use something like this on the login form:
$_SESSION['manager'] = false;
if ($AccountType == 2) {
$_SESSION['manager'] = true;
}
then later:
if ($_SESSION['manager']) {
// display manager-only options
} else {
// display user-only options
}
// Display options for everyone here

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