PHP - Use Glob To Display Contents Of A Folder Based On A Session Variable

Hi,
I want to store files to variable array using glob()

like
$files[0] = xyz.txt
$files[1] = pqr.txt
.
.
.
$files[n] = nfile.txt

I know how to list files from directory using glob()

Code: [Select]
<?php
foreach (glob("*.txt") as $filename) {
    echo "$filename size " . filesize($filename) . "\n";
}
?>



How can I do that ?

I have made a file explorer for my site so i can upload files, delete them and rename them. Everythings working fine except when i want to delete a folder. I realise you cant use ftp_delete, coz' that only works for files.

I have tried rmdir, and even though I am getting no errors, it seems to do nothing :s So what is the best way to delete a folder and its contents using PHP FTP commands?

Thanks guys

how do i make this part of the script delete the folder and contents.

Code: [Select]
<?php
$path = "../../gallery/gallery_files/gallery/";
if(isset($_POST['file']) && is_array($_POST['file']))
{
foreach($_POST['file'] as $file)
{
rmdir($path. "/" . $file) or die("Failed to delete file");
}

}
?>

I'm trying to extract the contents of a zip file to a folder.   I found the ZipArchive class and followed the examples to get it to work for the most part.  But I want to extract the files in the folder inside the zip file but leave the folder out.  So it should extract just the files to my given destination.  I found this on php.net.

Code: [Select]
If you want to copy one file at a time and remove the folder name that is stored in the ZIP file, so you don't have to create directories from the ZIP itself, then use this snippet (basically collapses the ZIP file into one Folder).

<?php

$path = 'zipfile.zip'

$zip = new ZipArchive;
if ($zip->open($path) === true) {
    for($i = 0; $i < $zip->numFiles; $i++) {
        $filename = $zip->getNameIndex($i);
        $fileinfo = pathinfo($filename);
        copy("zip://".$path."#".$filename, "/your/new/destination/".$fileinfo['basename']);
    }                  
    $zip->close();                  
}

?>
For some reason that 'copy' line is not working for me.  Obviosly I've changed the variables in the line to the correct variables.  Can someone help me out.
Thanks
Mike

Hey Guys,

I want to be able to upload a large amount of mp3's to a private area on our website and have a script that will scan the folder for all the filenames and then create a record in the db of all the filenames. It is to help with creating the playlist for my friends radio show. To manually insert all the filenames each month woudl be a nightmare!
I have been messing with a partial script I found on the web, and it kind of does something but not what I want it to.

heres the db

id      int(4)                  
title    varchar(250)            
mp3    varchar(250)            
sort    int(4)            
visible    int(1)            
added    datetime

heres the script i am trying to use:
 <?
 $listDir = array();
$dir = "./music";
         if($handler = opendir($dir)) {
             while (($sub = readdir($handler)) !== FALSE) {
                 if ($sub != "." && $sub != ".." && $sub !=  
"Thumb.db") {
                     if(is_file($dir."/".$sub)) {
                         $listDir[] = $sub;
                     }elseif(is_dir($dir."/".$sub)){
                         $listDir[$sub] = $this->ReadFolderDirectory($dir."/".$sub);
                     }
                 }
             }
             closedir($handler);
         }

$db = mysql_connect("localhost","gfab_ml","xxxxxxx");
mysql_select_db("gfab_ml",$db);
$colors=serialize($listDir); //takes the data from a post  
$sql="INSERT INTO ml_music (ID, title, added)  
VALUES('','$colors','')";
$result = mysql_query($sql) or die (mysql_error());



?>

it does insert the values into the db but what it does is creates 1 record and spits all the results in a really weird way like this. into the title field.

a:18:{i:0;s:23:"14 Mr Magic Premier.mp3";i:1;s:23:"track1.mp3";i:2;s:15:"track2.jpg";i:3;s:15:"track3.fxp";i:4;s:12:"track4.mp3";i:5;s:7:"track5.mp3";i:6;s:7:"track6.jpg";i:7;s:7:"track7.mp3";i:8;s:12:"track8.mp3"


Could anyone please help me to make it work? im a bit of a noob. but i do like to tinker!

Hi guys.

This is my first post here, so excuse me if i am posting this in the wrong forums.
I am making a website, and having a huge problem.

