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Deivas

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    Sorting

    in MySQL Help

    Posted

    Ok,

    When I do

    SELECT name, votes AS votes, @num := @num +1 AS Rank
FROM `servers` a, (

SELECT @num :=0
)b
ORDER BY votes DESC
LIMIT 0 , 10

    and when I go to the next page, it starts from 1 again?

    Sorting

    in MySQL Help

    Posted

    For example... I have:

     

    +------------------+---------+

    | ID____________| VOTES_ |

    +------------------+---------+

    |1_____________ | 55 ____|

    +------------------+---------+

    |2_____________ | 99 ____|

    +------------------+---------+

     

     

    ... and how can I make so it will be like this:

    +------------------+---------+

    | # ____________| VOTES_ |

    +------------------+---------+

    |1_____________ | 99 ____|

    +------------------+---------+

    |2_____________ | 55 ____|

    +------------------+---------+

    PHP ranking

    in PHP Coding Help

    Posted

    I do not understand your problem. Explain what this ranking does and the problem.

     

    $rank = 0;
while($info = mysql_fetch_array( $result )) {
    $rank++;
    Print "<tr>";
    Print "<td><a href=\"?id=" . $info['id'] . "\">" .$info['name'] . "</a></td> ";
    Print "<td>".$info['ip'] . " </td>";
    Print "<td>".$info['port'] . " </td>";
    Print "<td>".$info['shortdesc'] . " </td>";
    Print "<td>".$info['type'] . " </td>";
    Print "<td>".$info['username'] . " </td>";
    Print "<td>".$info['votes'] . " </td>";
}
echo "</tbody>";
Print "</table>";

     

    When I go to the next page $rank starts from 0 again.

    PHP ranking

    in PHP Coding Help

    Posted

    Ok,

    I have this: 

    $rowsPerPage = 50;

$pageNum = 1;

if(isset($_GET['page'])) {
    $pageNum = $_GET['page'];
}

$offset = ($pageNum - 1) * $rowsPerPage;

$type = $_GET['type'];

$query = " SELECT * FROM servers WHERE type = '$type'" .
        " LIMIT $offset, $rowsPerPage";
$result = mysql_query($query) or die('Error, query failed');


echo "
<table border=\"0\" id=\"hor-minimalist-b\">
<thead>
        <tr>
            <th scope=\"col\">#</th>
            <th scope=\"col\">Name</th>
            <th scope=\"col\">URL/IP</th>
            <th scope=\"col\">Port</th>
            <th scope=\"col\">Short description</th>
            <th scope=\"col\">Type:</th>
            <th scope=\"col\">Added by</th>
            <th scope=\"col\">Votes</th>
        </tr>
    </thead>
    <tbody>
";
$rank = 0;
while($info = mysql_fetch_array( $result )) {
    $rank++;
    Print "<tr>";
    Print "<td><a href=\"?id=" . $info['id'] . "\">" .$info['name'] . "</a></td> ";
    Print "<td>".$info['ip'] . " </td>";
    Print "<td>".$info['port'] . " </td>";
    Print "<td>".$info['shortdesc'] . " </td>";
    Print "<td>".$info['type'] . " </td>";
    Print "<td>".$info['username'] . " </td>";
    Print "<td>".$info['votes'] . " </td>";
}
echo "</tbody>";
Print "</table>";



$query   = "SELECT COUNT(*) AS numrows FROM servers WHERE type='rsps'";
$result  = mysql_query($query) or die('Error, query failed');
$row     = mysql_fetch_array($result, MYSQL_ASSOC);
$numrows = $row['numrows'];


$maxPage = ceil($numrows/$rowsPerPage);

$self = $_SERVER['PHP_SELF'];
$nav  = '';

for($page = 1; $page <= $maxPage; $page++) {
    if ($page == $pageNum) {
        $nav .= " $page ";
    }
    else {
        $nav .= " <a href=\"$self?page=$page&type=$type\">$page</a> ";
    }
}

if ($pageNum > 1) {
    $page  = $pageNum - 1;
    $prev  = " <a href=\"$self?page=$page&type=$type\">[Prev]</a> ";

