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  1. its a login form here is my code thats producing the error . I am really new to php. this is for a project . thanks for the help. function login($username , $password){ GLOBAL $dbc; //database $user_id= user_id_from_username($username); $username = sanatize($username); $password = md5($password); $query= mysqli_query($dbc,"SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username' AND `password ` = '$password'"); MYSQLI_ERROR($query); $check= mysqli_fetch_array( $query , MYSQLI_BOTH); // even if i add this set of code the query still gives the above error. }
  2. Hiya jees I'm having a nightmare if you can help me please you'd be saving my life! I'm using wamp and phpmyadmin. I created a database and then decided to add a password to it in phpmyadmin by going to users > root > and copy and pasted my password in twice. As soon as it saved I was kicked out of phpmyadmin. Since then I have been trying to use command prompt to get access to the database but the passwors is not being accepted. So I tried resetting the password by creating a mysql-init.txt using the code "UPDATE mysql.user SET Password=PASSWORD('worldpeace') WHERE User='root'; FLUSH PRIVILEGES;" Then I shut down mysql service and pathed in using "C:\wamp\bin\mysql\mysql5.5.24\bin mysqld --init-file=C:\\mysql-init.txt --console" This doesn't work for me! It just hangs at the end and never restarts the service, which is when the password is suppose to be changed. Please see screenshot of command prompt. Ok so after this attempt I have tried to get in by: mysql -u root -h 127.0.0.1 -P 3306 -p Enter password which I leave blank I then wanted to try and see the database so used show databases and was presented with ONLY the information_scheme and test! Surely I can't have deleted the database!!??? OMG I'm gona get fired. Additionally i have tried: mysql.exe -u=root -p=admin. I get an error "Error 1045 (28000) : Access denied for user '=root'@'localhost (using password: Yes) " If anybody can please please help me get my password reset you ill be saving my life! Thank you p.s. please note i have just used mysql.exe -u=root -p Enter password leave blank And from here I can show databases; and can see my database, so its not deleted! I just need to rest the password so I can get in.
  3. Hello, I am trying to setup an Apache, MySql, PHP combination on windows 2008 r2 machine and have gotten to the part where I am trying to setup PhpMyAdmin when i get an error stating "Call to undefined function mb_detect_encoding() in C:\...\...\gettext.inc on line 177" between google and this post: http://forums.phpfreaks.com/topic/42088-solved-scuttle-undefined-function/ It seemed likely to be that that the extension was commented out, so i verified the DLL was in the EXT directory and removed the comment for : extension=php_mbstring.dll then restarted the apache service. I then attempted to load the setup/index.php for PhpMyAdmin again, but it resulted in the same error. When i run php.exe -m It lists mbstring in the PHP Modules, however if i run <?php phpinfo(); ?> the resulting page lists NOTHING about mbstring... I would greatly appreciate any help regarding this issue. These are the various components and/or important info... OS: Windows 2008 R2 w/SP1 Apache: 2.4.4 win32 w/ ssl 0.9.8 PHP: v5.4.16 Win32 VC90x86 threadsafe MySql: 5.6.11.0 (presumably x86) PhpMyAdmin: 4.0.3 following setup guide: http://lifeofageekadmin.com/how-install-apache-2-4-php-5-4-and-mysql-5-5-21-on-windows-7/
  4. hey guys, i have a MySQL query that has multiple LIKE wildcard statement and multiple ORDER BY statements. "SELECT * FROM `content` WHERE `release_date` LIKE '%2013-06%' OR `release_date` LIKE '%2013-05%' OR `release_date` LIKE '%2013-04%' AND `release_country` = 'USA' AND `is_active_member` = 1 ORDER BY `votes' DESC, `release_date` DESC LIMIT 0 , 50" as you can see i have 3 wildcard's 5 total conditionals 2 ORDER BY fields The ISSUE: my wildcard statements work just fine and filters what i need, everything after ( release_country, is_active_member) dont seem to filter, also my ORDER BY only passes the first conditional. and release_date is not sorted. i have done plenty of research to see if there are other formats to this( which there is ) but they also don't seem to work. The data and field names are correct, and when i place the other fields first, they come out correctly. but i cant get this exact query to work. any suggestions as why this is happening ? thanks guys
  5. I have a MySQL database of addresses that has worked well up to now, however adding an Irish address with a.... ' .... symbol in it has thrown it off completely, I can see why this would happen, but is there a simple solution to avoid this problem in future? I was thinking this must be a common enough problem for there to be a function to fix inputs or something, I noticed phpMyAdmin doubles up the symbol and the entry is entered correctly...
