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PHP - 1 Unique Site Per User
the code im running is
Code: [Select] $result = mysql_query("SELECT count(*) FROM slinks where approval='1'"); $do = mysql_fetch_row($result); $random = mt_rand(0,$do[0] - 1); $result = mysql_query("SELECT * FROM slinks LIMIT $random, 1"); The above is finished off into a iframe Code: [Select] <iframe src="<?php while ($row = mysql_fetch_array($result)){ print $row["url"];}?> "frameborder="0" width="100%" height="100%" name="surfing"id="dframe"></iframe>resulting in a random site loaded ever time the php file is ran now all i need to figure out with a little help from the great users at phpfreaks is how i can get the random site to only be ran one time per user. example click = randomsite1.com click = randomsite2.com click = randomsite3.com click = randomsite4.com click = randomsite5.com click = randomsite5.com = <--SKIP done been called so lets move along click = randomsite6.com I guess somehow im going to have to store the urls that are loaded into a variable and compare the variable to the the random site to make sure we dont get a match. then again i guess i would have to some how figure out how to store the data per unique user as well . Any suggestions would be greatly appreciated Similar TutorialsHi everyone, I'm new here and really in need of some help with some script I've been trying to get right. Basically I'm trying to write a function that will interface with Gravity Forms and Wordpress but this is NOT a GForms or Wordpress issue it's pure PHP honest! The function will pull the post code a user has entered on the first page of a Gravity Form, and then take the spaces out of it and make sure it's in upper case. This much I've got working. Now I need it to add a hyphen and a three digit number starting at 001, so it'll look like AB12CD-001. Now it will need to check if this is username has already been used by calling the Wordpress function username_exists (http://codex.wordpress.org/Function_Reference/username_exists) and if it does increase the suffix number by one and recheck until it finds the next available number. This could result in the user living in the post code AB1 2CD ending up with the username AB12CD-021 if there are 20 others in the local area obviously. Now this will then post back to the Gravity Forms registration form on the next page and fill in the username for the user, this I can also do. So hopefully you can see it's the PHP bit I'm stuck on and I would really appreciate any help. Cheers guys Okay, so I have a database with user log on info, and unique ID's. How to I allow the user to save info from a form, and be able to log out and come back and log on to see/edit that info. Thanks! I've started a to-do app, and everything works fine. Every user has his own unique to-do items that he can add. The only issue is that when I log-in with a user, ALL the items of all users are displayed when I'm redirect to my main page. I want to make it as so every user has his own page where only the to-do items he has created are visible, not every item from every user. I'm a total beginner with PHP so I'm sorry if this is a stupid question. This is my todomain.php code
<?php
require 'db_conn.php';
require 'config.php';
session_start();
// If the session variable is empty, this
// means the user is yet to login
// User will be sent to 'login.php' page
// to allow the user to login
if (!isset($_SESSION['id'])) {
$_SESSION['msg'] = "You have to log in first";
header('location: login.php');
}
// Logout button will destroy the session, and
// will unset the session variables
// User will be headed to 'login.php'
// after loggin out
if (isset($_GET['logout'])) {
session_destroy();
unset($_SESSION['id']);
header("location: login.php");
}
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset= "UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<link rel="stylesheet" href="css/bootstrap-grid.min.css">
<link rel="stylesheet" href="css/bootstrap.min.css">
<link rel="stylesheet" href="style.css">
<a href="index.php?logout='1'" style="color: black;">
Click here to Logout
</a>
<?php if (isset($_SESSION['success'])) : ?>
<div class="error success" >
<h3>
<?php
echo $_SESSION['success'];
