PHP - How Can I Play With These Role In My Database?
How can i play with these role in my database table as you can see here //I have a database table CREATE TABLE users ( id int(10) NOT NULL PRIMARY KEY AUTO_INCREMENT, urole varchar(10) ); in the role column, I have 2 roles first role is only used once but the sound one can be assigned to more than one user, now I want to check my table if the first role is already registered, then we can't register user with that role, also the second role can exist two or more times //check role, a variable user role has been defined as $urole = $_POST['urole']; $check = $connect -> prepare('SELECT * FROM users WHERE urole = ?'); $check -> execute([$urole]); $checkfetch = $check -> fetch(); //I'm stacking here, I want to put $checkfetch['urole'] in rowCount() to be counted but I think this is not a correct way of using rowCount() if(($checkfetch['urole'] == 'MainAdmin') && ($checkfetch ->rowCount() == 1)) { echo 'This role can be used by only once!'; } //another role if(($checkfetch['urole'] == 'NormalAdmin') && ($checkfetch ->rowCount() < 4)) { echo 'This role can be used 4 times only!'; }
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Hi, I need code that reads from the roles database and then selects which file from these 3 which I want. For example, the user.php file would be loaded if the user has UName = user, Pass = 124, and Roles = User added to the database. But the admin.php and boss.php files would not appear to him.
<?php session_start(); if(!(isset($_SESSION['User']))) { header("Location: index.php"); exit(0); } ?> <!DOCTYPE html> <html> <body> <?php include "config.php"; ?> <!--show for User--> <?php include 'user.php';?> <!--show for Admin--> <?php include 'admin.php';?> <!--show for Boss--> <?php include 'boss.php';?> </body> </html>
Ok say i have a database with users who have a video url field for them to enter a video url to be submitted to the database, im trying to create a page where the videos will be displayed one at a time for example ' 1st video plays then it switches to the next random video from the database and so on' i want only to play so many seconds of the video before it switches as they will be presentations so i dont want viewers to watch them all in full, could anyone help me out with some code or feedback on this please? I'm not totally new to php but I'm no guru either. I'm building an intranet and have three seperate user roles, user - manager - admin. There are some menu items I don't want to allow simple users to see. This will expand later to give them different views of a page as well (some can view/others can edit). As it stands the login validation is holding they're user level in $SESSION. To give you an idea take a look at what I'm trying to do: Code: [Select] function usermenu($usermenu) { if($user_level=0) echo ("<ul id="gooeymenu2" class="solidblockmenu"> <li><a href="main.php">Home</a></li> <li><a href="forms.php">Forms</a></li> <li><a href="/support/index.php" target="_new">Support</a></li> <li><a href="documents.php">Documents</a></li> <li><a href="admin/index.php">Admin</a></li> <li><a href="logout.php">Logout</a></li> </ul> <script> gooeymenu.setup({id:'gooeymenu2', selectitem:1, fx:'swing'}) </script>" "); else($user_level=1,2) echo ("<ul id="gooeymenu2" class="solidblockmenu"> <li><a href="main.php">Home</a></li> <li><a href="forms.php">Forms</a></li> <li><a href="/support/index.php" target="_new">Support</a></li> <li><a href="documents.php">Documents</a></li> <li><a href="new.php">New Adviser</a></li> <li><a href="admin/index.php">Admin</a></li> <li><a href="logout.php">Logout</a></li> </ul> <script> gooeymenu.setup({id:'gooeymenu2', selectitem:1, fx:'swing'}) </script>" "); Any help would be great, thanks in advance. Jason Hi all,
I am working on a project where i need to implement rbac control. I sthere any library available in codeigniter to extend the functionality. I have started working on codeigniter. i want to implement this in codeigniter. Please some on e guide how to achieve that. how to check roles and permssions.
Code what i made so far. Your comments at what should i do differently.