  6. I've been working on a project for the past few weeks and it's been a huge learning curve so I apologise in advance. The idea is to choose ingredients and you will retrieve recipes that contain them. Once the results show up the user can click to view in more detail. The user will then be taken to a page which will show the recipe name, image, ingredients and steps of that recipe. That is where I am currently stuck at. These are the results from a search, when you click look you will be directed to this page. I don't know if this is right but this is the what I used for the form action. echo '<form action="recipe.php?id=<?php echo $recipe_id;?>" method="POST">'; echo '<input id="search" name="Look" type="Submit" value="Look" >'; echo '</form>'; On the recipe page I have this php code at the top <?php mysql_connect("10.168.1.56", "cupboard_test", "***") or die("Connection Failed"); mysql_select_db("cupboard_test")or die("Connection Failed"); $result = mysql_query("SELECT `name` FROM `recipe` WHERE `id` = ".$_GET['id']." LIMIT 1"); ?> And I am trying to echo the name further down the page. <?php echo $result['name'].'<br />'; ?> I know I have probably made quite a few mistakes here and I do apologise but it's all part of the learning experience.
  7. I have a folder in my ftp called recipeimages and it contains all images for different recipes. In my database I have a recipe table and an image path but I think I may have done it wrong and I was wondering how I would display that image to my webpage. <?php $path = "#"; ?> <img src=<?php echo $src; ?> /> I don't know whether or not this is the correct code I would use. The ideal way would be to display an image assigned to a certain recipe. This is the recipe table, this is probably wrong but you get the idea of what I want to accomplish. This is the table and the fields used. I have the imagepath set as varchar as I don't want to directly store the images in the database. I have never done anything like this before so I would love a bit of help. Thank you in advance.
  8. Currently I am able to pull the recipe name from the recipe table but I want to be able to grab the ingredients required from the ingredients table. I know it's to do with JOINS but I'm new to JOINS. This is the ingredients table This is the recipeingredients table, this has two primary keys so I am able to assign multiple ingredients to one recipe This is the recipe table This is the search script <?php $query = $_GET['query']; // gets value sent over search form $min_length = 3; if(strlen($query) >= $min_length){ $query = htmlspecialchars($query); $query = mysql_real_escape_string($query); // makes sure nobody uses SQL injection $raw_results = mysql_query("SELECT * FROM recipes WHERE (`recipename` LIKE '%".$query."%') OR (`ingredients` LIKE '%".$query."%')") or die(mysql_error()); if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following while($results = mysql_fetch_array($raw_results)){ // $results = mysql_fetch_array($raw_results) puts data from database into array, while it's valid it does the loop echo "<p>Recipe:".$results['recipename']."</p><p>Ingredients:".$results['ingredients']."<p>Instructions:".$results['instructions']."</p>"; // posts results gotten from database(title and text) you can also show id ($results['id']) } } else{ // if there is no matching rows do following echo "No results"; } } else{ // if query length is less than minimum echo "Minimum length is ".$min_length; } ?> This is the sample data from the ingredients table This is the sample data from the recipeingredients table This is sample date from the recipe table
