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PHP - Get Children From Hierarchical Database
id Name Owner 1 Root 0 2 Child 1 3 Child 1 4 Level-2-Child 2 So, I want to get an element, then get all elements that are either direct or indirect children of it. So - getChildren(2) should return 2 and 4, because 4 is a direct child of 2. And getChildren(1) should return 1, 2, 3, 4 because all items are a child of root. I'm pretty sure it needs to be done with a recursive function. Can someone help me out? Similar TutorialsI have got a script which is able to create hierarchical categories infinitely and display them as items in a ul. What I need now is to get these categories to appear in a hierarchical dropdown i.e. the user clicks a menu and the dropdown drops down, then the categories that don't have any children will simply be displayed whereas the categories that do have children will have an arrow pointing to the right and once hovered over all of their children will be displayed in a new menu with their children and so on a nd so forth. Very similar to the way the windows menu bar works except with more levels. Basically I want it to function sort of like a spry menu bar except it can go down then across etc etc and any further categories and sub categories that are added in will be shown. Can anybody help me out with this or perhaps point me in the direction of a tutorial that helps to avhieve this effect. Heres my code that creates them in an unordered list:-- <?php $link = mysqli_connect('localhost','',''); mysqli_select_db($link,'mydb'); $query = mysqli_query($link,'SELECT * FROM nested_categories'); while ( $row = mysqli_fetch_assoc($query) ){ $menu_array[$row['id']] = array('name' => $row['name'],'parent' => $row['parent_id']); } function generate_menu($parent){ $has_childs = false; global $menu_array; foreach($menu_array as $key => $value){ if ($value['parent'] == $parent){ if ($has_childs === false){ $has_childs = true; echo '<ul id="categories">'; } echo '<li>'. $value['name']; generate_menu($key); echo '</li>'; } } if ($has_childs === true) echo '</ul>'; } generate_menu(0); ?> I need to build a hierarchical product category table with data from another table. The original product info table has a basic structu ID | SKU | Category1 | Category2 | Category3 1 | 001 | Apparel | Boots | Work Boots 2 | 002 | Apparel | Shirts | Long Sleeve The categories table this is going into: ID | Name | Parent_ID 1 | Apparel | 0 2 | Boots | 1 3 | Work Boots | 2 4 | Shirts | 1 5 | Long Sleeve | 4 I've been trying to use PHP to increment ID's by: SELECT DISTINCT Category1 SELECT DISTINCT Category1,Category2 SELECT DISTINCT Category1, Category2, Category3 Code: [Select] $i='0'; while ($row = mysql_fetch_assoc($result)) { $i++; echo $i . " - ". $row['Category1'] . "<br>"; } One problem is that I can't get the parent ID that was generated in this method. Another is that Category2 and 3 are not uniquely named at all. So my question is, how do I distinctly select all three categories, and get each parent ID also? how do i find out if a parent has any children? for example id - name - parent_id 1 - title1 - 0 (parent) 2 - title2 - 0 (parent) 3 - link1 - 1 (child) 4 - link2 - 1 (child) 5 - link3 - 2 (child) 5 - link4 - 2 (child) the above has 2 parent with 2 links each. so.. if an id has no "parent_id" pointing to it must be a child if an id has "parent_id" pointing to it must be a parent i have a list Code: [Select] <ul> <li><a href="">'.$link.'</a></li> <li><a href="" class="subTitle">'.$title.'</a> <ul class="subLink"> <li><a href="">'.$link.'</a></li> </ul> </li> </ul> so i want to do something like if id has no "parent_id" pointing to it, it is a $link if id has "parent_id" pointing to it, it is a $title any thoughts thanks Code: [Select] $domdoc=new DOMDocument(); $domdoc->formatOutput=TRUE; $empty_cart_xml= '<Order> <Cart> <Items> <Item>1</Item> <Item>2</Item> <Item>3</Item> </Items> </Cart> </Order>'; $domdoc->loadXML($empty_cart_xml); print $domdoc->saveXML()."