if profile id = 1 include no profile.php? - php

can someone help i have users on my site and when a user clicks on their profile it takes the user to profile.php?id=(whatever id number)
I am trying to make it so that if a user tries to go to profile id= '1' then this will include noprofile.php basically telling the user that profile doesnt exist.
I have been doing this which seemed to work but now for whatever reson after a couple of days its stopped working and im not sure why?
Can someone show me where im going wrong thanks.
code in profile.php:
<?php
$admin_account = admin_account();
while ($admin = mysql_fetch_array($admin_account)) {
include ('includes/mod_profile/mod_noprofile.php');
}
?>
function:
function admin_account() {
global $connection;
global $profile_id;
$query = "SELECT user_id
FROM ptb_profiles
WHERE user_id =\"$profile_id\"
AND ptb_profiles.user_id='1'";
$admin_account = mysql_query($query, $connection);
confirm_query($query, $connection);
return $admin_account;
}

Do a test before you execute the query to see if the id==1
if isset($_GET["id"]==1){
include ('includes/mod_profile/mod_noprofile.php');
}
else {
//do query and get user profile information
}

Related

PHP SQL IF statement problems

Ok so I'm trying to pull some data from my SQL database and use it in an IF statement.
I have a database called DB_Default
and I have a table called Users
Inside Users I have all the normal columns such as id, username, password but I also have a column called isadmin.
I want to be able to do a mysql query through PHP selecting all users with the username $_SESSION[username] and isadmin = 1.
What I aim on doing is including a few navigation links for escalated privileged users only. So as the page that the code is going to be put into is for logged in users only I thought right, lets use the already defined username session i,e if my sessions username was set to Admin.
the statement should be
$result = mysql_query("SELECT * FROM users WHERE user_name='" . $_SESSION["user_name"] . "' and isadmin = '". $admin."'");
I have set $admin = "1"; so that it should only be able to return results if the user logged in has isadmin set to 1.
Instead I'm either having the navigation links show to any user regardless of their isadmin status or not at all.
To be honest the code is very messy as it's 5:40am and I haven't been coding for a while so quite rusty so I'm more than aware of how easy this should be of a fix.
I'm 99% sure it has to do with the fact I just set $admin = "1"; but for the life of me can't figure out where I've gone wrong. Going to give it a rest for today and come back tomorrow. Hopefully by then someone will have posted a resolution and obviously I'll vote up the best answer otherwise I'll give the code a lookover and see if I can't fix it myself!
Thanks!
UPDATE - Included code
<?php
$admin = 1;
$conn = mysql_connect("localhost","root","password");
mysql_select_db("DB_Default",$conn);
$result = mysql_query("SELECT * FROM users WHERE user_name='" . $_SESSION["user_name"] . "' and isadmin = '". $admin."'");
$row = mysql_fetch_array($result);
if(is_array($row)) {
$_SESSION["isadmin"] = $row[isadmin];
} else {
}
if($_SESSION["isadmin"] == '1'){
//has admin privs
} else {
//does not have admin privs
}
I have not yet set up the navigation links as I haven't gotten to that stage yet however the links would be inside the if statement admin links in the admin part and not in the non admin part.
My guess right now is going to be the following: You set the $_SESSION variable when you log in as Admin. After that the $_SESSION is never changed. Which pretty much means that if you log in as Admin and then re-log in as someone else, $_SESSION['isadmin'] will already be set to 1 thus providing full access. So what you will need to do is change the else part to:
if(is_array($row))
{ $_SESSION["isadmin"] = $row[isadmin]; }
else
{ $_SESSION['isadmin'] = 0; }

Problems updating MySQL, "username" in a table using PHP

I'm probably not using the best method to create a user system, but it doesn't need to be fancy. I also know that I'm not the most organized
The logins and everything are alright, but I'm having a problem updating the credentials.
For example, I'm allowing users to change their username. I have the "Change Username" (Not that name) form to submit to update-username.php.
I already have mysql_real_escape_string, in the function "cleanString" in another page. My textarea submitting already has the old text in it, so you can change and view it before hand.
$user_id = "";
if(isset($_POST['id']))
{
$user_id = $_POST['id'];
}
$query = "SELECT username,email,display_name,access,password FROM users WHERE user_id='$user_id'";
$results = mysql_query($query);
if(!$results) { //Check to see if query failed
die(mysql_error());
}
$resultsfetch=mysql_fetch_array($results);
$username = $resultsfetch['username'];
$usernamenew = $_POST['usernameinput'];
if(isset($_POST['usernameinput'])) {
$usernamenew = cleanString($_POST['usernameinput']);
}
if($usernamenew !=$username){
$submit = "UPDATE users SET username = '$usernamenew' WHERE user_id = '$user_id'";
mysql_query($submit);
if(!$submit) { //Check to see if query failed
die(mysql_error());
}
}
It's probably something stupid or simple that I missed, or something really huge. Mainly because I am absent minded.
$submit = sprintf("UPDATE users SET username = '%s' WHERE user_id = %d",mysql_real_escape_string($usernamenew),mysql_real_escape_string($user_id));
If the page is loaded, $user_id will be NULL so noting will be updated! Make sure that this page loads, by sending $_POST['id'] . if these things are correct, check this.
"Did the database user have any permission to update the table? "
I have re-arranged your code. added comments where i changed. Try this
if (isset($_POST['id'], $_POST['usernameinput'])) { // Check if both POST id and usernameinput is available
$user_id = (int)$_POST['id']; //assuming this is an integer
$query = "SELECT username,email,display_name,access,password FROM users WHERE user_id='$user_id'";
$results = mysql_query($query);
if (!$results) {//Check to see if query failed
die(mysql_error());
}
if (mysql_num_rows($result) > 0) { //verify if there is really a user with such id
$resultsfetch = mysql_fetch_array($results);
$username = $resultsfetch['username'];
$usernamenew = cleanString($_POST['usernameinput']);
if ($usernamenew != $username) {
$submit = "UPDATE users SET username = '$usernamenew' WHERE user_id = '$user_id'";
if (!mysql_query($submit)) {//Check to see if query failed
die(mysql_error());
}
}
}else{
die("no such user with userid=$user_id");
}
}
Warning: mysql_ function is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used.
So, I guess I figured it out. It's an issue with my code carrying over to the next page.
The code I had been shown only broke the page, whether it be missing an integer, or something else. I'm not 100% sure.
Thanks for all the help guys, but now I know the issue.
EDIT:
I had forgotten to echo the $user_id in my hidden field.