  9. So I have this code that checks if a user if banned on the website or not (Through a MySQL database) However, this code that DID IN FACT work, is not working anymore. Whenever I try to check if a player is banned, that I know for sure are banned, they show up as not banned. I have no clue what happened to change the way it works, but all I know is that is doesn't work anymore. Here is what I have: config.inc.php: <?php session_start(); error_reporting(1); mysql_connect("localhost", "root", "<hidden>")or die("DB connection failed: " . mysql_error()); mysql_select_db("bridge")or die("DB selection failed"); ?> index.php: <?php require_once("config.inc.php"); error_reporting(0); $action = $_GET["a"]; if ($action=="checkban"){ $player = $_GET["p"]; $username = mysql_real_escape_string($_GET["p"]); $result = mysql_query("SELECT * FROM player_bans WHERE UPPER(player_bans) = UPPER('$username') LIMIT 1"); $row = mysql_fetch_array($result); if ($row["banned"] == 1){ echo("success"); //send(array($row)); } else { echo("failiure<br />"); echo($row["banned"]); } } ?> Here is the player_bans table as well:
  10. Schweppers

    SQL

    Hi, I'm trying to make a forum for my website and I keep getting this error when trying to create a new topic Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 10 I don't know what page this could be referring to. I have 2 pages to connect to two different DB's one being: config.inc.php the other being: connect.php I don't know what all more information you need to help with this, comment and I'll tell you more if needed.
  11. I didn't want to have to post this question. I need someones professional help with connecting to my school phpmysql server. I have my code here.</p> If someone sees that there might be something wrong please please help me and at least point me in the right direction.<br /> What I need to do is extract data from a table that I have created employee_directory that I need to display in a table on a responsive site. <!DOCTYPE HTML> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <title>Phone Directory</title> <!--Mobile specific meta goodness --> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <!--google fonts--> <link href='http://fonts.googleapis.com/css? family=PT+Sans:400,700|PT+Serif:400,700,400italic' rel='stylesheet' type='text/css'> <!--css--> <link rel="stylesheet" type="text/css" href="main.css" /> <!--[if lt IE 9]> <script src="http://html5shim.googlecode.com/svn/trunk/html5.js"></script> <![endif]--> <!--[if !IE]><!--> <style> /* This query will take effect for any screen smaller than 760px and also iPads specifically. */ @media only screen and (max-width: 760px), (min-device-width: 768px) and (max-device-width: 1024px) { /* Force table to not be like tables anymore */ #employeedb { border:1px solid #000; border-collapse:collapse; font-family:Vollkorn; font-size:12px; text-align:right; } #employeedb th { border: 1px solid #333; padding: 3px 6px; color: #e9edd5; } #employeedb td { border:1px solid #999; padding:3px 6px; } table, thead, tbody, th, td, tr { display: block; color: #c6f104; font-family: Vollkorn; margin-right: auto; margin-left: auto; padding-left: 5px; } /* Hide table headers (but not display: none;, for accessibility) */ thead tr { position: absolute; top: -9999px; left: -9999px; } tr { border: 1px solid #ccc; } td { /* Behave like a "row" */ border: none; border-bottom: 1px solid #eee; position: relative; padding-left: 50%; } td:before { /* Now like a table header */ position: absolute; /* Top/left values mimic padding */ top: 6px; left: 6px; width: 45%; padding-right: 10px; white-space: nowrap; } /* Label the data */ td:nth-of-type(1):before { content: "ID"; } td:nth-of-type(2):before { content: "Last Name"; } td:nth-of-type(3):before { content: "Cell Number"; } td:nth-of-type(4):before { content: "Email"; } } /* Smartphones (portrait and landscape) ----------- */ @media only screen and (min-device-width : 320px) and (max-device-width : 480px) { body { padding: 0; margin: 0; width: 320px; } } /* iPads (portrait and landscape) ----------- */ @media only screen and (min-device-width: 768px) and (max-device-width: 1024px) { body { width: 