<hr/>"; //works up to this point $xpath=new DOMXPath($domdoc); $items=$xpath->query('Order/Cart/Items'); foreach($itemses AS $items) { $items->appendChild($domdoc->createElement('Item','4')); } print $domdoc->saveXML(); All I want to do is to add a new Item to Items. What am I doing wrong? in a table cell i have a select statement with options from 1 to 10. Onchange (if they select 3), i want to display a subform in the same table cell where the person can add the for the three children name (Textfield) sex (select: M/F) date of birth (select for day, select for month and select for year) currently, i've tried something with javascript but it's only saves the child_id and parent_id and does not save the rest. Moreover i have no clue how to populate the date of birth. Please if you have any solutions, tell me! Any help is very much appreciated! Hi all, I need some help here. Currently I am using 2 functions. 1) To call out the 'academic levels' 2) To call out the 'subjects' which matches to the specific 'academic level' However, I am stucked in the codes. Please view my picture attachment, my question is, how do I assigned 'red box (Pre-School)' to 'red box (subjects)' and so on...What are the codes or tweaks that I should make so that it will look like 'what do I want to achieve.jpg' My process and execution $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) or die(mysqli_error($dbc)); //Run query $tutor_id = mysqli_real_escape_string($dbc, $_GET['tutor_id']); $query = "SELECT tl.level_id, tl.level_name, ts.subject_id, ts.subject_name, tsl.subject_level_id, IF(tosl.tutor_id='{$tutor_id}', 1, 0) as checked FROM tutor_level AS tl INNER JOIN tutor_subject_level AS tsl USING (level_id) INNER JOIN tutor_subject AS ts USING (subject_id) LEFT JOIN tutor_overall_level_subject AS tosl ON tosl.subject_level_id = tsl.subject_level_id AND tosl.tutor_id = '{$tutor_id}' ORDER BY tl.level_id, ts.subject_name"; $sql = mysqli_query($dbc, $query) or die(mysqli_error($dbc)); $query1 = "SELECT tl.level_id, tl.level_name, IF(tslvl.tutor_id='{$tutor_id}', 1, 0) as checked FROM tutor_level AS tl LEFT JOIN tutor_selected_level AS tslvl ON tslvl.level_id = tl.level_id AND tslvl.tutor_id='{$tutor_id}' ORDER BY tl.level_id, tl.level_name"; $sql1 = mysqli_query($dbc, $query1) or die(mysqli_error($dbc)); //Process the results $checkboxes = ''; //Create variable for output $current_level_id = false; //Flag to check when records change level while($data = mysqli_fetch_array($sql)) //Iterate throug the DB results { if($current_level_id != $data['level_id']) //Determine if this level is different from the last { print_r ($data); $subjectCheckboxes .= createSubjectCheckboxes($subject_data, 5); $current_level_id = $data['level_id']; $subject_data = array(); } //Add the current record to the $level_data array $subject_data[] = $data; } //$checkboxes .= createLevelCheckboxes($subject_data, $level_data, 5); while($data1 = mysqli_fetch_array($sql1)) { print_r ($data1); $levelCheckboxes .= createLevelCheckboxes($level_data, 5); $level_data = array(); $level_data[] = $data1; } //Call the createLevelCheckboxes() function to generate the HTML for the LAST level records $levelCheckboxes .= createLevelCheckboxes($level_data, 5); $subjectCheckboxes .= createSubjectCheckboxes($subject_data, 5); My functions function createLevelCheckboxes($levelArray, $columns) { if(count($levelArray)==0) { return false; } $htmlOutput = ''; foreach($levelArray as $data1) { //Display level header row $levelID = $levelArray[0]['level_id']; $levelName = $levelArray[0]['level_name']; $checked = ($data1['checked'] == 1) ? ' checked="checked"' : ''; $htmlOutput .= "<tr>\n"; $htmlOutput .= "<th colspan=\"{$columns}\" style=\"text-align:left;padding-top:15px;\">"; $htmlOutput .= "<input name=\"level[]\" type=\"checkbox\" id=\"level_{$levelID}\" type=\"checkbox\" {$checked} value=\"{$levelID}\">"; $htmlOutput .= "<span class=\"zone_text_enlarge\"><label for=\"level_{$levelID}\">{$levelName}</label></span>"; $htmlOutput .= "</th>\n"; $htmlOutput .= "</tr>\n"; } return $htmlOutput; } function createSubjectCheckboxes($subjectArray, $columns) { //Display each subject $recordCount = 0; foreach($subjectArray as $data) { //Increment counter $recordCount++; //Start new row if needed, 1/5 = R1 --> So create a new row if ($recordCount % $columns == 1) { $htmlOutput .= "<tr>\n"; } //Display the record $subjectID = $data['subject_level_id']; $subjectName = $data['subject_name']; $checked = ($data['checked'] == 1) ? ' checked="checked"' : ''; $htmlOutput .= " <td>\n"; $htmlOutput .= " <input name=\"subject_level[]\" class=\"subject_a\" type=\"checkbox\" {$checked}"; $htmlOutput .