    $first = " <a href=\"$self?page=1&type=$type\">[First Page]</a> ";
}
else {
    $prev  = ' ';
    $first = ' ';
}

if ($pageNum < $maxPage) {
    $page = $pageNum + 1;
    $next = " <a href=\"$self?page=$page&type=$type\">[Next]</a> ";

    $last = " <a href=\"$self?page=$maxPage&type=$type\">[Last Page]</a> ";
}
else {
    $next = ' ';
    $last = ' ';
}

echo $first . $prev . $nav . $next . $last;

     

    When I go to the next page, the $ranks starts over again.

    How can I make it to continue counting.

    Edit website live

    in PHP Coding Help

    Posted

    Hey,

    is it possible to edit your website live (like when you enter: 

    javascript:document.body.contentEditable='true'; document.designMode='on'; void 0

    into your address bar)?

    I need this because I can't afford to buy DreamViewer.

    If not,

    how can I add javascript:document.body.contentEditable='true'; document.designMode='on'; void 0 to my website so I will not have to keep entering it every time i edit my website?

     

    I don't mean to sound offensive if I am, but are you 12? You cannot edit your web page with this code, You can manipulate it but refreshing will remove anything you have done.

     

    Dreamweaver is expensive and pointless, you may as well get something free/open source and good such as NVU

     

    I was trying to edit the post but it didn't met me.

     

    I use 

    javascript:document.body.contentEditable='true'; document.designMode='on'; void 0

    this when I make like news pages, so I can see what it will look like on the page,

    I was just wondering if there is a free program like "DreamWeaver".

     

    Array not producing what I intended

    in PHP Coding Help

    Posted 

    <?php
$Question["1"] = '1-A';
$Question["2"] = '2-A';
$Question["3"] = '1-A';
$Question["4"] = '2-A';
$Question["5"] = '1-B';



echo "1 - (" . $Question["1"] . ")<br />";
echo "2 - (" . $Question["2"] . ")<br />";
echo "3 - (" . $Question["3"] . ")<br />";
echo "4 - (" . $Question["4"] . ")<br />";
echo "5 - (" . $Question["5"] . ")<br />";

?>

    or 

    <?php
$Question["1"] = '1-A';
$Question["2"] = '2-A';
$Question["3"] = '1-A';
$Question["4"] = '2-A';
$Question["5"] = '1-B';



echo "1(" . $Question["1"] . "),";
echo "2(" . $Question["2"] . "),";
echo "3(" . $Question["3"] . "),";
echo "4(" . $Question["4"] . "),";
echo "5(" . $Question["5"] . ")";

?>

    Edit website live

    in PHP Coding Help

    Posted

    Hey,

    is it possible to edit your website live (like when you enter: 

    javascript:document.body.contentEditable='true'; document.designMode='on'; void 0

    into your address bar)?

    I need this because I can't afford to buy DreamViewer.

    If not,

    how can I add javascript:document.body.contentEditable='true'; document.designMode='on'; void 0 to my website so I will not have to keep entering it every time i edit my website?

    $_GET

    in PHP Coding Help

    Posted

    Could you explain you problem a little more? What are you trying to do?

    there's $_GET[] to get a variable from the url

    and there's $_POST[] to get something from a form the page before

     

    I have a lot of links in my website, and on most of them I need to do this (this might be wrong) : 

    if($_GET['type'] != null){
                                                if($_GET['order'] == null){
                                            $nav .= " <a href=\"$self?page=$page&&type=$type\">$page</a> ";
                                                }
                                                if($_GET['order'] != null){
                                            $nav .= " <a href=\"$self?page=$page&&type=$type&&order=$order\">$page</a> ";
                                                }
                                            }
                                            if($_GET['type'] == null){
                                            $nav .= " <a href=\"$self?page=$page\">$page</a> ";
                                            }

    and etc ,

    I am asking if it is possible to do it faster?

