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cs.punk

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Posts posted by cs.punk

    Quick Questin: Should I escape / as \/ when outputting to xhtml?

    in Javascript Help

    Posted

    Unless they state some special reason for it, you shouldn't.

     

    Even with the CDATA element in place, validators get confused by forward

    slashes (like the one in the </em> tag). Whenever you want to use a / character

    in JavaScript strings, precede it with a backslash (<\/em>). Doing so helps

    the validator understand that you intend to place a slash character at the next

    position.

     

    er

    cell 'rollover' in table.. Ay ideas?

    in CSS Help

    Posted

    Ok well i have a dynamic table with data and want two submin buttons. Edit and delete.

     

    The problem i am having is how to create a 'rollover' image in a table cell? I want it filling  the whole cell? Although I don't really know how to go about it with CSS.

     

    My other option would be to use plain images?

     

    My table:

    <table id='navlist'>
<tr>
<th>Title</th>
<th width='80'>Picture</th>
<th width='80'>ID</th>
<th width='80'>Edit</th>
<th width='80'>Delete</th>
</tr>

<tr>
<td>Test</td>
<td>yes</td>
<td>1</td>
<td id='edit'><a href='/admin/delete.php?id=1'>Edit</a></td>
<td id='delete'><a href='/admin/delete.php?id=1'>Delete</a></td>
</tr>

</table>

     

    And my CSS:

    table#navlist {
margin: 0 0 0 0px;
padding: 0;
width: 100%;
border: 2px solid #234EC4;
background-color: #FFFFFF;
}

#navlist td {
list-style-type: none;
background-color: #6AA6FF;
color: #000000;
text-align: left;
padding: 5px;
border: none;
font-family: Verdana, Arial, Helvetica, sans-serif;
font-size: 14px;
}

#navlist th {
list-style-type: none;
background-color: #6AA6FF;
color: #000000;
text-align: left;
padding: 5px;
border: 1px solid #000066;
font-family: Verdana, Arial, Helvetica, sans-serif;
font-weight:bold;
font-size: 14px;
}

     

    I was thinking making an ID for delete an edit and use a background image? Er somehow...

     

    Any suggestions on how to go about this?

    XP to Windows 7

    in Miscellaneous

    Posted

    Right. From that article:

     

    To upgrade your PC from Windows XP to Windows 7, you'll need to select the Custom option during Windows 7 installation. A custom installation doesn't preserve your programs, files, or settings. It's sometimes called a "clean" installation for that reason.

     

    lmao

    PHP not printing varible in echo command

    in PHP Coding Help

    Posted

    I've never heard of the curly brace thing so I'd be interested to know if it works.

     

    If not try:

     

    str_replace('$dir', $dir, $whateveryourdatafromMySQLwascalled);

     

     

    Yeah it worked and guess thats the only option.. Still see it as a waste of resources but oh well.

     

    Oh and the {} never worked so..

    PHP not printing varible in echo command

    in PHP Coding Help

    Posted

    Ok I'll try and explain again..

     

    Database entry: "You are $x years old"

     

    In the actual database there is for example varible $x.

    In my php code i set varible $x as 50.

     

    I then echo out the database entry.

    But instead of saying "You are 50 years old"(which is what i want) Instead it prints a literal  $x "You are $x years old".

     

    :) Hope i explained myself better...

    PHP not printing varible in echo command

    in PHP Coding Help

    Posted

    From the database

    game_src

    <object classid="clsid:D27CDB6E-AE6D-11cf-96B8-444553540000" codebase="http://download.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=6,0,29,0" width="550" height="400">
<param name="movie" value="[color=red]$dir[/color]">
<param name="quality" value="high">
<embed src="[color=red]$dir[/color]" quality="high" pluginspage="http://www.macromedia.com/go/getflashplayer" type="application/x-shockwave-flash" width="550" height="400">
</embed>
</object>

     

    PHP:

    	<?php
	$sql = "SELECT * FROM games";
	$query = mysqli_query($con, $sql) or die;

	while ($row = mysqli_fetch_assoc($query))
	 {$dir = $row['path'];
	   $game_src = $row['game_src'];
	   echo "<h2>{$row['title']}</h2>
	 			$game_src";
	 }
	?>

     

    HTML Source shows:

    <object classid="clsid:D27CDB6E-AE6D-11cf-96B8-444553540000" codebase="http://download.macromedia.com/pub/shockwave/cabs/flash/swflash.cab#version=6,0,29,0" width="550" height="400">
<param name="movie" value="$dir">
<param name="quality" value="high">
<embed src="$dir" quality="high" pluginspage="http://www.macromedia.com/go/getflashplayer" type="application/x-shockwave-flash" width="550" height="400">
</embed>
</object>

     

    Where am i going wrong?

    simple role based login

    in PHP Coding Help

    Posted

    hi. i am trying to make a simple login system in flash php mysql.I have the database table users and the user is either admin or simple user. i want a php form that will check the role entry and return different result to process it in flash. i have the code for admin login without role which works:

    <?php

    include_once("settings.inc.php");

    include_once("functions.inc.php");

    $password = MD5($_GET['userPassword']); // md5()

    $query = "SELECT * FROM user WHERE username = '" . $_GET['userName'] . "' AND password = '$password'";

    $result = @mysql_query($query);

    if($result){

        if(mysql_num_rows($result) == 1){

            echo "status=ok";

        }

    else{

            fail("The user name and password could not be validated.");

        }

    }else{

        fail("There was an error getting information on the user.", mysql_error());

    }

    ?>

     

    i guess a line is missing to check the roles but i can't think of it!!!! help please.

