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newbtophp

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Posts posted by newbtophp

    PHP rating

    in PHP Coding Help

    Posted

    OK, bare with me - i've added a rating system (i've defined the mysql row etc., just havent added any way to calculate) to user uploads at my site, so users can vote:

     

    0-5 and its out of 5, but the problem is, im not sure how to calculate the score (how many votes until it gets to 5/5?) , what are you suggestions - thanks

    help with variable

    in PHP Coding Help

    Posted

    I've got a problem, i want to echo some images depending on the $id, however for me to get that id i have to extract it from a query. So i've done that, but i want to extract all id's (not just 1), so i can echo images for all them ids (not just echo the images based on 1 id), so I did a while() loop, however doesnt seem to work  :-\

     

    My code:

     

    <?php
include "connect.php";
$user = "Peter";
$sql = mysql_query("select * from game_main where user = '$user'");

while($row = mysql_fetch_array($sql)){
$id = $row['id'];
}
//do i have to do foreach $id as  something??

$views = mysql_num_rows(mysql_query("SELECT userip FROM game_views WHERE sub_id = '$id'"));
$sql3 = mysql_query("select monthstamp from game_monthlyfeatured where first = '$id'");
$num3 = mysql_num_rows($sql3);
$sql4 = mysql_query("select weekstamp from game_weeklyfeatured where first = '$id'");
$num4 = mysql_num_rows($sql4);
$sql5 = mysql_query("select daystamp from game_dailyfeatured where first = '$id'");
$num5 = mysql_num_rows($sql5);
$sql6 = mysql_query("select date from site_featured where id = '$id'");
$num6 = mysql_num_rows($sql6);
if($num3 == 1) {
?>
	<img src="../images2/game/game_awards_1monthly.gif" width="20" height="20" />
<?php
}
if($num4 == 1)	{
?>
<img src="../images2/game/game_awards_topweekly.gif" width="20" height="20" />
<?php
}
if($num5 == 1) {
?>
	<img src="../images2/game/game_awards_topdaily.gif" width="20" height="20" />
<?php
}
if ($views > 1000) {
?>
	<img src="../images2/game/game_awards_1000views.gif" width="20" height="20" />
<?php
}
if ($views > 25000) {
?>
	<img src="../images2/game/game_awards_25000views.gif" width="20" height="20" />
<?php
}
if ($num6 > 0) {
?>
<img src="../images2/game/game_awards_modpick.png" width="16" height="16" />
<?php 
} 

?>

     

    Hope someone can help, thanks.

    Get image size help

    in PHP Coding Help

    Posted

    Im using getimagesize to detect the size of an avatar which is stored in a db. However getimagesize doesn't always seem to work (for e.g. if the image url's location is down) - it displays the following error, instead of the avator:

     

    Warning: getimagesize(http://imageurl.com/img.png) [function.getimagesize]: failed to open stream: HTTP request failed! HTTP/1.1 404 Not Found in

     

    Heres my code:

     

    <?php

if(!empty($cmnt['user_avatar'])) {

list($avx, $avy) = getimagesize($cmnt['user_avatar']);

if(($avx + $avy) > 200) {

$avx = 80;

$avy = 80;

} 	

} else {
echo '<img src="../images/noavatar.gif" width="80" height="80" alt="No Avatar" />';
?>

     

    Is their a way to resolve this issue, or perhaps echo a message instead of the warning displaying :/

    preg_replace trouble

    in Regex Help

    Posted

    Do you really need a regular expression?

     

    $variable = str_replace(";echo '\n';", "?>\n<?php", $variable);

     

    (If the string contains \r\n style newlines, use that instead of just \n.)

     

    Thanks mate!!

     

    :D

     

    works perfect.!

    preg_replace trouble

    in Regex Help

    Posted

    Hmm im having abit of trouble, how would i replace:

     

    ;echo '
';

    (contains a new line after opening quote).

     

    with:

     

    ?>
<?php

     

     

    Example:

     

    <?php
include("../config.php");
;echo '
';
$variable = "test";
echo $variable;
?>

     

    Would turn into:

     

    <?php
include("../config.php");
?>
<?php
$variable = "test";
echo $variable;
?>

     

    :-\

    preg_replace everything before

    in Regex Help

    Posted

    Do you need the pattern matching capabilities of regular expressions or would basic string functions work for you? See strstr.

     

    echo strstr($code, 'error_reporting(E_ALL ^ E_NOTICE);');

     

    Both would work - equally fine in this case.

