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PHP - How Do I Create A 'admin Approved' Entry?
Hello. I'm not a php programmer. I just play one on the internet. Point being that I know enough to make small things happen and can slop some code together for little ideas. For my newest idea, I'm trying to have a user submitted form that requires admin approval. I know how to create a form and do the mysql insert, but how I do create a "hold" on the data being sent from a form, so that I can require approval? I'm looking for the easiest way.. If I need to approve directly through mysql (instead of creating an admin.php page), then that would be fine. I don't anticipate a lot of submissions. Just wasn't sure how this is usually handled.
Thanks! Similar Tutorialsplease i need tutorial to learn me how to create admin panel like jommla or other >>> please dont late to answer this is my first topic Hi all, firstly apologies as this is a cross post from another forum and we have hit a block.. I am hoping that opening this up to another set of gurus we can get a resolution. What I am trying to achieve is this... I have 2 tables Main and FinancialYear. Main holds all data which I use a form to post the data to it..(all works fine). I use this code to create a drop down in the insert.php form. again this works. Code: [Select] <tr><td>Financial Year: xxxx/xxxx</td><td> <!-- pulls the data from the table variable to populate the dropdown menu --> <?php $database = 'Projects_Main'; $fintable = 'FinancialYear'; if (!mysql_connect($db_host, $db_user, $db_pwd)) die("Can't connect to database 'cos somethin' is wrong"); if (!mysql_select_db($database)) die("Can't select database"); $result = mysql_query("SELECT FinancialYear_id, FinancialYear FROM {$fintable} order by FinancialYear"); $options=""; while ($row=mysql_fetch_array($result)) { //$id=$row["FinancialYear_id"]; $thing=$row["FinancialYear"]; $options.="<OPTION VALUE=\"$thing\">".$thing.'</option>'; } ?> <SELECT NAME="FinancialYear"> <OPTION VALUE=0>Choose</OPTION> <?=$options?> </SELECT> </td></tr> What I have done is built another form which list all records in the database and creates an update url for every record that passes the field Project_id where i use $_get to retrieve the Project_id to retrieve the relevant data into the update.php form. I am able to populate the form with all the correct information BUT I am looking to introduce some dropdowns to aid updating the data and provide consistency to the data. . Code: [Select] // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db('Projects_Main')or die("cannot select DB"); // get value of id that sent from address bar $Project_id=$_GET['Project_id']; //define vars $FinancialYear=$_POST['FinancialYear']; // other vars defined here also.. about 30 // Retrieve data from database $sql="SELECT * FROM Main WHERE Project_id='$Project_id'"; $result=mysql_query($sql); $rows=mysql_fetch_array($result); ?> //update_record_ac.php posts the data to the dbase. <form name="form1" method="post" action="update_record_ac.php"> <center> <table> <tr><td><b>Store Details<b></td></tr> <tr><td>Financial Year:</td><td> // takes the data from $rows and present to form <input name="FinancialYear" type="text" id="FinancialYear" value="<?php echo $rows['FinancialYear']; ?>"> // this is where I need to create the drop down.. see my other comments in the post..... </td></tr> the financialYear table consists if the following; financialyear_id - pri, auto inc. ---- data format is 2010/2011, 2011/2012.... financialyear the main table contains 30 fields .. won't list em all... Project_id - pri, auto inc financialyear I need the drop down to pull the data from the financialyear table and then to present or focus on the currently stored data... so if the store value in the table Main is 2010/2011 if Ii was to select the update url in the list_record.php it will pull all the relevant data into update_record.php form. the financialyear field in the form should be a dropdown with all the financial years listed but the 2010/2011 is selected or focused. I still need to be able to change the entry and post this back to the table Main..... So the dropdown contains the list of years from the financialyear table but when the record is pulled from table main the year that is stored in table Main should be highlighted in the dropdown and I should be able to select a new record and post back to the table Main.. any thoughts... please don't slate for the cross post, I haven't sanatised the data at any stage. I know i'm open to injection attacks. and yes my code is a little dirty... all these will be rectified as i finalise the process and ensure the consept works. Thanks for taking the time to read and hopefully you are able to understand the requirement and are able to assist. thanks Balgrath Alright, wasn't quite sure how to summarize this in the title, but I want to: Check if a user status is "active" or not based on the UserName input. I have a table witch holds: Code: [Select] VarChar Username Var CharPassWord int Active Ted TedsPW 1 something like the above(assuming it formatted correctly. In my php script I will want to input a variable for Username to check for: inputUN in this example would be "Ted". $UserNameToCheck = $_GET['inputUN']; Then I want to check for that UserName in the database, if it exists, I want pull the value for the "Active" field for just that UserName and echo it. Thanks for any help. I am using php to upload a file to my server, and at the same time inserting the files name and url into my mysql database.
