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chriscloyd

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Posts posted by chriscloyd

    [SOLVED] shoing errors using implode or explode

    in PHP Coding Help

    Posted

    http://www.chaoslegionclan.net/CEGL-Work/newsite/index.php?p=register

     

    yes that link i showed u shows you the form even when i eneter info into the required ones

    which r

    username password confirm password email confirm email i still get the errors heres my updated code

     

    <?php
include("config.php");
//check required fileds 
//username password c_password email c_email
$reason = array();
$username = $_POST['username'];
$password = $_POST['password'];
$c_password = $_POST['c_password'];
$email = $_POST['email'];
$c_email = $_POST['c_email'];
if (empty($username)) {
$reason[] = '-You did not submit a username.';
}
if (empty($password)) {
$reason[] = '-You did not submit a password.';
}
if (empty($c_password)) {
$reason[] = '-You did not submit a email address.';
}
if (empty($email)) {
$reason[] = '-You did not submit a password.';
}
if (empty($c_email)) {
$reason[] = '-You did not submit a confirmation email address.';
}
if (!empty($reason)) {
$reasons = implode("<br>",$reason);
header("Location: ../index.php?p=register&error=".$reasons);
}
?>

    [SOLVED] shoing errors using implode or explode

    in PHP Coding Help

    Posted 

    <?php
include("config.php");
//check required fileds 
//username password c_password email c_email
if ($_POST['username'] == '') {
$error = '1';
$reason = '-You did not submit a username.';
}
if ($_POST['password'] == '') {
$error = '1';
$reason = '-You did not submit a password.';
}
if ($_POST['c_password'] == '') {
$error = '1';
$reason = '-You did not submit a email address.';
}
if ($_POST['email'] == '') {
$error = '1';
$reason = '-You did not submit a password.';
}
if ($_POST['c_email'] == '') {
$error = '1';
$reason = '-You did not submit a confirmation email address.';
}
if ($error) {
//return to register page and display reasons
}
?>

    say if they did not fill in any of those fields I want to go back and show all the errors when i do it, it was only showin 1 error

    random

    in PHP Coding Help

    Posted

    Okay im trying to do something that seems very hard that i cant even think of how to do it as of right now.  Okay.

    I have 80 teams in a league i created im trying to run a script to make each team get matched up to 8 random teams

    how could i go about doing this

    [SOLVED] echo problem

    in PHP Coding Help

    Posted

    okay heres my problem im get a var $for_id

    when i echo it by its self it shows but when i echo it in a link it wont show

     

    heres my code

     

    <link href="cl.css" rel="stylesheet" type="text/css">
<?php
include("config.php");
//viewtopic.php
$topic_id = $_GET['id'];
$topic_info = mysql_query("SELECT * FROM cl_topics WHERE top_id = '$topic_id'");
$topic = mysql_fetch_array($topic_info);
$old_top_views = $topic['top_views'];
$top_views = $topic['top_views'] + 1;
mysql_query("UPDATE cl_topics SET top_views = '$top_views' WHERE top_id = '$topic_id' ");
$forum_id = $topic['for_id'];
$forum_info = mysql_query("SELECT `for_title` FROM cl_forum WHERE for_id = '$forum_id'");
$forum = mysql_fetch_array($forum_info);
$mod_all = mysql_query("SELECT * FROM cl_moderators WHERE for_id = '$forum_id'");
$mod = mysql_fetch_array($mod_all);
$mod_num = mysql_num_rows($mod_all);
if ($mod_num > 1) {
$show_mod = 'Moderator: ';
} else {
$show_mod = 'Moderators: ';
}
echo '<a href="forums.php" class="headerright">Chaos Legion Forums</a><br>
-> <a href="viewforum.php?id="';
echo $topic['for_id'];
echo '" class="headerright">'.$forum['for_title'].'</a><br>
--> <span class="headerright">'.$topic['top_title'].'</span> ';
?>
<a href="javascript:void();" class="style1" onClick="show_hide('topic')">+/-</a>
<?
if ($mod['username'] == $_SESSION['cluser']) {
echo '<img src="images/topic_delete.gif" width="20" height="18" border="0" /> <img src="images/topic_move.gif" width="20" height="18" border="0" /> <img src="images/topic_lock.gif" width="20" height="18" border="0" />';
}
echo '<br><br><br>Start Test for $topic[for_id]<br>';
echo $topic['for_id'];
echo '<br>End Test for $topic[for_id]<br><br><br>';
//get all topics and post numbers and what not
echo '<div id="topic">';
echo '</div>';
echo '<br><br><a href="posttopic.php?id='.$forum_id.'"><img src="images/newtopic.gif" width="82" height="25" border="0" alt="Post new topic"></a> <a href="postreply.php?id='.$topic_id.'"><img src="images/reply.gif" width="82" height="25" border="0" alt="Post reply"></a>';
?>

