|
PHP - Admin Part Not Inputting Into Database
This is the second script I’ve installed on the same site and admin has been able to delete from the database, but somehow still can’t insert into database? My config is set right, but can’t understand why two different scripts are having the same issue? Similar TutorialsHi! So I'm fairly new to PHP, and am currently making a website from scratch including as much PHP functionality as I can for learning purposes. As I have just finished making a blog function, I started to wonder. Is it normal to store every part of a website in a the database? The blog part of the website I just created has of course all the differet blog entries stored in a table in the db, but is it also normal practice to store about, contact etc. in tables too? Or should categories like these be created using brand new .php documents? I am in the progress of making a page controller and thought I should clearify things first, before I go any further. This question may seem stupid, but I don't know alot about what the normal practices of doing things like these in php are. Anyways, hope someone can help me out! Hi, I'm trying to type in a name of a song into an input field, for example: I'll Be Missing you This field is captured through $_POST and set to a variable $title I then update the table with this new title. Once it is updated, all that is shown in the data is: I The single quote, and anything after it is gone completely. Here is my query. How can I change this so it includes the single quote and everything after it? $sql = "UPDATE sheets SET artist = '$artist', title = '$title', active = '$activestatus' WHERE id = $value"; $result = mysql_query($sql) or die(mysql_error().'<br>'.$sql); If more code is required to understand what I'm talking about, let me know. Code: [Select] <? $out = preg_replace('/^(.{701}[^.]*).*/i','$1.',$detrsltnewsrow[news_desc]); echo $out; ?> </td></tr><tr><td colspan="2" class="para" style="padding-left:10px;"> <?= substr(stripslashes(trim($detrsltnewsrow[news_desc])),701) ?> </td></tr> I have the above snippet.. The first php statement, basically grasp the first 701 characters with the closet next stop "." character and out puts it. then out puts the HTML tags I have a problem with the second statement. I want to output anything after what has been outputted by: Code: [Select] <? $out = preg_replace('/^(.{701}[^.]*).*/i','$1.',$detrsltnewsrow[news_desc]); echo $out; ?> So need the correct syntax for Code: [Select] <?= substr(stripslashes(trim($detrsltnewsrow[news_desc])),701) ?> Currently it breaks at exactly the 701 character, want it to continue from the sentence the first code ended in. My query gets the results and orders by one of the fields. Once I get the MySQL results I would like to find the first entry that has a letter as the first character of the same field that the list was ordered by, then split the results in to two parts and swap them. So that the results that have a letter at the start of the same sorted field are as the begining and the results that have the numbers as the start at the end of the array. But also so that the array works the same way as the original results string, so i can use say $results['mysqlfield'] hello. I need your help please. I'm building logistics website with user panel and admin panel. I've done all login and register forms. now I want to : admin can add package with: tracking number , weight , cost , and declaration form. user can fill declaration form after admin add package to user panel. then admin can see the declared form. is it possible in php? thank you in advance Hello, Do you know where I can download a nice looking PHP admin dashboard for free? Thanks in advance for the help I get this error: Code: [Select] Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\user\user.php on line 5 code: user.php: Code: [Select] <?php $get = (isset($_GET['id'])); //this means that user.php?id=1 would mean $get = 1. Note: This is not SQL Inject protected. $users = mysql_query("SELECT * FROM users WHERE id='".$get."'"); while ($row = mysql_fetch_array($users)) { echo ' Id = '.$row['id'].' Name = '.$row['name'].' Username = '.$row['username'].' Password = '.$row['password'].' Reg. on = '.$row['date'].' '; } ?> <html> <body> <form action='user.php' method='GET'> Username: <input type='text' value=''> <input type='submit' value='submit'> </form> <?php //what goes here? ?> </body> </html> Hi, I am new here 🙂 I have been learning PHP and am currently working on my own OOP MVC CMS. I am up to the stage where I would like to start working on the admin area, but I am not sure how best to organise things. Should I create admin specific Controllers and Models? In Views, should I create a sub directory Admin, and place all admin view templates within it? Are there any good books on OOP/MVC you would recommend?
Hallo I have a problem.