The script is used for orders of different services. And i am using sessions to store the information through serveral pages.
But my problem is that on my final page, where the script sends an email with i can not get it to view the contents of one session.
I am using the same session on a different page and it works like a charm. When i try debugging the session it won't print the contents, so i am guessing something is very wrong?

My code of the page is bellow, let me know if you need anything else:
http://pastebin.com/YHh3bGWS

Please help me 

Hello,

I have a form that ends up creating a file.  The file has one of two values in the file:
specials.php
specials8.php

Within several of my web pages, I have an image map, and I would like one of the href values to be the value that PHP would get from the underlying file.

How would I code this to make the image map function accordingly?

Notice the "Daily Specials" entry within the image map.  It is this href that I need to become the value from the file.

The unsuccessful code that I tried is:

<?php
$file38a = "test/activate.txt";
$file38b = file_get_contents("$file38a");
echo "<img src=\"sample2/sidebar1.jpg\" alt=\"Collegeville Diner\" width=\"94\" height=\"330\" border=\"0\" align=\"top\" usemap=\"#Map2\" longdesc=\"http://collegevillediner.com\" />";
echo "<map name=\"Map2\" id=\"Map2\">";
echo "<area shape=\"rect\" coords=\"8,102,88,152\" href=\"about.htm\" target=\"_self\" alt=\"About the Diner\" />";
echo "<area shape=\"rect\" coords=\"7,159,90,204\" href=\"menu.htm\" target=\"_self\" alt=\"Our Menu\" />";
echo "<area shape=\"rect\" coords=\"7,214,90,262\" href=\"$file38b\" target=\"_self\" alt=\"Daily Specials\" />;
<area shape=\"rect\" coords=\"0,275,86,320\" href=\"desserts.htm\" target=\"_self\" />";
echo "</map>";
?>

Hi all,
I have two PHP files, one in root and the other in a folder called admin.

If I set $_SESSION in a file in the admin folder and then view $_SESSION in any other file within the admin folder, everything is OK.

However, if I then go to a file in root $_SESSION is blank, and it is blank if I then go back to a file in admin.

This problem only exists on my hosting and doesn't happen on a local dev machine.  I've compared the php.ini session settings and there both the same.

Any ideas?

Files
admin/test.php
<?php
session_start();

if( empty($_SESSION) ) {
echo "Session not set<br />";
$_SESSION['test'] = "test";
}

echo "<pre>";var_dump($_SESSION);echo "</pre>";


/test.php
<?php
session_start();

echo "<pre>";var_dump($_SESSION);echo "</pre>";

I have 2 pages...1 one that has something like 500 checkboxes that list all of the active users in a database.  When any number of checkboxes are selected it passes the variables from the checkboxes to the next page, asking for confirmation.  

 

Right now I have the 2nd page displaying numbers for the people, but doesn't display the correct entries.

 

The code is below, hopefully it helps:

 

if(isset($_POST['Update'])) //if the update button was pressed
{
    if (isset($_POST['Player_number'])) //checks to see if any checkboxes are selected
    {
    
        //$Player_number = IMPLODE(',',$_POST['Player_number']); //putting the comma in between the array items
        echo $Player_number;
        ?>
        <input type="hidden" name="<?PHP $Player_number; ?>" value="<?PHP $Player_number; ?>">
$Player_number = IMPLODE(',',$_POST['Player_number']); //putting the comma in between the array items

$result = mysql_query("SELECT * FROM `players` WHERE `Player_number` = '$Player_number' ORDER BY Player_Last_Name") or die(mysql_error());

while ($row = mysql_fetch_array($result))
//while ($row = mysql_fetch_assoc($result))
 {
                // here is your data
 
                echo "FName: ".$row['Player_First_Name'];
                echo "lName: ".$row['Player_Last_Name'];
                echo "email: ".$row['player_email'];
}