    You have an error ...

    in MySQL Help

    Posted

    For example I have: 

    $data123 = mysql_query("SELECT * FROM servers WHERE type = '$type' LIMIT $offset, $rowsPerPage")
                                            or die(mysql_error());

     

    If the type is not selected nothing prints out. What can I do to make it if nothing is selected all results print out?

     

    FIXED.

    You have an error ...

    in MySQL Help

    Posted

    For example I have: 

    $data123 = mysql_query("SELECT * FROM servers WHERE type = '$type' LIMIT $offset, $rowsPerPage")
                                            or die(mysql_error());

     

    If the type is not selected nothing prints out. What can I do to make it if nothing is selected all results print out?

    $_GET

    in PHP Coding Help

    Posted

    How to find out if there is no $_GET method.

    At the moment I have an if statement:

     

    if($_GET['id']){
$link1 = mysql_connect("****","*****","*****");
mysql_select_db("*******", $link1);

$result1 = mysql_query("SELECT * FROM servers WHERE id = $id", $link1);
$num_rows1 = mysql_num_rows($result1);

$data = mysql_query("SELECT * FROM servers WHERE id = $id")
or die(mysql_error());

if($num_rows1 == 1){
while($info1 = mysql_fetch_array( $data ))
{
echo " 
<table border=\"1\">
<tr>
	<td><b>Name          </a></td>
	<td>" .$info1['name'] ."</td>
</tr>
<tr>
	<td><b>Type          </a></td>
	<td>" .$info1['type'] ."</td>
</tr>
<tr>
	<td><b>URL/IP:          </b></td>
	<td>" .$info1['ip'] ."</td>
</tr>
<tr>
	<td><b>Port          </b></td>
	<td>" .$info1['port']. "</td>
</tr>
<tr>
	<td><b>Description          </b></td>
	<td>" .$info1['desc']. "</td>
</tr>
</table>
<hr>
<h4>Other information</h4>
<table border=\"1\">
<tr>
	<td><b>Added by          </b></td>
	<td>" .$info1['username']. "</td>
</tr>
<tr>
	<td><b>ID          </b></td>
	<td>" .$info1['id']. "</td>
</tr>
<tr>
	<td><b>Date added (m/d/y)          </b></td>
	<td>" .$info1['date']. "</td>
</tr>
</table>

";
}
}

}

    What if statement should I do for normal stuff (just for index.php (not index.php?id=324234))?

    index.php?blah

    in PHP Coding Help

    Posted

    How do I make it so if someone clicks on a link and it will sort everything by votes descending and the link will change to something like : index.php?sort=votes

     

    Do I have to use $_GET method?

    How come?

    in PHP Coding Help

    Posted

    How come that if i do that at my homepage it prints out the username.

     

    if(isSet($_SESSION['username'])) {
           echo $_SESSION['username'];}else{
echo "lol";
}

     

    if I do that in another directory it prints out lol.

    password

    in PHP Coding Help

    Posted

    For example:

    My username is : Hello

    and my password is : bye.

     

    If I enter Hello and bye in the login it says "The submitted login info is incorrect.".

     

    But if I change my password to 123, it works.

     

    Can anyone tell me whats wrong?

     

    do_login.php

    <?php
if(empty($_POST)) exit;

include 'config.php';
include_once 'connect.php';

// declare post fields

$post_username = $_POST['username'];
$post_password = $_POST['password'];

$post_autologin = $_POST['autologin'];

  $query = "select username,password from app_members where username='$post_username' and '$post_password'";
  $result = mysql_query($query) or die("Could not query players");
  $result2 = mysql_fetch_array($result);
  if ($result2)
  {
    $_SESSION['username'] = $post_username;

// Autologin Requested?

if($post_autologin == 1)
{

setcookie ($cookie_name, 'usr='.$config_username, time() + $cookie_time);
}

exit('OK');
}
else
{
echo '<div id="error_notification">The submitted login info is incorrect.</div>';
}
?>

     

    login.php

    <?php
require_once 'config.php';