This is my code:
<?php include 'connect.php';
?>
<html>
<head>
<title>Admin Insert page!</title>
</head>
<body>
<?php
error_reporting(-1);ini_set('display_errors',1);
if (isset($_POST['submit'])){
$name = $_POST['name'];
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM users WHERE user='$name' AND password='$password'");
$num = mysql_num_rows($result);
if($num == 0){
echo "Bad login, go <a href='login.php'>back</a>";
}else{
session_start();
$_SESSION['name'] = $name;
header("Location: admin.php");
}
}else{
?>
<form action='login.php' methody='post'>
Username: <input type='text' name='name'/><br />
Password: <input type='password' name='password'/><br />
<input type='submit' name='submit' value='Login' />
</body>
</html>
I try to use console to find the problem but I didn't....
I know that there is some problem with $num
Can somebody help me?
Thank you.
Edited by Artur, 19 October 2014 - 12:11 PM. Hey, So i have an admin.php page that lists all of the users in the database and im wondering how i can add functions so the administrator can delete / ban the user from the webpage i'm not sure on how you would select the user?
Hello I am trying to add an IF statement to my login script so that if the username entered is 'admin' it directs to 'adminpage.php Here is my script: <?php include ("connection.php"); session_start(); // Collect data from form and save in variables //See if any info was submitted $Username = $_GET['Username']; //Clean data - trim space $Username = trim ( $Username); //Check its ok - if not then add an error message to the error string if (empty($Username)) $errorString = $errorString."<br>Please supply Username."; //See if any info was submitted $Password = $_GET['Password']; //Clean data - trim space $Password = trim ( $Password); //Check its ok - if not then add an error message to the error string if (empty($Password)) $errorString = $errorString."<br>Please supply Password."; // Query to search the user table $query= "SELECT * FROM Users WHERE Username='$Username' AND Password='$Password'"; // Run query through connection $result = mysql_query ($query); $row = mysql_fetch_assoc($result); // if rows found set authenticated user to the user name entered if (mysql_num_rows($result) > 0) { $_SESSION["authenticatedUser"] = $Username; $_SESSION['UserID'] = $row['UserID']; // Relocate to the logged-in page header("Location: loggedon.php"); } else // login failed redirect back to login page with error message { $_SESSION["message"] = "Could not connect as $Username " ; header("Location: login.php"); } ?> Thank you Any help would be greatly appreciated! <?php $host="localhost"; // Host name $username="user"; // Mysql username $password=""; // Mysql password $db_name=""; // Database name $tbl_name=""; // Table name mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $barcodeID=$_POST['barcode']; echo $barcodeID; $barcodeID = stripslashes($barcodeID); $barcodeID = mysql_real_escape_string($barcodeID); $sql="SELECT * FROM $tbl_name WHERE BarcodeID='$barcodeID'"; $result=mysql_query($sql); // Mysql_num_row is counting table row $count=mysql_num_rows($result); $count=mysql_num_rows($result); if($count==1){ $_SESSION['barcode'] = $barcodeSession; $_SESSION['userlevel'] = $row['Priority']; if($row['userlevel'] == "Admin") { header("location:AdminSection.php"); }else{ header("location:index.php"); } header("location:LoggedIn.php"); } else { header("location:index.php"); } ?> when the script has been run, I want it to redirect to either the user page or admin page depending on their priority level. if Priority field == "Admin" then go to admin page. Can you see anything missing? Thank You Hey guys, I've set up a database with a login and logout script for my site.. There is a TINYINT value called admin and it either equals 1 or 0 depending on whether the user is an admin or not.. The registration script works perfectly to create the table value and the login script works fine for the site.. The question I had was if I wanted to add a link to the bottom of every page that said: Go to Administration Panel and make it only viewable by ADMINS I figured this little script would work.. Here would be the end of the page: Code: [Select] <br /> <center>Copyright © 2010 <a href="http://www.website.com">www.WEBSITE.com</a></center> <?php include('includes/start_admincheck.php'); ?> <center><a href="<?php echo $homedir .'admin.php'; ?>">Go to Administration Panel</a></center> <?php include('includes/end_admincheck.php'); ?> </body> </html> Inside start_admincheck.php I have: (NOTE: $cUsername refers to a setcookie and $cAdmin does as well.. They are defined on my Variable page included at the top.) Code: [Select] <?php include('variables/variables.php'); ?