$sql = "UPDATE uploads SET name = '$name', url='$target_path'";
$statement = $dbh->prepare($sql);
$statement->execute();
This is working, however, when I upload a new file, rather than making a new entry in my database, it just overwrites the first one.I'm quite new at mysql so was wondering how I would make it add new entrys instead of overwriting the current one? 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 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 Hello, Do you know where I can download a nice looking PHP admin dashboard for free? Thanks in advance for the help 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 $end error it is the last line of the code and leaves me lost.. Something I'm missing guys? As always, thanks in advance Hallo I have a problem.
This is my code:
<?php include 'connect.php';
?>
<html>
<head>
<title>Admin Insert page!</title>
</head>
<body>
<?php
error_reporting(-1);ini_set('display_errors',1);
if (isset($_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. 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?
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> 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 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 Hi guys, Can anyone help me; I have created a registration form (can be use for create or modify) and login form (Admin). What I am trying to do is; once the admin log in he/she can create / register a new user which contains: -Firstname -Surname -Address -Mobile -Dept Name -Username -Password -Repeat Password My DB will look like this: Table PERSONS: id, firstname, surname, address, mobile, dept_id, username, password. Table USER: id, username, password Table DEPT: id, dept_name Can anyone help me how am I going to related the USER table into the PERSONS so when admin register a new user - the data will be created the into database as well as the data can be extracted for modification. Any suggestion? Here are my code: register.php <?php require 'includes/application_top.php'; if (!isset($_POST['name']) && isset($_GET['id'])) { $mode = "Modifying"; // Get data from DB $q = "SELECT * FROM `persons` WHERE `ID` = '".$_GET['id']."'"; $result = mysql_query($q) or die (mysql_error()); $row = mysql_fetch_array($result); $name = $row['firstname']; $surname = $row['surname']; $address = $row['address']; $dept = $row['dept_id']; $mobile = $row['mobile']; }else if (!isset($_POST['name']) && !isset($_GET['id'])) { $mode = "Register"; // Data is empty $name = $surname = $address = $dept = $mobile = ""; } else { $errors = array(); if ($_POST['name'] == "") $errors[] = "Name"; if ($_POST['surname'] == "") $errors[] = "Surname"; if ($_POST['mobile'] == "" || !is_numeric ($_POST['mobile'])) $errors[] = "Mobile No"; if (count($errors)) { $errormsg = "Please fill the blank info:<br/ >".implode('<br />',$errors); $mode = $_POST['mode']; $name = $_POST['name']; $surname = $_POST['surname']; $address = $_POST['address']; $dept = $_POST['dept']; $mobile = $_POST['mobile']; } else { foreach ($_POST as $key => $val) { $_SESSION[$key] = $val; } header("Location: confirmPage.php"); } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Modify Document</title> </head> <body> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <?php if (isset($errormsg)) echo "<div id=\"error_message\" style=\"color:red;\">$errormsg</div>"; ?> <div align="center"> <table width="370" border="0"> <h1> <?php echo $mode; ?> A User </h1> <p><font color="orangered" size="+1"><tt><b>*</b></tt></font> indicates a required field</p> <tr> <th width="200" height="35" align="left" scope="row" >First Name <font color="orangered" size="+1"><tt><b>*</b></tt></font> </th> <td width="160"><input type="text" name="name" value="<?php echo $name;?>" size="25"/></td> </tr> <tr> <th height="35" align="left"> Surname <font color="orangered" size="+1"><tt><b>*</b></tt></font> </th> <td> <input type="text" name="surname" value="<?php echo $surname; ?>" size="25"/></td> </tr> <tr> <th height="35" align="left"> Address</th> <td> <input type="text" name="address" value="<?php echo $address; ?>" size="25"/></td> </tr> <tr> <th height="35" align="left"> Choose a username <font color="orangered" size="+1"><tt>*</tt></font></th> <td> <input name="username" type="text" maxlength="100" size="25" /> </td> </tr> <tr> <th height="35" align="left"> Choose a password <font color="orangered" size="+1"><tt><b>*</b></tt></font> </th> <td> <input name="password" type="password" maxlength="100" size="25" /> </td> </tr> <tr> <th height="35" align="left"> Repeat your password <font color="orangered" size="+1"><tt><b>*</b></tt></font> </th> <td> <input name="repeatpassword" type="password" maxlength="100" size="25" /> </td> </tr> <tr> <th height="35" align="left">Department</th> <td> <select name="dept"> <option value="">Select..</option> <?php $data = mysql_query ("SELECT * FROM `dept` ORDER BY `id` DESC") or die (mysql_error()); while($row_dept = mysql_fetch_array( $data )) { ?> <option value="<?php echo $row_dept['id'] ;?>" <?php if($row_dept['id']==$dept){echo ' selected="selected"';}?>> <?php echo $row_dept['dept_name'] ;?> </option> <?php } ?> </select> </td> </tr> <tr> <th height="35" align="left">Mobile</th> <td><input type="text" name="mobile" value="<?php echo $mobile; ?>" size="25"/></td> </tr> <tr> <td align="right" colspan="2"> <hr noshade="noshade" /> </td> </tr> </table> <br/> <a href="index.php"> <input type="button" name="back" value="Back" /></a> <input type="hidden" name="id" value="<?php echo isset($_GET['id']); ?