unset($_SESSION['success']);
echo $_SESSION["username"]
?>
</h3>
</div>
<?php endif ?>
<title>TODO App</title>
<script src="https://kit.fontawesome.com/d72928d9b9.js" crossorigin="anonymous"></script>
</head>
<body>
<div class="container m-5 p-2 rounded mx-auto bg-light shadow">
<!-- App title section -->
<div class="row m-1 p-4">
<div class="col">
<div class="p-1 h1 text-primary text-center mx-auto display-inline-block">
<i class="fa fa-check bg-primary text-white rounded p-2"></i>
<u>My Todo-s</u>
</div>
</div>
</div>
<!-- Create todo section -->
<form action="app/add.php" method="POST" autocomplete="off">
<div class="row m-1 p-3">
<div class="col col-11 mx-auto">
<div class="row bg-white rounded shadow-sm p-2 add-todo-wrapper align-items-center justify-content-center">
<div class="col">
<input class="form-control form-control-lg border-0 add-todo-input bg-transparent rounded" type="text" name="title" placeholder="Add new ..">
</div>
<div class="col-auto m-0 px-2 d-flex align-items-center">
<label class="text-secondary my-2 p-0 px-1 view-opt-label due-date-label d-none">Due date not set</label>
<i class="fa fa-calendar my-2 px-1 text-primary btn due-date-button" data-toggle="tooltip" data-placement="bottom" title="Set a Due date"></i>
<i class="fa fa-calendar-times-o my-2 px-1 text-danger btn clear-due-date-button d-none" data-toggle="tooltip" data-placement="bottom" title="Clear Due date"></i>
</div>
<div class="col-auto px-0 mx-0 mr-2">
<button type="submit" class="btn btn-primary">Add</button>
</div>
</div>
</div>
</div>
</form>
<?php
$todos=$conn->query("SELECT * FROM todos ORDER BY id ASC");
?>
<div class="row mx-1 px-5 pb-3 w-80">
<div class="col mx-auto">
<?php
while($todo=$todos->fetch(PDO::FETCH_ASSOC)){ ?>
<!-- Todo Item 1 -->
<div class="row px-3 align-items-center todo-item rounded">
<?php if($todo['checked']){ ?>
<input type="checkbox" class="check-box" data-todo-id="<?php echo $todo['id'];?>" checked />
<h2 class="checked"><?php echo $todo['title'] ?> </h2>
<div class="col-auto m-1 p-0 todo-actions">
<div class="row d-flex align-items-center justify-content-end">
</div>
</div>
<?php } else{ ?>
<input type="checkbox" class="check-box" data-todo-id="<?php echo $todo['id'];?>"/>
<h2><?php echo $todo['title'] ?></h2>
<?php } ?>
<div class="col-auto m-1 p-0 d-flex align-items-center">
<h2 class="m-0 p-0">
<i class="fa fa-square-o text-primary btn m-0 p-0 d-none" data-toggle="tooltip" data-placement="bottom" title="Mark as complete"></i>
</h2>
</div>
<div class="col-auto m-1 p-0 todo-actions">
<div class="row d-flex align-items-center justify-content-end">
<h5 class="m-0 p-0 px-2">
<i class="fa fa-pencil text-info btn m-0 p-0" data-toggle="tooltip" data-placement="bottom" title="Edit todo"></i>
</h5>
<h5 class="m-0 p-0 px-2">
<i class="remove-to-do fa fa-trash-o text-danger btn m-0 p-0" data-toggle="tooltip" data-placement="bottom" title="Delete todo" id="<?php echo $todo['id']; ?>"></i>
</h5>
</div>
<div class="row todo-created-info">
<div class="col-auto d-flex align-items-center pr-2">
<i class="fa fa-info-circle my-2 px-2 text-black-50 btn" data-toggle="tooltip" data-placement="bottom" title="" data-original-title="Created date"></i>
<label class="date-label my-2 text-black-50"><?php echo $todo['date_time'] ?></label>
</div>
</div>
</div>
</div>
<?php } ?>
</div>
<script src="https://code.jquery.com/jquery-3.3.1.slim.min.js" integrity="sha384-q8i/X+965DzO0rT7abK41JStQIAqVgRVzpbzo5smXKp4YfRvH+8abtTE1Pi6jizo" crossorigin="anonymous"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.14.3/umd/popper.min.js" integrity="sha384-ZMP7rVo3mIykV+2+9J3UJ46jBk0WLaUAdn689aCwoqbBJiSnjAK/l8WvCWPIPm49" crossorigin="anonymous"></script>
<script src="js/bootstrap.bundle.min.js"></script>
<script src="js/myjs.js" ></script>
<script src="js/jquery-3.3.1.min.js"></script>
<script>
$(document).ready(function(){
$(".remove-to-do").click(function(e){
const id = $(this).attr('id');
$.post('app/remove.php',
{
id: id
},
(data) => {
if(data){
$(this).parent().parent().parent().parent().hide(300);
}
}
);
});
$(".check-box").click(function(e){
const id = $(this).attr('data-todo-id');
$.post('app/checking.php',
{
id: id
},
(data) => {
if(data!='error'){
const h2= $(this).next();
if(data === '1'){
h2.removeClass('checked');
