codewalkz

[CODE]<?php mysql_connect("localhost", "db_username_here", "db_password_here") or die(mysql_error()); mysql_select_db("yourdbname") or die(mysql_error()); $username = $_COOKIE['ID_my_site']; $pass = $_COOKIE['Key_my_site']; $query = "SELECT ID FROM agents WHERE username = '$username'"; $result = mysql_query($query) or die('Error : ' . mysql_error()); $row = mysql_fetch_array($result); $owner = $row['ID']; $uploadDir = 'images/'; if(isset($_POST['submit'])) { $fileName = $_FILES['file']['name']; $tmpName = …

if you're actually in "design" phase then it does not need a dynamic page to use php etc. But if you're pulling the information from some kind of database and placing it in your custom design.... Still tell us what you really need hehe

The problem is because of an invalid close tag "{". The error is not in the code you provided obviously because each are closed. :(

This could be done by passing your form into a php or any dynamic page which will store the data submitted into a database or csv or whatever.

[CODE]$result = mysql_query("SELECT games.*, console.name AS gconsole , genre.name AS ggenre , company.name as gcompany, status.name as gstatus from games as games left join console on (games.console = console.id) left join genre on (games.genre = genre.id) left join company on (games.company = company.id) left join status on (games.status = status.id)"); …

[CODE]<?php //DB Info // connect to db mysql_connect("$host", "$username", "$password")or die("cannot connect server "); mysql_select_db("$db_name")or die("cannot select DB"); $result = mysql_query("SELECT * FROM table ORDER BY year") or die(mysql_error()); while($row = mysql_fetch_array( $result )) { $xml_output2 = "<li id=\""; $xml_output2 .= $row['year']."\">"; $xml_output2 .= "<img src=\"/images/".$y.".png\" width=\"256\" height=\"256\" />"; $xml_output2 …

explain what you need more.

What exactly are you trying to do? Where is the image? How do you include it in the process?

[QUOTE=stbuchok;1671941]Please, if this is a business site that is supposed to be professional, don't just learn something over the next few weeks and slap something together. If you don't know how to do it, hire a professional. You are going to have huge gaping security holes all over the place. …

jQuery is your friend on this :> Submitform without reloading via callback

Dude you better check this simple one: [url]http://www.phpeasystep.com/phptu/29.html[/url]

I currently have this function [CODE] processfn('sponsor = 1', 1, '', 0, 'buyer = 0'); function processfn($sponsor, $level, $sad, $sadista, $bumili) { $result = mysql_query("SELECT id FROM ahentes WHERE $sponsor") or die(mysql_error()); if (mysql_num_rows($result) > 0) { $sad = mysql_num_rows($result); while ($row = mysql_fetch_array($result)){ $sila[] = "buyer = '".$row['id']."' "; …

It's like a tree pyramid. I want to get the members sponsored by the current user. The result will be used to get its sponsored members and so on until everyone is read. How can I achieve this? I currently have this code but I can only go to level …

I am currently making a binary tree and I was able to create the tree. What I am facing now is " How to get the total number of children in my left side and total number in my right side ". The structure is: [CODE] 1 2 3 4 …

Hi, I need something like recursion wherein the query repeats itself using it's result to start the NEXT - SAME - QUERY. I made 1 but it just simple reecho the same result. [CODE]// KUNIN ANG ID NI KURIMAW $root = $_GET['id']; $result = mysql_query("SELECT COUNT(*) as count, id, sponsor …

Hi I currently have date in database in this format: 11-7-2009 But I think I need to echo this in this format: 2009-11-07

I would like to ask how can I check if a sample date is earlier than another specified one. I have this: $Monday = 1-18-2010 $enroll = 12-29-2009 [CODE] $enroll = $row['enroll']; if($enroll >= $Monday){ echo "<br>$enroll is earlier than $Monday this returns TRUE ( 1 )"; }else{ echo "<br>$Monday …

i need to know if there is a member whose enrollment date is earlier than a specified date. If this case, the specified date = $Monday [CODE] $Monday = 12-30-2009; $result = mysql_query("SELECT COUNT(*) FROM agents WHERE (sponsor = '$parent') and (lft between $leftlft AND $leftrgt) and (enrollment_date >= '$Monday')"); …

Hi, i am working on a member database. I need to get all the registrations except the current week. I have a "week" field in my database. This one is fine and echos all the records from week 1 to 8: [CODE]SELECT COUNT(*) FROM agents WHERE week < 9 and …

You can check this Pagination script: [url]http://www.phpeasystep.com/phptu/29.html[/url]

I am having this 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 ''3'' at line 1 The $get1 value is 3. I tried enough but to no avail. I don't see any …

I can't seem to limit the result to a desired number. I have this code: [code]//GET THE TOTAL OF LEVEL 2 $result = mysql_query("SELECT COUNT(*) FROM agents WHERE (sponsor = '1' or sponsor = '2' or sponsor = '3') LIMIT 0,1")or die(mysql_error()); while($row = mysql_fetch_array( $result )) { $lvl2parent = …

I believe you are asking on how to create a binary tree. Are on MLM Project? I was able to get what you want. You can refer here bro: [url]http://articles.sitepoint.com/article/hierarchical-data-database/1[/url] Goodluck! It needs a bit of patience and understanding ^_^ It's the best tutorial i saw when it comes to …

i must be able to extract the results of this query and use it as another value for "Parent" in the succeeding "THE SAME QUERY". Right here, the value of parent in WHERE clause is "1". Now, I want to create the same query but the value of the parent …

I am currently working on an MLM System which supports Unilevel with Dynamic Compression. So far, though the below code looks really messy, it was able to get what I want which is to get the Unilevel 8 Levels Deep. The system should run this way: 1 is sponsor of …

RewriteEngine on RewriteRule ^([^/\.]+)/?$ index.php?j=$1 [L] This makes index.php?j=sample look like [url]http://www.domain.com/sample[/url] Now, i want to except [url]http://www.domain.com/account[/url] from the rewrite rule. How can I proceed? Thanks a lot!

[CODE]<?php mysql_connect("localhost", "db_username_here", "db_password_here") or die(mysql_error()); mysql_select_db("yourdbname") or die(mysql_error()); $username = $_COOKIE['ID_my_site']; $pass = $_COOKIE['Key_my_site']; $query = "SELECT ID FROM agents WHERE username = '$username'"; $result = mysql_query($query) or die('Error : ' . mysql_error()); $row = mysql_fetch_array($result); $owner = $row['ID']; $uploadDir = 'images/'; if(isset($_POST['submit'])) { $fileName = $_FILES['file']['name']; $tmpName = …

Anyone who knows how to apply this in php scripting? the scripts in this site can be xecuted in mysql console but I can't get it working in php script in the website im studying. I want to create a tree like this: [URL="http://sitepointstatic.com/graphics/sitepoint_numbering.gif"]http://sitepointstatic.com/graphics/sitepoint_numbering.gif[/URL]

mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("test") or die(mysql_error()); $root = 1; function display_tree($root) { // retrieve the left and right value of the $root node $result = mysql_query('SELECT lft, rgt FROM tree '. 'WHERE name="'.$root.'";'); $row = mysql_fetch_array($result); // start with an empty $right stack $right = array(); // now, …

Hi, I have a related problem: [url]http://www.daniweb.com/forums/thread231989.html[/url] I have a binary tree and I would like to get the number of people in each side of a specific user. For example: 1 is upline of 2 and 3 I now want to get the total number of people under under …