// Is the user already logged in? Redirect him/her to the private page

if(isSet($_SESSION['username']))
{
header("Location: private.php");
exit;
}
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
  <TITLE>AJAX Login Form</TITLE>

  <META name="Author" Content="Bit Repository">
  <META name="Keywords" Content="ajax, login, form, mootools">
  <META name="Description" Content="A nice & simple AJAX Login Form">

  <script type="text/javascript" src="js/mootools-1.2.1-core-yc.js"></script>
  <script type="text/javascript" src="js/process.js"></script>

  <link rel="stylesheet" type="text/css" href="style.css" />
</HEAD>

<BODY>

<center>

<div id="status">

<fieldset><legend align="center">Authentication</legend>

<div id="login_response"><!-- spanner --></div>

<form id="login" name="login" method="post" action="do_login.php">
<table align="center" width="300" border="0">
<tr>
<td width="80">Username</td><td><input id="name" type="text" name="username"></td>
</tr>
<tr>
<td>Password</td>
<td><input type="password" name="password"></td>
</tr>
<tr>
<td> </td>
<td><input type="checkbox" name="autologin" value="1">Remember Me</td>
</tr>
<tr>
<td> </td>
<td><input id="submit" type="submit" name="submit" value="Login">
<div id="ajax_loading"><img align="absmiddle" src="images/spinner.gif"> Processing...</div></td>
</tr>
</table>
</form>
</fieldset>

</div>
</center>
</BODY>
</HTML>

     

    if you will need anything else tell me.

    Variables?

    in PHP Coding Help

    Posted

    You need a few things. A database with the table to hold the data for the "websites". A form to add "websites" to the table. A page to display the data from the database table.

     

    I would start with the database/table and populate it manually with a few rows. Then I would move on to a page to display it. Once that works I would make another page that has a form and the code to add the data to the database/table. Then I would think about some type of registration/administration page for the whole thing.

     

    Then seeing as this seems to be your first website/application I would figure out all the things it does and doesnt do and how I would make it better and then start over again.

     

     

    HTH

    Teamatomic

     

    I have a database, a page to add websites, and a website to show the websites is http://darkdrift.net/lol/htdocs/StuffOnline/servers.php , as you can see it says "Please choose a type:" you choose a type from the select box and click "Continue" and then a list of websites show up. But when someone clicks on "Continue" I want the link to change to http://darkdrift.net/lol/htdocs/StuffOnline/servers.php?type=websites or something, and when I will finish that I will make it so if someone clicks on the name it will redirect to a page and it will show then information about the website, and the site will be something like: http://darkdrift.net/lol/htdocs/StuffOnline/servers.php?type=websites;id=4523

    Variables?

    in PHP Coding Help

    Posted 

    <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
    <link rel="stylesheet" href="./style.css" type="text/css"/>
    <title>Online Servers | Top list</title>

    <body>

        <?php
        include './includes/banner.php';
        include './includes/menu.php';
        ?>

        <div id="content">
            <table id="content" width="86%" align="left">
                <tr>
                    <td><h1>Top List</h1><p>

                        <form name="form1" action="servers.php" method="post">
                            Please choose a type:
                            <select name="type">
                                <option value="website">Webistes</option>

                            </select><br/>
                            <input type="submit" value="Continue"/>
                        </form>
                        <hr/>
                        <?php
                        include 'connect.php';
                        $type = $_POST['type'];

                        $sitequery  = "SELECT * FROM site ORDER BY votes DESC";
                        $siteresult = mysql_query($sitequery);

                        if($type == website) {
                            return $site;
                            while($row = mysql_fetch_array($siteresult, MYSQL_ASSOC)) {
                                echo "
                                Name : {$row['sitename']}</a> <br>" .
                               "URL : {$row['siteurl']} <br>" .
                               "Short description : {$row['shortdetails']} <br>" .
                               "Votes : {$row['votes']} <br><hr><br>";
                            }

                        }
                        ?>

                </tr>
            </table>
        </div>
    </body>
</html>

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