495px; } } .wrapper .background-illusion h2 { color: #d4f24e; font-family: Vollkorn; font-size: 29px; } body table tr th h3 { color: #c6f104; } body div table tr th { color: #B0F104; } </style> </head> <body> <div class="wrapper"> <header class="background-illusion"> <h2 align="center">Phone Directory</h2> <p align="center">for Student Admin Department at WTCC<br> WEB 000 | Assignment 7<br> </p> </header> </div> <nav class="nav right"> <ul> <li class="current"><a href="index.php">Home</a></li> <li><a href="#">Assignments</a></li> <li><a href="#">Discussion Board</a></li> <li><a href="#">Support</a></li> <li><a href="#">Contact</a></li> </ul> </nav> <table width="85%"> <tr> <th>ID</th> <th>Last Name</th> <th>Cell</th> <th>Email</th> </tr> <tr> <?php $dbhost = '198.86.244.3/phpmyadmin/'; $dbuser = 'XXXX_dbXXXXXX'; $dbpass = 'DxXxXxX'; // Connect to database server $conn = mysql_connect("$dbhost", "$dbuser", "$dbpass") or die (mysql_error ()); echo "You are now connected to MySQL<br /><br />"; // Select database $dbname = 'employee_directory'; mysql_select_db($dbname); // Get data from the database depending on the value of the id in the URL $query = "SELECT * FROM employee_directory ORDER BY lastname"; $result = mysql_query($query) or die(mysql_error()); ?> <?php while($row = mysql_fetch_array($result)); ?> <td><?php echo $row['id']; ?></td> <td><?php echo $row['lastname']; ?></td> <td><?php echo $row['cellphone']; ?></td> <td><?php echo $row['email']; ?></td> </tr> </table> <footer class="background-illusion"> <div class="container"> <div align="center"> <p>Copyright © 2013 | <a href="http://dianamagers.com">XXXXXXX</a> | Spring Semester 2013 </p></div> </div> </footer> </div> </body> </html> Of course, I couldn't provide my username and password ..... Sorry. I don't need to know that. This is what it looks like on the server now. http://wcet2.waketec...es(1)/index.php
  12. Hello! I've recently been having trouble with my basic membership website. It is currently being tested with mamp on localhost. So I have two pages, login.php, and register.php. My register.php file is working fine. One of the things about my register file is the password, which is the thing I'm having serious trouble with. Here is the register.php code (Php code only) $query = "INSERT INTO users (username,password,email) VALUES ('$username',md5(md5('$password'),'$email')"; if (!mysql_query($query)){ die('ERROR: ' . mysql_error . ''); } echo "You have been registered successfully"; As far as I can see, the register.php file successfully inserts the data into my database, since I can see it through phpMyAdmin. Now here's my login.php file. <?php $checkusername = $_POST['username']; $checkpassword = $_POST['password']; if (strlen($checkusername) <= 0){ echo "<p id='error'>You Need to enter a username!</p>"; }else{ if (strlen($checkpassword) <= 0){ echo "<p id='error'>You need to enter a password!</p>"; }else{ $query = mysql_query("SELECT * FROM users WHERE username='" . $checkusername ."'"); if (mysql_num_rows($query) == 1){ $query = mysql_query("SELECT * FROM users WHERE password='" . md5(md5($checkpassword)) . "'"); if (mysql_num_rows($query) == 1){ echo "Welcome, " . $checkusername . "! You are now logged in!"; $_SESSION['user'] = $checkusername; $_SESSION['pass'] = $checkpassword; }else{ echo "<p id='error'>Wrong Password!</p>"; } }else{ echo "<p id='error'>Wrong Username!</p>"; } } } ?> So, the issue here is when I fill out my php login form, I get a wrong password error. At first, I thought that I wasn't properly connecting to my database table, but I got no error. So I then decided to remove all of the md5 encryption. Without the encryption, my password got entered into the database, and the login file worked. So I think that my problem is the md5(). Now, how do I fix it? I'm sorry if it turns out that the issue was some sort of basic mistake. I'm not the most experienced php coder.