}else{
h2.addclass('checked');
}
}
}
);
});
});
</script>
</body>
</html>
I have a mysql table which will store users email addresses (each is unique and is the primary field) and a timestamp. I have added another column called `'unique_code' (varchar(64), utf8_unicode_ci)`. What I would very much appreciate assistance with is; a) Generating a 5 digit alphanumeric code, ie: 5ABH6 b) Check all rows the 'unique_code' column to ensure it is unique, otherwise re-generate and check again c) Insert the uniquely generated 5 digit alphanumeric code into `'unique_code'` column, corresponding to the email address just entered. d) display the code on screen. What code must I put and where? **My current php is as follows:** Code: [Select] require "includes/connect.php"; $msg = ''; if($_POST['email']){ // Requested with AJAX: $ajax = ($_SERVER['HTTP_X_REQUESTED_WITH'] == 'XMLHttpRequest'); try{ if(!filter_input(INPUT_POST,'email',FILTER_VALIDATE_EMAIL)){ throw new Exception('Invalid Email!'); } $mysqli->query("INSERT INTO coming_soon_emails SET email='".$mysqli->real_escape_string($_POST['email'])."'"); if($mysqli->affected_rows != 1){ throw new Exception('You are already on the notification list.'); } if($ajax){ die('{"status":1}'); } $msg = "Thank you!"; } catch (Exception $e){ if($ajax){ die(json_encode(array('error'=>$e->getMessage()))); } $msg = $e->getMessage(); } } I have a site where I want another user that has a password fills out a form & it then downloads to my server. I want them to be able to then download that file from my site at the same time the form is submitted. I've tried adding this code to the bottom of the php file that the form points to but it just displays the file on the screen instead of downloading to the user's computer.
<?php
$file = '/site/downloadfile';
if (file_exists($file)) {
header('Content-Description: File Transfer');
header('Content-Type: application/octet-stream');
header('Content-Disposition: attachment; filename="'.basename($file).'"');
header('Expires: 0');
header('Cache-Control: must-revalidate');
header('Pragma: public');
header('Content-Length: ' . filesize($file));
readfile($file);
exit;
}
?>
if I put the same code in a separate file it works perfectly but i don't want to add another file if i can help it. Any help appreciated
I've got a site that requires the user to follow a well-defined workflow, corresponding to a sequence of pages. You'll have the right idea if you think of a store where the user selects items, then chooses a delivery method, then pays (although the site's actual purpose is somewhat different than that). I just tracked down a puzzling logical failu a user followed the workflow up to a certain point, then opened a new browser window, pointed it to the site, and started a second workflow. The site stores information about the user's activities in session variables, and the two windows shared a single session, so the site got tremendously confused about what the user was doing. Given the site's design, visiting the site in two windows at once is a legitimate thing for a user to do. I'm darned if I see how I can accommodate it, though. Other designers must have dealt with this problem many times. What ways of solving it have been found effective? In this case, the simplest and most effective solution would be to create a new session for each window in which the user visits the site. Is there any way to do that? Hi all,
I'm working on a hobby project to improve my skills and learn PHP and MySQL. Don't know if this is the right place to ask this question, so please let me know if it isn't.
I already created a single user site. It has a simple PHP login and after login information (like personal information and messages) is pulled from the database and shown. The personal information is stored in a table called settings and the messages are stored in a table called messages.
I like to go to the next level and create a multi-user login, but I have no idea were to start. I think I have to switch to another login type were both PHP and MySQL are used. How has the database be designed? Can I use the same structure as in the single user site I created? How can the right (belonging to the user) information be retrieved from the database?