> <?php mysql_connect("$mysql_hostname", "$mysql_username", "$mysql_password") or die(mysql_error()); mysql_select_db("$mysql_database") or die(mysql_error()); if(isset($cUsername)) { $check = mysql_query("SELECT * FROM users WHERE username = '$cUsername'")or die(mysql_error()); while($info = mysql_fetch_array( $check )) { if (($cAdmin == 1) && ($info['admin'] == 1)) { ?> And this is the end_admincheck.php Code: [Select] <?php include('variables/variables.php'); ?> <?php } else die(); } } else die(); ?> ?> I get this Parse error thrown at the bottom of the page: Code: [Select] Parse error: syntax error, unexpected $end in /*******/includes/start_admincheck.php on line 15 Naturally I would checkout line 15 in start_admincheck.php, but normally when I get an $end error it is the last line of the code and leaves me lost.. Something I'm missing guys? As always, thanks in advance Hey, in a nutshell the only thing in admin.php is the ability to moderate unapproved images, however, once approved, the "Approve Delete" links are still on screen. How it works is a user uploads an image, the filename is added to mysql and the image is added to uploads/ once I Approve an image, the image is then moved to img/ to display on the index.php (to prevent porn and anything that doesn't belong to the general public). I know what's happening, because I've got while loops to display the image while looping through the mysql database, so once the image is moved, the links are still on screen, displaying an "Approve Delete" for every image in the database. Also another thing that happens is the images on index.php are blank until approved. How can I work around this? Here is the index.php when an image hasn't been approved: http://www.xodiac.net/1.png And here is the admin.php displaying Approve and Delete once an image has been approved: http://www.xodiac.net/2.png So i got my login down and the cookies, kinda set up my problem is how do i make the admin panle save the true/false in the string in settings.php id like do do it with a drop down menu to enable/disable it. any help? Code download i want that when someone post some comment on post that don't insert in table directly before inserting i can check and after approving then insert in table for which what i doo ?any IDea about this I am using the following to check that the user is logged on before he/she views pages on my site can I adapt what is here so that only some pages can be viewed by admin only? <?php include("../php/dbconnect.php"); //connects to the database //session code session_start(); //Check if user is authenticated if(!isset($_SESSION['username'])){ //User not logged in, redirect to login page header( "Location: http://webdev/schools/hhs/psy_bookings/" ); } else { //User is logged in, contiue (use session vars to diplay username/email) //echo "'Welcome, {$_SESSION['username']}. You are still logged in. <br />'"; // echo "'Your email address is: {$_SESSION['email']}.'"; }//end of session code ?> Ok.. I have made an admin section for a couple of sites, and it works fine, alough i'm almost certain that the way i have done it is not the "proper way". What I have done is had a form with a password field that sends the password via POST to another script that checks the password is correct via variable, something like this: Code: [Select] <?php $pass = $_POST["pass"]; $correct = "imapassword"; if($pass == $correct){ //display contents of page else{ //return user to login screen } ?> And then it saves the password into a cookie, that dies either over time or when the browser closes. And then on each page it tests wether that cookie is still there and is correct. When the user wants to log out it just destroys the cookie. This seems like a really hashed up way of doing it, could anybody let me know the bare essentials for making a similar system, but the "right way". Thankyou in advance I am trying to check for an admin user to access the admin panel. I have been playing around try different things and this what I have ended up with in my database table I have a column called usergroup and i do the follow to check for admin user. Code: [Select] $checkAdmin = mysql_query("SELECT * FROM `users` WHERE email='$email' , usergroup = 'admin'"); $adminUser = mysql_num_rows($checkAdmin); if ($adminUser == 0) { echo count($adminUser); die ('You do not have permissions to access this area'); } I do the select statement through phpmyadmin and it comes back with one row. which is basically hat i want to check for. I do have a variable called $email which is getting a value from the email cookie. currently $adminUser Return a value of 10. All of the count() functions is for testing purposes only. |
|||||||