     

    to view what im talking about and the actual code check here

    try clicking on general forum thats what im trying to echo it in

    http://www.chaoslegionclan.net/viewtopic.php?id=1

    [SOLVED] My output from mysql is messing up

    in PHP Coding Help

    Posted

    okay heres my problem

    when i have my output echoed its showing L

    heres my code

    <?php
$dbhost = 'localhost';
$dbuser = '******';
$dbpass = '******';
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');
$dbname = '******';
mysql_select_db($dbname);
//news//
$get_news = mysql_query("SELECT * FROM `cl_news` ORDER BY `news_id` DESC LIMIT 0 , 5");
if ($get_news) {
while ($news = mysql_fetch_assoc($get_news)) {
	$title = $news['news_title'];
	$news = $news['news_message'];
	$author = $news['news_author'];
	$date = $news['news_date'];
	$id = $news['news_id'];
	?>
	<table width="100%" border="0" cellspacing="1" cellpadding="0">
	<tr>
	<td colspan="2" class="headerright"><?php echo $title; ?>
	<a href="javascript:void();" class="style1" onclick="show_hide('<?php echo $id; ?>')">+/-</a></td>
	</tr>
	</table>
	<table id="<?php echo $id; ?>" width="100%" border="0" cellspacing="1" cellpadding="0">
	<tr>
	<td colspan="2" class="contentright"><?php echo $news; ?></td>
	</tr>
	<tr>
	<td width="68%" class="contentright">Posted on : <?php echo $author; ?><br>
	On : <?php echo $date; ?><br>
	</td>
	<td width="32%" class="contentright"><a href="view_news.php?id=<?php echo $id; ?>">Comments (0)</a> </td>
	</tr>
	</table>
	<?php	
}
} else {
echo 'Theres no news to display right now. Sorry for the inconvenience.';
}
?>

     

    but on phpmyadmin i did the query because theres only 1 news article in there this is what shows up

     

    SQL query: SELECT * FROM `cl_news` LIMIT 0, 15 ;

    Rows: 1

    news_id news_title news_author news_date news_views news_message

    1 This is a test news haqshot Wednesday, February 28, 2007 2:52pm 0 Lorem ipsum dolor sit amet, consectetuer adipiscin...

    call a function from php class

    in PHP Coding Help

    Posted

    Okay i have some functions in my php class but im tyring to call a function from inside my class like below but it does not work heres my code

    [code]<?php
    class getproject
    {
    private $id;
        function __construct() {
    if (!isset($_GET['project'])) {
    $get_project = mysql_query("SELECT * FROM projects ORDER BY id DESC");
    $pid = mysql_fetch_assoc($get_project);
    $this->id = $pid['id'];
    } else {
    $this->id = $_GET['project'];
    }
        }
    function name() {
    if ($result = mysql_query("SELECT `title` FROM projects WHERE id = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['title'];
    }
    }
    }
    function type() {
    if ($result = mysql_query("SELECT `type` FROM projects WHERE id = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['type'];
    }
    }
    }
    function start() {
    if ($result = mysql_query("SELECT `startdate` FROM projects WHERE id = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['startdate'];
    }
    }
    }
    function enddate() {
    if ($result = mysql_query("SELECT `enddate` FROM projects WHERE id = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['enddate'];
    }
    }
    }
    function file1() {
    if ($result = mysql_query("SELECT `location` FROM project_screenshots WHERE pid = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['location'];
    }
    }
    }
    function file2() {
    if ($result = mysql_query("SELECT `location` FROM project_screenshots WHERE pid = '{$this->id}' AND id != '{$project->file1()}")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['location'];
    }
    }
    }
    }
    ?>[/code]

    php fetch array

    in PHP Coding Help

    Posted

    heres my error im getting

    Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\includes\liveprojects.php on line 12

    heres the code i have

    [code]<?php
    $get_num_project = mysql_query("SELECT * FROM projects WHERE status = 'Completed'");
    $num_project = mysql_num_rows($get_num_project);
    if ($num_project <= 5) {
    $maxlimit = 5;
    $minlimit = 1;
    } else {
    $maxlimit = rand(6,$num_project);
    $minlimit = ($maxlimit - 5);
    }
    $project_get = mysql_query("SELECT * FROM projects WHERE status = 'Completed' LIMIT '$minlimit','$maxlimit'");
    while ($p = mysql_fetch_array($project_get)) {
    echo '<table width="95%" border="0" cellspacing="0" cellpadding="0">
      <tr>
        <td width="85%">'.$p['title'].'</td>
        <td align="center" width="15%"><a href=protfolio.php?project='.$p['id'].'">View</a></td>
      </tr>
    </table>';
    }
    ?>[/code]