Thanks in advance !
I would appreciate your assistance, there are tons of login scripts and they work just fine. However I need my operators to login and then list their activities for the other operators who are logged in to see and if desired send their clients on the desired activity. I have the login working like a charm and the activities are listed just beautifully. How do I combine the two tables in the MySQL with PHP so the operator Logged in can only make changes to his listing but see the others. FIRST THE ONE script the member logges in here to the one table in MSQL: <?php session_start(); require_once('config.php'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } $login = clean($_POST['login']); $password = clean($_POST['password']); if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'"; $result=mysql_query($qry); if($result) { if(mysql_num_rows($result) == 1) { session_regenerate_id(); $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['member_id']; $_SESSION['SESS_FIRST_NAME'] = $member['firstname']; $_SESSION['SESS_LAST_NAME'] = $member['lastname']; session_write_close(); header("location: member-index.php"); exit(); }else { header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?> ................................................. ................................ Now I need the person who logged in to the table above to be able to make multiple entries to the table below <? $ID=$_POST['ID']; $title=$_POST['title']; $cost=$_POST['cost']; $activity=$_POST['activity']; $ayear=$_POST['aday']; $aday=$_POST['ayear']; $seats=$_POST['special']; $special=$_POST['seats']; mysql_connect("xxxxxx", "xxx350234427", "========") or die(mysql_error()); mysql_select_db("xxxx") or die(mysql_error()); mysql_query("INSERT INTO `activity` VALUES ('ID','$title', '$cost','$activity', '$aday', '$ayear', '$special', '$seats')"); Print "Your information has been successfully added to the database!" ?> Click <a href="member-profile.php">HERE</a> to return to the main menu <?php ?> hi, i have made a website where people resgister their details of them and products. they have to enter the following details in form Name of company name of the product company address email id password mobile number contact and brief details about their company
user can then login with email id and pwd. now after login ..user will get a page where he can upload the photos of products images and their price, so now my question is that when he finishes uploading (|by clicking on upload button) the product images and price text box ..then on final uploaded webspage it should show all other things which he registerd before (company name , mobile number etc) along with images and price...hence the main question that user does not need to enter mobile and address while uploading images and filling proce ..but on the final page it should show mobile and address along with price and images..as user is not going to enter mobile and address again and again as he will have multiple products to upload.
Actually, what i want to do is to use the email to fetch the $email,$password and $randomnumber from database after Hi, so far I have managed to set up a somewhat basic login website with a mysql database backend. Once they have logged on they go to a "main menu" page. What I need to define is that user A sees button A but only that button, etc. (Then of course that same rule would have to apply if they tried to directly go to the page, but I am guessing I can do that in the same way that I currently do to force a login). If anyone has any tutorials or sample code I would much appreciate it. Thanks, Hi guys, I am trying to put together a little system that allows users to log onto my website and access there own personal page. I am creating each page myself and uploading content specific to them which cannot be viewed by anyone else. I have got the system to work up as far as: 1/ The user logs in 2/ Once logged in they are re-directed to their own page using 'theirusername.php' Thats all good and working how I need it too. The problem I have is this. If I log onto the website using USER A details - I get taken to USER A's page like I should but - If I then go to my browser and type in USERBdetails.php I can then access USER B's page. This cannot happen!! I need for USER A not to be able to access USER B profile - there is obviously no point in the login otherwise! If you are not logged in you obviously cannot access any secure page. That much is working! Please find below the code I am using: LOGIN <?php session_start(); function dbconnect() { $link = mysql_connect("localhost", "username", "password") or die ("Error: ".mysql_error()); } ?