My configs.php
<?php $userQuery = 'SELECT * FROM users WHERE id = :id'; $user = $db->prepare($userQuery); $user->bindParam(':id', $_SESSION['userId'], PDO::PARAM_INT); $user->execute(); $userInfo = $user->fetch(PDO::FETCH_ASSOC); ?>functions.php <?php function loginCheck(){ global $db; if(isset($_SESSION['userId'], $_SESSION['loginString'])){ $query = 'SELECT username FROM users WHERE id = :id'; $user = $db->prepare($query); $user->bindParam(':id', $_SESSION['userId'], PDO::PARAM_INT); $user->execute(); $row = $user->fetch(PDO::FETCH_ASSOC); if($user->rowCount() == 1){ if(hash('sha512', $row['username'].$_SERVER['HTTP_USER_AGENT']) == $_SESSION['loginString']){ return true; }else{ return false; } }else{ return false; } }else{ return false; } } function checkUserRole(){//can be user, admin and moderator global $userInfo; if($userInfo['userRole'] == 'admin' or $userInfo['userRole'] == 'moderator'){ return true; }else{ return false; } } ?>shoutbox.php Can this be done with one query? global $db, $userInfo; $sbQuery = 'SELECT * FROM shoutbox ORDER BY dateCreated DESC LIMIT 30'; $sb = $db->query($sbQuery); $usersQuery = 'SELECT * FROM users WHERE shoutBoxBan = "yes"'; $users= $db->query($usersQuery); $usersRow = $users->fetch(PDO::FETCH_ASSOC); $hiddenAction = ''; while($sbRow = $sb->fetch(PDO::FETCH_ASSOC)){ if(loginCheck() and checkUserRole()){ $hiddenAction = " <a href=\"javascript:;\" onClick=\"deleteMessage('".$sbRow['id']."')\" class=\"shoutBoxDelete\" title=\"Delete\">x</a>"; if($usersRow['username'] == $sbRow['username']){ $hiddenAction .= " <a href=\"javascript:;\" onClick=\"unBan('".$sbRow['username']."')\" class=\"shoutBoxBan\" title=\"Unban\">u</a>"; }else{ if($userInfo['username'] != $sbRow['username']){//admin and moderator cant ban themselves. $hiddenAction .= " <a href=\"javascript:;\" onClick=\"banUser('".$sbRow['username']."')\" class=\"shoutBoxBan\" title=\"Ban\">o</a>"; $hiddenAction .= " <a href=\"javascript:;\" onClick=\"tempBanUser('".$sbRow['username']."')\" class=\"shoutBoxBan\" title=\"Temp Ban\">ø</a>"; } } } .................................... Hello all! So glad I found this forum. I would appreciate some assistance please. I'm working on a filter for a Custom Post Type . I need it to filter the list depending on the user's role. The way this should work is the following...
* Users in roles "formusers1" and "formusers2" can post. Users can only see their own posts. So far I can filter by roles "formusers1" and "formusers2" using `$query->set('author', $current_user->ID);` . However, when try to filter the list for role "formchecker1" I see posts from all roles. What am I doing wrong? Here's the rest of the code. Thanks for checking out!
```
function filter_posts_list($query) {
//MY VARIABLES
//FILTERING
if (current_user_can('formchecker2') && ('edit.php' == $pagenow) && $typenow == 'mycustomcpt' ) {
if ((current_user_can('formusers1') || current_user_can('formusers2')) && ('edit.php' == $pagenow) && $typenow == 'mycustomcpt') { I am trying to populate a custom field called "Customer Type" current user role. The custom field is displayed on my checkout page. I tried the below in my functions.php of my child theme and thought it would work but it does nothing. Can anyone tell me what I might be doing wrong?