     

    Thanks for that, strstr() is pretty nifty :)

    preg_replace everything before

    in Regex Help

    Posted

    How would i remove everything before the first error_reporting(E_ALL ^ E_NOTICE); ?

     

    <?php
if ($variable==false){print <<<EOF
Error
EOF; 
die();
} 
?>
<?php
error_reporting(E_ALL ^ E_NOTICE);
define('NAME', 'Email');
define('DIR', 'includes/cl_plugins/');
?>

     

    Would turn into:

     

    error_reporting(E_ALL ^ E_NOTICE);
define('NAME', 'Email');
define('DIR', 'includes/cl_plugins/');
?>

     

    My attempt:

     

    preg_replace('~.+?error_reporting\(E_ALL ^ E_NOTICE\);~s', '');

     

    :/

    Not sure how to do this.

    in PHP Coding Help

    Posted

    I tried simply doing the following:

     

    <?php
$file = file_get_contents('test.php');


$file = preg_replace("~base64_decode\('([^']+)'\)~s", base64_decode($1), $file);


highlight_string($file);
?>

     

    but that returns me with an error:

     

    Parse error: syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$' in /home/user/public_html/file.php on line 6

     

    Uploaded file not retaining extension

    in PHP Coding Help

    Posted 

    if ($file != "")
   {
$filename = $_FILES['file']['name'];
$goodfiles = array("jpg","pdf","doc","zip","xls","txt");
$ext = strtolower(end(explode('.',$filename)));
if(in_array($ext,$goodfiles))

{ 
      // copy the pic to the current directory and delete the temporary file
      if (!copy($_FILES['file']['tmp_name'], $filepath . $filename)){
            $msg = "There was an error uploading your file.<br>";
         }
         
      unlink($_FILES['file']['tmp_name']);

      // update file in db
      $qry = "UPDATE table SET title='$file_title',url='$file_url' WHERE event_id = $event_id";
         mysql_query($qry) or die ("Error during query!");

} else {

$msg = "This file type is not allowed.";

}

   }

    Not sure how to do this.

    in PHP Coding Help

    Posted

    Does not make sense. Please elaborate.

     

    Heres my attempt:

     

    <?php
//Don't work - just used my attempt as example
$input =  file_get_contents('test.php');

function recursive64($input)
{

    $regex = "/base64_decode\('([^']+)'\)/Umis";
    
    preg_match("/base64_decode\('([^']+)'\)/Umis", $input, $var);

    $input = base64_decode($var[1]);

    return preg_replace_callback($regex, 'recursive64', $input);
}

$output = recursive64($input);

echo($output);
?>

     

     

    Lets say I have this code (within file.php):

     

    <?php

$var = "test file";

$var2 = base64_decode('ZXhhbXBsZQ==');

$test = array(base64_decode('ZXhwZXJpbWVudA==') => 'example');

?>

     

    Theirfore when its echo'd (ran throught my function) the output would be:

     

    <?php

$var = "test file";

$var2 = 'example';

$test = array('experiment' => 'example');

?>

     

    My aim is, to replace each base64_decode('string') with 'decodedstring', without effecting the actual file (so only the base64 is replaced).

     

    Not sure how to do this.

    in PHP Coding Help

    Posted

    By creating a new array

    <?php
$test = array('ZXhwZXJpbWVudA==' => 'example');
$new = array();
foreach($test as $key => $val) {
$new[base64_decode($key)] = $val;
}
print_r($new);
?>

     

    Thanks, how would i integrate that so it replaces the original encoded with the base64_decoded within the actual file (without placing the decoded in an array)?  :-\

    Not sure how to do this.

    in PHP Coding Help

    Posted

    Lets say I have a file:

     

    <?php

$var = "test file";

$var2 = base64_decode('ZXhhbXBsZQ==');

$test = array(base64_decode('ZXhwZXJpbWVudA==') => 'example');

?>

     

    How would i replace each encoded base64 with its decoded (loop it with php?) so it would become:

     

     

    <?php

$var = "test file";

$var2 = 'example';

$test = array('experiment' => 'example');

?>

     

    All help is greatly apreciated.

    preg_replace space

    in Regex Help

    Posted

    You could possibly replace all instances of only single spaces using look ahead/behind to ensure it is only a single space if thats what your after.

     

    Yeps thats what i was trying to do, - solved it.  :D

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