  13. Hi there i have several users created in my membership db table but since I added in my validateUsers function to my controller to make sure non registered users cannot sign into my site, registered users cannot log in either. Why is this? Controller: class Login extends CI_Controller { function Login() { parent::__construct(); $this->load->model('membership'); } function loguserin() { $this->load->helper(array('form', 'url')); $this->load->library('form_validation'); $this->form_validation->set_rules('username', 'Username', 'required|min_length[4]|max_length[12]|callback_validateUser|trim'); $this->form_validation->set_rules('password', 'Password', 'required|md5|trim'); $username = $this->input->post('username'); $password = $this->input->post('password'); if ($this->form_validation->run()) { $this->validateUser($username, $password); $this->session->set_userdata('status', 'OK'); $this->session->set_userdata('username', $username); redirect('home'); } else { $this->session->set_userdata('status', 'NOT_OK'); $this->load->view('shared/header'); $this->load->view('account/logintitle'); $this->load->view('account/loginview'); $this->load->view('shared/footer'); } } function validateUser($username, $password) { $this->db->select('*')->from('membership'); $this->db->where('username', $username); $this->db->where('password',MD5($password)); $query = $this->db->get(); if ($query ->num_rows ==1) { return true; } else { var_dump($username, $password);// this only seems to throw back the username, not the password, don't know why $this->form_validation->set_message('validateUser', 'Invalid username/password'); return false; } } function index() { $this->load->view('shared/header'); $this->load->view('account/logintitle'); $this->load->view('account/loginview'); $this->load->view('shared/footer'); } Thanks for the help
  14. this is the scrip i currently have but i get "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'range,groups,b_m,product_code,desc_1,desc_2,qty,nett_price,goods_value) VALUES (' at line 1" <?php if (isset($_POST['submit'])) { if(is_uploaded_file($_FILES['filename']['tmp_name'])){ $handle = fopen($_FILES['filename']['tmp_name'], "r"); while(($data = fgetcsv($handle,1000,",")) !==false) { $ST=$data[0]; $date=$data[1]; $actual_date=$data[2]; $account=$data[3]; $branch=$data[4]; $customer_name=$data[5]; $i_c=$data[6]; $invoice_num=$data[7]; $category=$data[8]; $order_no=$data[9]; $order_item=$data[10]; $range=$data[11]; $groups=$data[12]; $b_m=$data[13]; $product_code=$data[14]; $desc_1=$data[15]; $desc_2=$data[16]; $qty=$data[17]; $nett_price=$data[18]; $goods_value=$data[19]; $query = "INSERT INTO sales_orders (ST,date,actual_date,account,branch,customer_name,i_c,invoice_num,category,order_no,order_item,range,groups,b_m,product_code,desc_1,desc_2,qty,nett_price,goods_value) VALUES ('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]','$data[7]','$data[8]','$data[9]','$data[10]','$data[11]','$data[12]','$data[13]','$data[14]','$data[15]','$data[16]','$data[17]','$data[18]','$data[19]')"; mysql_query($query) or die(mysql_error()); } } } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR...nsitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title><?php echo COMPANY_NAME; ?> | Sales Figures Upload Screen</title> <link href="<?php echo ROOT; ?>css/login.css" type="text/css" rel="stylesheet" /> <link type="text/css" href="<?php echo ROOT; ?>css/start/jquery-ui-1.8.7.custom.css" rel="stylesheet" /> <script type="text/javascript" src="<?php echo ROOT; ?>js/jquery-1.4.4.min.js"></script> <script type="text/javascript" src="<?php echo ROOT; ?>js/jquery-ui-1.8.7.custom.min.js"></script> </head> <body> <div id="box"> <div id="content"> <h1>Sales CSV Update</h1> <p>Please use a valid CSV file extracted from Exant Exporter.</p> <form method="post" action="uploadsale.php" enctype="multipart/form-data"> <p> <input type="file" name="filename"/> </p> <input name="submit" type="hidden"/> <button tabindex="4" type="submit">Upload Data</button> </form> </div> </div> </body> </html> <?php mysql_close($connection); ?>
  15. Hello I have 2 MySQL databases on the same server with same structure and same data. Occasionally data in one database are changed for 7 days and then should be back to the same value as in first database. Through PhpMyAdmin I am trying to write statement that should do thi but something I am missing. I have databases: db1_local and db2_local Table gitem column gbaseprice (should be synchronized from db1 to db2) column gitemcode (has same value in both databases and synchronizing should be followed with this parameter so that id's don't mess up) Appreciate any help Here is statement e.g. SQL statement that run in db2 UPDATE gitem SET gbaseprice=db1_local.gbaseprice FROM db1_local.gitem(`gbaseprice`) INNER JOIN db1_local.gitemtable ON db1_local.gitem(`gitemcode`)=db2_local.gitem(`gitemcode`)
  16. Hello All, I was wondering what is the point of having multiple databases? Why not just have X amount of tables inside of one database? If you have multiple databases that means you have to open and close more than one database on one page. I am wondering this because I was thinking about putting all of my uploaded photos into their own database to make a more searchable and well structured database. Does it make it faster? Thanks in advance!