= " id=\"subject_level_{$subjectID}\" value=\"{$subjectID}\"/>\n"; $htmlOutput .= " <label for=\"subject_level_{$subjectID}\" class=\"subject_1\">{$subjectName}</label>\n"; $htmlOutput .= "</td>\n"; //Close row if needed, 5/5 = 0 --> So close the row if ($recordCount % $columns == 0) { $htmlOutput .= "</tr>\n"; } } //Close last row if needed if ($recordCount % $columns != 0) { $htmlOutput .= "</tr>\n"; } return $htmlOutput; } At the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_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 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. 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;'>$type</span></input><br/>"; } } } echo '</td></tr></table>'; ?> </div> I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
Hi guys, I was just wondering if anyone could help me. I've got a My_SQL database containing articles, a summary for the article and a date. I have a basic CMS system set-up, but I want to create a script that when users sign up to a mail list it forwards the summary and dates of the articles database. If that makes sense? But I only want it to forward the most recent 5 rows. I'm pretty new to PHP and I've been mostly following tutorials thus far, but this is quite specific. Thanks in advance! hello I want query from one table and insert in another table on another domain . each database on one domain name. for example http://www.site.com $con1 and http://www.site1.com $con. can anyone help me? my code is : <?php $dbuser1 = "insert in this database"; $dbpass1 = "insert in this database"; $dbhost1 = "localhost"; $dbname1 = "insert in this database"; // Connecting, selecting database $con1 = mysql_connect($dbhost1, $dbuser1, $dbpass1) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname1) or die('Could not select database'); $dbuser = "query from this database"; $dbpass = "query from this database"; $dbhost = "localhost"; $dbname = "query from this database"; // Connecting, selecting database $con = mysql_connect($dbhost, $dbuser, $dbpass) or die('Could not connect: ' . mysql_error()); mysql_select_db($dbname) or die('Could not select database'); //query from database $query = mysql_query("SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1"); while($row=mysql_fetch_array($query)){ $result=$row[0]; $text=$row[1]."</br>Size:(".$row[4].")"; $alias=$row[2]; $link = '<a target="_blank" href='.$row[3].'>Download</a>'; echo $result; } //insert into database mysql_query("SET NAMES 'utf8'", $con1); $query3= " INSERT INTO `jos_content` (`id`, `title`, `alias`, `) VALUES (NULL, '".$result."', '".$alias."', '')"; if (!mysql_query($query3,$con1)) { die('Error: text add' . mysql_error()); } mysql_close($con); mysql_close($con1); ?> the second database found on the cloud
i try to get JSON data but how to insert and update them to another online database with the same table my php script to return json data <?php include_once('db.php'); $users = array(); $users_data = $db -> prepare('SELECT id, username FROM users'); $users_data -> execute(); while($fetched = $users_data->fetch()) { $users[$fetchedt['id']] = array ( 'id' => $fetched['id'], 'username' => $fetched['name'] ); } echo json_encode($leaders);
i get
{"1":{"id":1,"username":"jeremia"},"2":{"id":2,"username":"Ernest"}} Edited March 24 by mahenda <?php $image_url = 'images/'; //User defined variables for page settings $rows_per_page = 2; $cols_per_page = 4; //Master array of ALL the images in the order to be displayed $images = array( 'image1.jpg', 'image2.jpg', 'image3.jpg', 'image4.jpg', 'image5.jpg', 'image6.jpg', 'image7.jpg', 'image8.jpg', 'image9.jpg', 'image10.jpg', 'image11.jpg', 'image12.jpg', 'image13.jpg', 'image14.jpg', 'image15.jpg', 'image16.jpg', 'image17.jpg', 'image18.jpg', 'image19.jpg' ); //END USER DEFINED VARIABLES //System defined variable $records_per_page = $rows_per_page * $cols_per_page; $total_records = count($images); $total_pages = ceil($total_records / $records_per_page); //Get/define current page $current_page = (int) $_GET['page']; if($current_page<1 || $current_page>$total_pages) { $current_page = 1; } //Get records for the current page $page_images = array_splice($images, ($current_page-1)*$records_per_page, $records_per_page); //Create ouput for the records of the current page $ouput = "<table border=\"1\">\n"; for($row=0; $row<$rows_per_page; $row++) { $ouput .