>"> <input type="hidden" name="mode" value="<?php echo $mode; ?>"> <input type="submit" value="<?php echo ($mode == "Register") ? 'Register' : 'Modify'; ?>"/> </div> </form> </body> </html> Hello, I have a problem with my website, Admin login page (http://www.tranceprofile.com/storeadmin/admin_login.php I can not login to my Admin controle panel. Login information: Username: Mitch Password: schuur111 Username: Admin Password: poopoo Can someone help me ? Here is my admin_login.php source code. If you need some other source code in my /storeadmin folder please tell Code: [Select] <?php session_start(); if (isset($_SESSION["manager"])) { header("location: index.php"); exit(); } ?> <?php // Parse the log in form if the user has filled it out and pressed "Log In" if (isset($_POST["username"]) && isset($_POST["password"])) { $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); // filter everything but numbers and letters $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); // filter everything but numbers and letters // Connect to the MySQL database include "../storescripts/connect_to_mysql.php"; $sql = mysql_query("SELECT id FROM admin WHERE username='$manager' AND password='$password' LIMIT 1"); // query the person // ------- MAKE SURE PERSON EXISTS IN DATABASE --------- $existCount = mysql_num_rows($sql); // count the row nums if ($existCount == 1) { // evaluate the count while($row = mysql_fetch_array($sql)){ $id = $row["id"]; } $_SESSION["id"] = $id; $_SESSION["manager"] = $manager; $_SESSION["password"] = $password; header("location: index.php"); exit(); } else { echo 'That information is incorrect, try again <a href="index.php">Click Here</a>'; exit(); } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Admin Log In </title> <link rel="stylesheet" href="../style/style.css" type="text/css" media="screen" /> </head> <body> <div align="center" id="mainWrapper"> <?php include_once("../template_header.php");?> <div id="pageContent"><br /> <div align="left" style="margin-left:24px;"> <h2>Please Log In To Manage the Store</h2> <form id="form1" name="form1" method="post" action="admin_login.php"> User Name:<br /> <input name="username" type="text" id="username" size="40" /> <br /><br /> Password:<br /> <input name="password" type="password" id="password" size="40" /> <br /> <br /> <br /> <input type="submit" name="button" id="button" value="Log In" /> </form> <p> </p> </div> <br /> <br /> <br /> </div> <?php include_once("../template_footer.php");?> </div> </body> </html> I need to know how to allow admin to essentially be "all users" in otherwords, edit everyones profile and not just their own. Tell me what codes would be helpful and i will send them on in. By the way, i do have a script that allows members to edit own profile. Hey all, I'm using cakephp and it asks me a question and I'm not sure what to put in, because I don't necessarily know the consequences of what I put in: Code: [Select] Would you like to create the methods for admin routing? (y/n) [y] > y You need to enable Configu :write('Routing.admin','admin') in /app/config/core.php to use admin routing. What would you like the admin route to be? Example: www.example.com/admin/controller What would you like the admin route to be? [admin] > Thanks for response hi guys how you doing? i new here so take it easy on me . basically just need some quick help and i thought this would be the best place to ask. ive been working on a admin login script but cant seem to get it right, i mean i can login in with random passwords :/ and also everytime i go to the index.php it shows the information i dont want it without being logged in. ive got the script running live just incase anyone wants to see what i mean its at http://www.lukerodham.co.uk/admin heres the code. Thanks in advance. index.php Code: [Select] <?php require_once("login.php"); $adminuser = $_SESSION['user']; ?> <html> <head> <title>hoonigans.co.uk</title> </head> <body> <h3 align="center">Welcome to the admin page.</h3> <span class="maintext"><br /> <p align="center">If you would like to post some news please <a href="news/post.php">click here</a>.<br /> To logout please <a href="logout.php">click here</a></p> </body> </html> login.php Code: [Select] <?php function loginpage($error){ echo " <html> <body> <div align='center'> <form method='post' action='".$_SERVER['REQUEST_URI']."'> <label>username: <input type='text' name='username' id='username'><br> <label>password: <input type='password' name='password' id='password'><br> </label> <label> <input type='submit' name='submit' id='submit' value='submit'> </label> </form> </div> </body> </html> "; } $username = $_POST['username']; $password = $_POST['password']; $login = $_post['login']; $host = *********; $dbuser = *********; $dbname = *********; $dbpass = *********; mysql_connect("$host","$dbuser","$dbpass"); mysql_select_db("$dbname"); session_start(); if($_SESSION['user'] != $username){ if(!$submit){ loginpage(false); } elseif($submit){ $get = mysql_query("SELECT * FROM users WHERE username='$username'"); while ($row = mysql_fetch_assoc($get)){ $admin = $row['admin']; $passwordmatch = $row['password']; if ($passwordmatch==$password&&$admin==1){ $_SESSION['user']="$username"; echo "this worked"; } else{ die("Sorry wrong information."); } } } } ?> |
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