  17. Hello friends, i need help on setting up remote connection for phpmyadmin.. currently i have setuped phpmyadmin on a server..it works fine connecting via localhost.. or using the ip address of the server ex.. 192.168.0.222... then im using port forwarding on a router to point this ip address to a public ip address... can you guide me or give me some instructions on how is the correct way to access this mysql-phpmyadmin via remote connection as i cannot connect on the phpmyadmin remotely. thank you.. if you can provide me a diagram with settings.. it would be great.. Thanks again
  18. Hi Everyone, I am fairly new in PHP Development but have used Wordpress for the last 2 years. I am currently modifying a plugin to suit the client's requirements. It is a Custom To-DO list plugin. You may check out my progress here : [Page on meshfun.com] The plugin I am modifying is WordPress › TaskFreak! Free " WordPress Plugins It is almost all the required features, just the feature of Creating Groups and assigning is absent. When a new task is created,an option of assigned to is present but that can assign task to ONLY ONE Person. It should be able to assign to multiple users. The whole data is stored in the database in 8-9 tables. The main table DO has a field called as user_ID as you can see in the image below. It is updated when a new task is created. It contains a number (User ID of person who is assigned the task) or left blank in case the task is not assigned to anyone. : I would like the ability to assign to multiple users instead of one. What would be the ideal way ? Create new table or store the number of people assigned as array ? If a new table is created, what will be the fields to be created ? Note I was not able to add images so please check out here to see couple of images I wanted to show you : http://www.quora.com/Need-help-in-Modifying-a-Wordpress-Plugin I understand that you guys are quite busy but if you guys can atleast guide me on how to proceed, it would be great ! Thanks, Shalabh Here is the code which displays dropdown menu in the front end (taskfreak/inc/views/front/edit.php) <li> <label for="tfk_user_id"><?php _e('Assigned to', 'taskfreak') ?></label> <select name="user_id" id="tfk_user_id"> <option value="">—</option> <?php foreach ($this->users as $user): ?> <option value="<?php echo $user->ID ?>"<?php if ($this->pid && $this->data->get('user_id') == $user->ID) { echo " selected"; } ?>><?php echo $user->display_name ?></option> <?php endforeach; ?> </select> <?php echo $this->data->errors['user_id'] ?> </li> UPDATE : One gentleman suggested me to create a table with the following SQL COde : CREATE TABLE `task_user` ( `task_id` int(11) unsigned NOT NULL, `user_id` int(11) unsigned NOT NULL, `status` tinyint(4) NOT NULL DEFAULT '1' COMMENT 'status field is optional', UNIQUE KEY `task_user` (`task_id`,`user_id`,`type`) USING BTREE, KEY `task_id` (`task_id`), ) CHARSET=utf8 But I am still not able to convert the display into multiple checkboxes.
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