= "<tr>\n"; for($col=0; $col<$cols_per_page; $col++) { $imgIdx = ($row * $rows_per_page) + $col; $img = (isset($page_images[$imgIdx])) ? "<img src=\"{$image_url}{$page_images[$imgIdx]}\" />" : ' '; $ouput .= "<td>$img</td>\n"; } $ouput .= "</tr>\n"; } $ouput .= "</table>"; //Create pagination links $first = "First"; $prev = "Prev"; $next = "Next"; $last = "Last"; if($current_page>1) { $prevPage = $current_page - 1; $first = "<a href=\"test.php?page=1\">First</a>"; $prev = "<a href=\"test.php?page={$prevPage}\">Prev</a>"; } if($current_page<$total_pages) { $nextPage = $current_page + 1; $next = "<a href=\"test.php?page={$nextPage}\">Next</a>"; $last = "<a href=\"test.php?page={$total_pages}\">Last</a>"; } ?> <html> <body> <h2>Here are the records for page <?php echo $current_page; ?></h2> <ul> <?php echo $ouput; ?> </ul> Page <?php echo $current_page; ?> of <?php echo $total_pages; ?> <br /> <?php echo "{$first} | {$prev} | {$next} | {$last}"; ?> </body> </html> This current code provides an easy way of uploading pictures and having the other pictures move automatically without the need of a database. But I must wonder, if doing this on a database would be faster. Would it be? I mean the example adobe shows I'm only using 19 pictures, but if I ever reach say 1000 pictures, would the php file be too big? Should I just make a database now or it doesn't matter? Also, the code above, which I took from a phpfreak member, has the pagination not working. Anyone care to do a diagnosis? Thanks! When I do this: $matches++; $sql_update_matches = mysql_query("UPDATE Live_Event SET matches='$matches' WHERE id='$show_id' AND promo_id='$promotion_id'"); But when it should update the database from 0 to 1, it update it to U. What is the problem? Hi there, *little heads up: English isn't my native language, so excuse me for any mistakes in spelling etc.* I've got a register/login script with activation mail, I secure my pages with this bit of code: Code: [Select] } if (!$HTTP_SESSION_VARS["ingelogd"]) { header("Location: login.php"); } Because if someone logs in, the session is made: Code: [Select] // bekijk of de gegevens juist zijn if (mysql_num_rows($res) >= 1) { // registreer sessie (of cookie) $ingelogd = mysql_result($res, 0); session_register(ingelogd); // setcookie("ingelogd", "$ingelogd", time() + 1 * 86400); echo "<p style=\"font-family: Arial, Helvetica, sans-serif; font-size: 13px; color: #000000;\"> U bent succesvol ingelogd.</p>"; echo "<p style=\"font-family: Arial, Helvetica, sans-serif; font-size: 13px; color: #000000;\"> Klik hier om <a href=\"home.php\">verder</a> te gaan.</p>"; } My question: When someone registered, they entered their 'real' name, which is then placed into the database in the same row as the username, password etc. I would like to state on the homepage which is secured 'Welcome 'Name'!' I do I get this 'Name' value out of the database and into my page? Thanks in advance! Sander Hi, Just need to know if I am on the right lines, basically I am about to create two tables; one to hold customer information and another to hold information the customer gave me. Both tables will be populated with information that they give me at the same time (in other words from a single form). Just want to know, if I have a field called ID set to a primary key in both the tables, will that be enough to link them both together? Furthermore, what would the mySQL code look like if I wanted to view all the information referencing to specific ID? so info from the customer table and the other. Any and all help would be great. Lee Not sure if I can post this here, or if I should go to a mySQL forum but... I'm trying to keep track of my user's "score". Each time they do a certain thing they get a preset amount of points, depending on what they did. The way I see it would be having a table that is joined to the user's ID, and have a column for each task. Then when they do a task the number in that column will go up. Then I will multiply that number by amount the task is worth. I can't just adjust there score because they need to be able to see what they did. So I hope that made sense. And if you know a better way to deal with this please let me know! |
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