     

    Firstly use POST rather than GET. It is more secure. Look into mysql_real_escape_string to make your inputs database safe...

    Consider if I typed in "test"; die;"..

     

    You might have magic quotes turned on though. This automatically escapes all POST/GET data.

     

    Oh i forget!

    I would add a 'rank' column to your table. And do something like this:

     

    <?php
include_once("settings.inc.php");
include_once("functions.inc.php");

$password = MD5($_GET['userPassword']); // md5()

$query = "SELECT * FROM user WHERE username = '" . $_GET['userName'] . "' AND password = '$password'";
$result = @mysql_query($query);
if($result){
    if(mysql_num_rows($result) == 1){
        echo "status=ok";

    $row = mysql_fetch_row($result);
    echo "status=ok Rank={$row['3']}"; //        Which ever is the colunm your rank is starting from 0,1,2,3 etc
    } else{
        fail("The user name and password could not be validated.");
    }
}else{
    fail("There was an error getting information on the user.", mysql_error());
}
?>

    Using foreach to echo error. But error is given when not set...

    in PHP Coding Help

    Posted

    Uhm Im kinda stuck...

     

     

     

    <?php
if (isset($error))
 {echo "<p class='error'>The following errors were found:</p>
 			";
	print_r ($error);

  // missing fields	  
  $error['empty'] = array();
  foreach ($error['empty'] as $v)
  	{echo "<p class='error'>$v</p>";
	}

  // invalid size
  foreach ($error['size'] as $v)
  	{echo "<p class='error'>$v</p>";
	}

  // invalid format
  foreach ($error['format'] as $v)
  	{echo "<p class='error'>$v</p>";
	}
 }


// Coming from the xhtml form
   if (!empty($_POST['title']))
   	{$rdata['title'] = mysqli_real_escape_string($con, $_POST['title']);
}
   else
    {$error['empty'][] = "You have not entered a title."; 
}

   if (!empty($_POST['desc']))
   	{$rdata[''] = mysqli_real_escape_string($con, $_POST['desc']);
}
   else
    {$error['empty'][] = "You have not entered a descreption."; 
}
?>

     

    But every time no error from one array is found but an error is found form another is still uses foreach on it.. and since its not set as an array it gives an error. I cant use 

    $error['empty'] = array();

    because it would blank it when there is an error...

    Warning: Invalid argument supplied for foreach() in D:\Chris\reco\xampplite\htdocs\admin\admin.php on line 164

     

    A isset for each array? Is there an easier way?

    Please check this code out

    in PHP Coding Help

    Posted

    Here is the code

    			<td width="3%"><span class="headerLinkColor">»</span></td>
			<td align="left">
			<a target="leftFrame" href="<?=ADMIN_FOLDERS_PATH?><?=$folder_name?>/?sid=<?=$sid?>" class="headerLinkColor">
			<?=$cat_name?></a>
			</td>

		</tr></table>

	<?PHP   if( $flag  == 4){
			echo "</td>";
			$flag=1;
		}else{
			$flag++;
		}		
	 }// for loop ?></tr>
	</table>				 
	</td>

     

    And am getting links like this

     

    http://admin/credit_mgmt/?sid=532bb180e67ea7234d73502205321bfe

     

     

    i need the link to be site url/admin/credit_mgmt/?sid=532bb180e67ea7234d73502205321bfe

     

    any help? thanks

     

    Where is ADMIN_FOLDERS_PATH coming from? Or what is it set as?

    Help Downloading Files from Database

    in PHP Coding Help

    Posted

    Hey everyone, I need help with this simple script. Okay, I have files stored in a database that need to be downloaded. Here is the script I have so far:

     

    <?php

//Connect to the database
require_once('/mysql_connect.php') ;

$query = "SELECT enrollment_form FROM students where student_id = {$_GET['student']}" ; 
$result = mysqli_query($dbc, $query) or die('There was an error with the query. ' . mysqli_error($dbc)) ;
$num = mysqli_num_rows($result) ;

if ($num == 1) {

//Query was successfull
//Retrieve the file
$data = mysql_result($result, 0, "enrollment_form") ;
Header("Content-type: $data") ; 
echo $data ;

} else {

//The record does not exists
echo '<p>The record does not exists.</p>' ;

}

?>

     

    The error I am getting is this:

     

    Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/content/l/i/n//html/Payments/download_enrollment.php on line 14

     

    Warning: Cannot modify header information - headers already sent by (output started at /home/content/l/i/n//html/Payments/download_enrollment.php:14) in /home/content/l/i/n//html/Payments/download_enrollment.php on line 15

     

    Any help on this is greatly appreciated.

     

    Why use mysql_result? Taken from the PHP manual:

    When working on large result sets, you should consider using one of the functions that fetch an entire row (specified below). As these functions return the contents of multiple cells in one function call, they're MUCH quicker than mysql_result().

     

    Try mysqli_fetch_row... or better yet mysqli_fetch_assoc

     

    while ($row = mysqli_fetch_row($result))
{echo "Name: {$row['1']} Address {$row['2']}";
}

    Need help in making a commen.php file.

    in PHP Coding Help

    Posted

    I am looking for a way to make it easier to call events so i can just

    call it from the commen .php so i don't have to keep re-writing out the

    sql = every time i need it.

     

    Whats the best way to go about doing this.

    I am trying to go with php call function, still learning so it's going slow lol.

     

    My issue is, i have a database connect, to table users, when i call that

    function nothing happens and no error message.

    there was an error.

    include "includes/common.php";

function users(){
        return ($users);
    }

     

    I will try again after i get home tonight.

     

    Why not show your 'users' function to us?

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