    [SOLVED] a mysql_fetch error using php

    in PHP Coding Help

    Posted

    heres my error

    [quote]Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/srxstud/public_html/test22/includes/liveprojects.php on line 7
    [/quote]
    heres my code

    [code]<?php
    $get_num_project = mysql_query("SELECT * FROM projects WHERE status = 'Completed'");
    $num_project = mysql_num_rows($get_num_project);
    $maxlimit = rand(1,$num_project);
    $minlimit = $maxlimit - 5;
    $project_get = mysql_query("SELECT * FROM projects WHERE status = 'Completed' LIMIT '$minlimit','$maxlimit'");
    while ($p = mysql_fetch_assoc($project_get)) {
    echo '<table width="95%" border="0" cellspacing="0" cellpadding="0">
      <tr>
        <td width="85%">'.$p['title'].'</td>
        <td align="center" width="15%"><a href=portfolio.php?project='.$p['id'].'">View</a></td>
      </tr>
    </table>';
    }
    ?>[/code]

    create image thumbnail

    in PHP Coding Help

    Posted

    Hi there,
    Right now i have a script that uploads an image to my server.  Im having trouble makeing the script to make it a thumbnail ,  so pretty much a problem creating two copies of the file im uploading and making one a thumbnail heres my script so far can someone help me out?

    [code]
    <?php
    include("../config.php");
    $id = $_GET['id'];
    $get_project = mysql_query("SELECT * FROM projects WHERE id = '$id'");
    $project = mysql_fetch_assoc($get_project);
    $pclient = $project['clientid'];
    $get_client = mysql_query("SELECT * FROM users WHERE id = '$pclient'") or die(mysql_error());
    $client = mysql_fetch_array($get_client);
    $pname = $client['first'].' '.$client['last'];
    $file_name = $HTTP_POST_FILES['screenshot']['name'];
    $random_digit = rand(0000,9999);
    $new_file_name = "$pname.$random_digit.$file_name";
    $path= "screenshots/".$new_file_name;
    $pid = $_GET['id'];
    copy($HTTP_POST_FILES['screenshot']['tmp_name'],$path);
    mysql_query("INSERT INTO project_screenshots (`pid`,`name`,`location`) VALUES ('$pid','$new_file_name','$path')");
    header("Location: ../projects.php?page=viewproject&id=$id");
    ?>
    [/code]

    php class

    in PHP Coding Help

    Posted

    heres the error im getting

    Parse error: syntax error, unexpected T_STRING, expecting T_OLD_FUNCTION or T_FUNCTION or T_VAR or '}' in /home/srxstud/public_html/test22/includes/porfolio_getfunctions.php on line 4

    heres my code


    [code]
    <?php
    class getproject
    {
    private $id;
        function __construct() {
    if (!isset($_GET['project'])) {
    $get_project = mysql_query("SELECT * FROM projects ORDER BY id DESC");
    $pid = mysql_fetch_assoc($get_project);
    $this->id = $pid['id'];
    } else {
    $this->id = $_GET['project'];
    }
        }
    function name() {
    if ($result = mysql_query("SELECT `title` FROM projects WHERE id = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['title'];
    }
    }
    }
    function type() {
    if ($result = mysql_query("SELECT `type` FROM projects WHERE id = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['type'];
    }
    }
    }
    function start() {
    if ($result = mysql_query("SELECT `startdate` FROM projects WHERE id = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['startdate'];
    }
    }
    }
    function enddate() {
    if ($result = mysql_query("SELECT `enddate` FROM projects WHERE id = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['enddate'];
    }
    }
    }
    function file1() {
    if ($result = mysql_query("SELECT `location` FROM projects_files WHERE pid = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['location'];
    }
    }
    }
    function file2() {
    if ($result = mysql_query("SELECT `location` FROM projects_files WHERE pid = '{$this->id}'")) {
    if (mysql_num_rows($result) > 0) {
    $row = mysql_fetch_assoc($result);
    return $row['location'];
    }
    }
    }
    }
    ?>
    [/code]
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