> <?php if(isset($_SESSION['loggedin'])) { header("Location:" . strtolower($username) . ".php"); if(isset($_POST['submit'])) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $mysql = mysql_query("SELECT * FROM clients WHERE username = '{$username}' AND password = '{$password}'"); if(mysql_num_rows($mysql) < 1) { die("Password or Username incorrect! Please <a href='login.php'>click here</a> to try again"); } $_SESSION['loggedin'] = "YES"; $_SESSION['username'] = $username; $_SESSION['name'] header("Location:" . strtolower($username) . ".php"); } ?> HEADER ON EACH PHP PAGE <?php session_start(); if(!isset($_SESSION['loggedin'])) { die(Access to this page is restricted without a valid username and password); ?> --------------------------------------------------- Am I right in thinking it is something to do with the "loggedin" part? The system I have here is adapted from a normal login system I have been using for years. The original just checks the details and then does a 'session start'. This one obviously has to re-direct to a user specific page. To do this I used the <<header("Location:" . strtolower($username) . ".php");>> line to redirect to a page such as "usera.php" or "userb.php" Any help would be greatly appreciated! Ta Hi, I am getting frustrated beyond belief at the moment with trying to get a very simple script to run, I am using PHP 5.3.3 and MySQL 5.1 on a Win2k8 server with IIS7.5. Basically my script is connecting to a local database, running a single select query, returning those rows and building up a string from them. The problem is that I am receiving complete BS responses from PHP that the access is denied for the user being specified. This is complete rubbish since the user can connect via mysql, sqlyog, ASP.NET MVC without issue but for some bizarre reason it is not working via PHP. The code for the script is here : Code: [Select] <?php $mysql = mysql_connect('127.0.0.1:3306', 'myuser', 'mypass', 'mydatabase'); if (!$mysql) { die(mysql_error()); $content = "<nobr></nobr>"; } else { $result = mysql_query('SELECT * FROM tblEventGroup'); $content = "<nobr>"; if ($result) { while($row = mysql_fetch_assoc($result)) { $content .= "<span>"; $content .= $row['GroupName']; $content .= "</span>"; $content .= "<a href=\"../Event/EventSearch?groupid="; $content .= $row['GroupId']; $content .= "\" target=\"_blank\">Book here</a> "; } } mysql_close($mysql); $content .= "</nobr>"; } ?> I cannot for the life of me understand what the problem is, the return error is Access denied for user 'myuser'@'localhost' (using password: YES) Hello PHP freaks,
my codes dont allow me to log-in to Student Home with unique id. It says "Invalid Login or Password"
this is my form
<tr bgcolor="#E1E1E1" class="stylesmall">
<td width="35%" align="left" class="style7 style1">Learner Id : </td>
<td width="65%" align="left"><input name="learner_id" type="text" id="learner_id" action="Student_Home.php" method="post"></td>
</tr>
<tr bgcolor="#E1E1E1" class="stylesmall">
<td align="left" class="style7 style1">Password:</td>
<td align="left"><input name="student_password" type="password" id="student_password"><
/td>
and this is my handler.
<?php
session_start();
include 'Connect.php';
$flag = "";
$learner_id = $_POST['learner_id'];
$student_id = $_POST['student_id'];
$student_password = $_POST['student_password'];
$query = "select last_login_date from student_information where student_id='$student_id' and student_password='$student_password'";
$result = mysql_query($query,$link_id);
if(mysql_error() != null){
die(mysql_error());
}
if($date = mysql_fetch_array($result))
{
$lastdate = $date['last_login_date'];
$date2 = date("d-m-Y h:i A",strtotime($lastdate));
$_SESSION["student_id"] = $_POST["student_id"];
$_SESSION["lastlogin"] =$date2;
$_SESSION["type"] = "Student";
mysql_query("UPDATE student_information SET last_login_date=now() where student_id='$student_id'",$link_id);
if(mysql_error() != null){
die(mysql_error());
}
header("location: Student_Home.php?id={$student_id}");
die();
}
else
{
$flag = "invalid";
header("location:Student_login.php?flag=$flag");
die();
}
?>
PLease help me PHP friends to correct my codes.