$user = wp_get_current_user(); $fields['customertype'] = $user;
return $fields; add_filter( 'woocommerce_checkout_fields', 'onboarding_update_fields' ); Edited April 11 by JayXHiya! Is it possible to play an avi or mpg video file with php? If not is there any other recommeded scripts out there be it jquery, ajax, js ect that will do this? Many thanks, James. Hi guys, My first post here. I would like to seek you guys help about playing video from mysql. I wanted to upload video to database and also play it inside my webpage using the media player format. Thank in advance Hi, How to play any website's video using php.I searched many website.All are mention code for play video using a particular website like youtube only or VEOH only or Vimo only and etc. I need play video using any website's URL when we give input as valid url.Send me reply quick please......... Hi, tried google'ing for an answer but found very little help, so decided to post here. I need a way to play a 2 second wav or mp3 file in a shell script. Is there any way to do this? I was hoping for something as simple as this if possible if ($GetHit == 'true') { $PlaySound; echo "Sound Played\n\n"; } Thanks I thought to relax a bit and while i saw something and an idea came to my mind, lets play noobs, it will be fun... we ask noob questions here lolzzz
so my question is
Hello,
I have seen that there are always 3 users /* I know these are bots */ Google, Yahoo and Bing always online and we cannot see their profiles. They must be very professional hackers who have known how to hide their identities.. right??
This topic has been moved to Other. http://www.phpfreaks.com/forums/index.php?topic=320387.0 Here's my script: <?php $ch=curl_init('http://www.pictureinthesky.net/bltest/method1.php'); curl_setopt($ch,CURLOPT_COOKIEJAR,'cookie.txt'); curl_setopt($ch,CURLOPT_COOKIEFILE,'cookie.txt'); curl_setopt($ch,CURLOPT_RETURNTRANSFER,true); curl_setopt($ch,CURLOPT_HEADER,false); curl_setopt($ch,CURLOPT_FOLLOWLOCATION,true); curl_setopt($ch,CURLOPT_POST,true); curl_setopt($ch,CURLOPT_POSTFIELDS,'crime=1&vercode=ABC'); $page=curl_exec($ch); if (curl_errno($ch)) { echo 'ERROR'; } curl_close($ch); echo $page; ?> I'm trying to get that to call the URL in the script to check the anti XSS code in my other script is working but all it does is read the page in without submitting the form -I should get an error message! I've checked and the cookie.txt file is being created and contains: Code: [Select] # Netscape HTTP Cookie File # http://curl.haxx.se/rfc/cookie_spec.html # This file was generated by libcurl! Edit at your own risk. www.pictureinthesky.net FALSE / FALSE 0 PHPSESSID 7268ef0b4b6adae605156ac177bdd43e EDIT: The above script can be tried: http://www.pictureinthesky.net/curl/readpage.php I think it might be failing because cURL hasn't been told the name of the submit button and in my code I'm checking for it by name. This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=327233.0 This topic has been moved to Other. http://www.phpfreaks.com/forums/index.php?topic=307175.0 I have a very tricky php problem here. At least for me. On a page that is editing a job entry. I have a database generated group of checkboxes some of which have been checked when the job was entered. So the script has to do two things generate the complete list of checkboxes and check the ones that have already been selected in the job request. The below script is what I've done trying to figure it out. Description of what I'm trying to do: I'm generating the the boxes with a call to my database for the list of checkboxes. Then I'm building an array with all the possible items that could be checked to compare against. Ok here is where I think my first problem is. I need to make a second query to another table "worklog" in the same database. This is where the list of checked items are stored. What is the best way to do this? A second query seems wrong "or at least not efficient" Code: [Select] <div class="text">Job Type (<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div> <div class="field"> <?PHP $connection=mysql_connect ("localhost", "user", "password") or die ("I cannot connect to the database."); $db=mysql_select_db ("database", $connection) or die (mysql_error()); $query = "SELECT type FROM typelist ORDER BY type ASC"; $sql_result = mysql_query($query, $connection) or die (mysql_error()); $i=1; echo '<table valign="top"><tr><td>'; while ($row = mysql_fetch_array($sql_result)) { $type = $row["type"]; if ($i > 1 && $i % 26 == 0) echo '</td><td>'; else if ($i) echo ''; ++$i; $aTypes = array ("Spec Ad Campaign", "100 x 100 Logo", "100 x 35 (Featured Developer/Broker)", "100 x 40 (Featured Lender)", "120 x 180 (Home Page Auto)", "120 x 45 (Half Tile - Jobs)", "120x60 (Section Sponsor)", "125 x 40 (Profile Page logo)", "135 x 31 (Autos)", "135 x 60 (Autos)", "135 x 60 (Logotile - Jobs)", "150 x 40 (Auto Logo)", "160 x 240 (monster tile)", "160 x 400 (Skyscraper)", "160 x 600 (Tower)", "170x30 (Section Sponsor)", "240 x 180 JPG/GIF Auto Video", "240 x 180 size FLV video ", "300 x 250 (Story Ad)", "300 x 600 (Halfpage)", "468 x 60 (Banner Jpg/Gif)", "728 x 90 (Leaderboard)", "95 x 75 (Sweeps Logo) ", "Agent Profile Page", "Contest", "Creative Change", "Employer Profile", "Holiday Shop", "Home of the Week", "HP Half Banner (234 x 60 Static)", "Mobile Ads (320x53 *300x50*216x36*168x28)", "Newsletter", "Peelback", "Pencil Ad", "Print", "Real Deals", "Resize ad campaign", "Roll-Over Skyscraper", "Site Sponsor logo (170x30)", "Splash Page Production", "Travel Deals Update", "Video Ad", "Web Development", "Web Maintenance", "Web Quote (or Design)"); $dbTypes = explode(',',$row['type']); foreach ($aTypes as $type){ if(in_array($aType,$dbTypes)){ echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type' CHECKED><span style='color:#000;'>$type</span></input><br/>"; }else { echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>"; } } } echo '</td></tr></table>'; ?> </div> hello everyone, I am about to start coding my pages to display results from a database but before i do i want to know information about the following : Is it best to upload images to a database?and display them accordingly? or is it best to use images from a directory? What is most commonly used and or more reliable? Another topic i have trouble finding information on is actually positioning the output from you mysql database, is this practice done with tables?fields and rows? What is this method called? And is there more then one way to go about controlling result layout on your page? Sorry about the 1001 questions , but i am unable to find a clear answer on the topic ..especially question two . Thanks in advance. At the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");} HTML FORM - Price: <$50 PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");} The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. after cloasing connection of database i still got the values form database. Code: [Select] <?php session_start(); /* * To change this template, choose Tools | Templates * and open the template in the editor. */ require_once '../database/db_connecting.php'; $dbname="sahansevena";//set database name $con= setConnections();//make connections use implemented methode in db_connectiong.php mysql_select_db($dbname, $con); //update the time and date of the admin table $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4"; //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time $link= mysql_query($update_time); // mysql_select_db($dbname, $link); //$con=mysql_connect('localhost', 'root','ijts'); $result="select * from admin where username='a'"; $result=mysql_query($result); mysql_close($con); //here i just check after closing data baseconnection whether i do get reselts but i do, why? echo "after the cnnection was closed"; if(!$result){ echo "cont fetch data"; }else{ $row= mysql_fetch_array($result); echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4]; } // echo "<html>"; //echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>"; // echo "<thead>"; // echo"<tr>"; // echo "<th>"; // echo ID; // echo"</th>"; // echo" <th>";echo Username; echo"</th>"; // echo"<th>";echo Password; echo"</th>"; // echo"<th>";echo Last_logged_date; echo "</th>"; // echo "<th>";echo Last_logged_time; echo "</th>"; // echo" </tr>"; // echo" </thead>"; // echo" <tbody>"; //while($row= mysql_fetch_array($result,MYSQL_BOTH)){ // echo "<tr>"; // echo "<td>"; // echo $row[0]; // echo "</td>"; // echo "<td>"; // echo $row[1]; // echo "</td>"; // echo "<td>"; // echo $row[2]; // echo "</td>"; // echo "<td>"; // echo $row[3]; // echo "</td>"; // echo "<td>"; // echo $row[4]; // echo "</td>"; // echo "</tr>"; // } // echo" </tbody>"; // echo "</table>"; // echo "</html>"; session_destroy(); session_commit(); echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin']; ?> session is destroyed but database connection is not closed. thanks |