Get user info from mysql database

i am making my own php game. So far i have made almost everything. Now to finish it, i need to get id from user who is logged in. I'm not so familiar with the functions and sessions. Please help.
This is what i made so far:
In my index page people login. then they are redirected to this.
So $_POST['username'] is where user type his user name in index.
<?php
$username = $_POST['username'];
include("Files/config.php");
$connect = #mysql_connect(DB_SERVER, DB_USER, DB_PASSWORD);
if($connect) {
if(mysql_select_db(DB_NAME)) {
$sql = mysql_query("SELECT * FROM users WHERE `username`='$username'") or die(mysql_error());
$gatherinfo = mysql_fetch_array($sql);
global $getid;
$getid = $gatherinfo['id'];
echo $getid;
function getuid() {
$_SESSION['getuid'] = $getid;
echo $getid;
}
}
}
else{ echo "Can not connect";}
?>
I searched other scripts for this, i found on one it says just $session->uid and it shows his id from mysql.
In mysql database i have table users with info about them
Id, username, password (password is hashed), email,...
Please help me if you can :D
At the beginning of index file (where your user logging in) start named session (be careful to avoid echo or print any values before session_start:
<?php //index.php
session_name('SAMPLESESSION');
session_start();
then when you will get the logged User ID, write this value to the session variable, like this:
.....
$_SESSION['uid'] = $getid;
.....
in the script you was redirected by your index file start session with the same name and get your user ID:
<?php //redirectedfromindex.php
session_name('SAMPLESESSION');
session_start();
echo $_SESSION['uid'];
....
If I right understand you, these that you need.

This is my php for viewing user profile

My Profile php
<?php
//profile.php
require_once 'includes/global.php';
//check to see if they're logged in
if(!isset($_SESSION['logged_in'])) {
header("Location: login.php");
}
// finding user and viewing it
$tools = new FindUser();
$user = $tools->get($_REQUEST['userID']);
?>
This is my php for viewing user profile.
http://mywebsite.com/profile.php?userID=5 its working fine in this way.
i want my code to check if user is available in database for example if i add ?userID=10 which is not present in database it gives out mysql error or even if i use http://mywebsite.com/profile.phpthen also it give error.
so now i want if user is not available in database it should give that user is not available and when we use simple http://mywebsite.com/profile.php it should give auto add it to userID=1 OR REDIRECT it to home.php
If there is other way of doing this please let me know. well im very newbie in this field
Thanks for looking my question and answering :)
Solved
<?php
//profile.php
require_once 'includes/global.php';
//check to see if they're logged in
if(!isset($_SESSION['logged_in'])) {
header("Location: login.php");
}
$UserID = $_GET['userID'];
$CheckQuery = mysql_query("SELECT * FROM users WHERE id='$UserID'");
$CheckNumber = mysql_num_rows($CheckQuery);
if ($CheckNumber !== 1)
{
header("Location: index.php");
}
// finding user and viewing it
$tools = new FindUser();
$user = $tools->get($_REQUEST['userID']);
?>
You shouldn't use MySQL As it's depreciated,
If you really wish to use MySQL You could check at the start of the script if there is a row count for the User ID, Example:
<?
$UserID = $_GET['UserID'];
$UserID = mysql_real_escape_string($UserID);
$CheckQuery = mysql_query("SELECT * FROM users WHERE userID='$UserID'");
$CheckNumber = mysql_num_rows($CheckQuery);
if ($CheckNumber !== 1)
{
// Do something If user is Not Found
// Redirect to Another Page OR Something
}
?>
than check that query give with result if it wont found data in database than redirect
$result = mysql_query(...);
if(mysql_num_rows($result) !=1){ //
header("Location:signup.php");
exit();
}
You shouldn't use MySQL As it's depreciated, either use PDO or mysqli

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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