Howdy, The following code wil geenrate a dynamic pull-down list based on column values. Can someone help me tweak it so that is will NOT display a duplicate record value. That is, if the sponsors values in the table were Hart, Michaels, Michaels, Morella ...it would only display Hart, Michaels, Morella in the pull-down menu. <form name="form0"> <? $result = @mysql_query("select distinct sponsor from table ORDER BY sponsor ASC"); if (mysql_num_rows($result) > 0) { print "<select name=\"link\">"; ?> <option <?php if(empty($_GET['sponsor'])){ echo "selected=\"selected\""; } ?> value="<? echo "$page_name" ?>">SELECT A SPONSOR</option> <? while ($row = mysql_fetch_array($result)) { print "<option "; if($_GET['sponsor'] == $row['sponsor'] ){ echo "selected=\"selected\""; } print " value=\"index?sponsor=" . $row['sponsor'] . "\">" . $row['sponsor'] . "</option>\n"; } print "</select>"; } ?> </form> Thank you. ~Wayne Hi i'm making a new website based on PHP. I bumped on a problem now... The problem is that I have a loop where i get all records out of my sql database and get them into a form. Every record has a differend form with a button to. Every button then has a unique ID. Now i tried everyting i could think of to get the value out of a button when a button is clicked. But nothing worked... here is the part of the code where everything is for the form and the POST. Code: [Select] while($record = mysql_fetch_object($result)){ $id = $record->id; echo"<form action='home.php' method='post'>"; echo"<b>name:</b>".$record->name." <b>Price: </b>".$record->price.""; echo'<input type="hidden" name="id" value='.$id.'>'; echo"<input type='submit' value='Buy' /><br />"; echo"</form>"; } if($_POST) { $ID = intval($_POST['id']); $query1 = "SELECT price FROM items WHERE id = ".$ID.""; $result1 = mysql_query($query1); list($cost) = mysql_fetch_row($result1); $gold = getStat('gc',$userID); if($gold > $cost) { setStat('gc',$userID,($gold - $cost)); $queryw = "INSERT INTO users_inventory(item_id,user_id, quantity) VALUES ('".$ID."','".$userID."','1')"; $resultw = mysql_query($queryw); } else { echo"You can't afford this weapons."; } } The only thing that doesn't seem to work is this part Code: [Select] $ID = intval($_POST['id']); is always gives me this error: Notice: Undefined index: id in shop.php on line 26 Please help me out here i have no idea what to do from here on... ok so i have an array which has duplicate values in it. i used array_unique() to remove duplicate values. the array had values 1,2,2,3. after the following code: $unique = array_unique($suggest); $size = count($unique); for ($x=0;$x<=$size;$x++) { echo $unique[$x]; echo "<br>"; } it prints out 1,2 and then gives an error, "Notice: Undefined offset: 2 in E:\wamp\www\Copy of pro2\Classes.php on line 1074" and then finally 3. any ideas how i can solve this. im assuming that array_unique() assigns the index 0,1,3 of the duplicate value array to unique array at the same indexs 0,1,3. since the index 2 of duplicate array is a duplicate value it is ommited but the unique array does not have an array at index 2. Hi: Not sure if this is where to ask this (maybe it's a mySQL Issue)?? I am working on a zip code locator. The field/column "zip" (which holds the zip codes) is set as a UNIQUE Key in the database. Problem is: I want to allow people to enter a new city if there is one they want to add. The only way I can do this is to remove the UNIQUE Key from the column, or I get an error. Some cities - like Malvern, PA and Frazer, PA - have the same zip code: 19355 This works fine, but the issue that now happens is the zip locator on the frontend will no longer find that zip code. Does anyone know what a solution might be? I can post the zip locator code, and maybe there's a chunk of code I can remove or alter to fix this? Not sure how to address this one. Whats the easiest way to rename uploaded files to a unique name that would never be duplicated? Need a bit of help with some php and a few kind members helped me in the past. My code is: $query = "SELECT names FROM table WHERE type='$type'"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)) { foreach(unserialize($row['names']) as $key => $value){ echo $value . '<br />'; } } Basically, a field on my table is called names. The information in this field is serialized (so many names are together in a single field, for a single row). I want to be able to search the whole table (where type='$type' - so many rows) for all names and then show unique names only. My code above unserializes the names from each field and then lists them, but doesn't deal with any duplicates (which I need removing). I have looked at functions like "unique_array()", and have tried using it in different places but it isn't doing the job. Any help appreciated. |
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