foreach($_POST['Player_number'] as $row)
		{	
			$result = mysql_query("SELECT `Player_First_Name`,`Player_Last_Name`,`player_email`, `Player_number` FROM `players` WHERE `Player_number` = '$Player_number' ORDER BY Player_Last_Name") or die(mysql_error());
			$Player_number = IMPLODE(',',$_POST['Player_number']); //putting the comma in between the array items
			while($rows=mysql_fetch_array($result))
			{
				echo "Player Number: ".$Player_number;
				//echo "pNumber: ".$row;
				//echo $rows['Player_First_Name'];
	?>
				<tr bgcolor='<?PHP echo $bkcolor; ?>'>
				<td width = '20%' height="20"><div align="center"><input type="hidden" name="Player_number[]" value="<?php echo $row; ?>"></div></td>
				<td width= '20%' headers="20"><div align="center"><?php echo $rows['Player_First_Name']; ?></div></td>
				<td width= '20%' headers="20"><div align="center"><?php echo $rows['Player_Last_Name']; ?></div></td>
				<td width= '20%' headers="20"><div align="center"><?php echo $rows['player_email']; ?></div></td>						
			<?PHP
			}//close of the while loop	
		}//close of the foreach loop
	}//close of if (isset($_POST['Player_number']))		
} //close of (isset($_POST['Update']
                  
                  

  My issue that it goes through the For loop that is listed above, but list the player first name X amount of times.
 

My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user  and $priv are  undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working.

The code I'm using:

<?php
session_start();
function verify() {
	//verify that the user is logged in via the login page. Session_start has already been called. 
	if (!isset($_SESSION['loggedin'])) {
	header('Location: /index.html'); 
	exit;
	}
	//if user is logged in, we then lookup necessary privleges. $_SESSION['name'] was written with the login name upon login. Privleges   // are written in db as a single-digit integer of of 0 for users, 1 for administrators, and 2 for special users. 
        $user === $_SESSION['name'];
    //Connect to Databse
	$link = mysqli_connect("127.0.0.1", "database user", "password", "database");
	if (!$link) {
  	  echo "Error: Unable to connect to MySQL." . PHP_EOL;
    	echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    	echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    	exit;
        }
        //SQL Statement to lookup privlege information. 
       if ($result = mysqli_query($link, "SELECT su FROM accounts WHERE username = $user", MYSQLI_STORE_RESULT)) {
         //LOOP TO CYCLE THROUGH SQL RESULTS AND STORE Privlege information as vairable $priv.
            while ($row = $result->fetch_assoc()) {
           $priv === $row["su"];
            }
        }
        // close SQL connection.
        mysqli_close($link);

 // Verify privleges and take action. Only a privlege of "1" is allowed to view this page. A privlege of "2" indicates special 
//accounts used in other scripts that have certain indermediate additional functions, but are not trusted administrators. 

        if ($priv !== 1) {
        echo $_SESSION['name'];
        echo "you have privlege level of $priv";
        echo "<br>";
        echo 'Your account does not have the privleges necessary to view this page';
        exit;
        }
        
}
verify();
?>

 

I am doing simple program to display text entered by user..

Following is my code but its display 'error msg' instate of txtbox contents

<html>
<head><title>Date</title></head>
<body>
<form action="" enctype="multipart/form-data">
Enter Date:
<input type="text" name="txtdate"/>
<input type="submit" value="Display"/>
</form>
</body>
</html>

<?php
if(isset($_POST['txtdate']))
{
   $display =$_POST['txtdate'];
echo 'You have entered'.$display;
}
else
{
echo 'Error msg';
}
?>

Hi friends, I have to transform this url domain.com/main/folder1/folder2/folder3 to domain.com/get.php?var1=folder1&var2=folder2&var3=folder3   How can i do it with htaccess.   Thank u
Edited by exarhis, 08 July 2014 - 04:42 AM.

Hey guys - I have some code that pulls an image out of a folder and displays it on the page. The problem is, there won't always be an image to display in which case I'd rather the code not even display.

Here's my code so far: Code: [Select]
<tr><td><a href="images/uploads/<?php echo $row['loc_id']; ?>_1.jpg"><img id="morephoto" src="images/uploads/thumbs/<?php echo $row['loc_id']; ?>_1thb.jpg" /></a></td></tr>
The images are renamed on upload to have the id number for that row appended to the front of the file name and that is how I'm calling them back in.

I know I need to write an if statement that contains the code from the <tr> to the </tr> to display if the image exists, the problem is, since there isn't a field for this in the database, I don't know how to check it? I'm still a noob so I appreciate any help that is offered.

Thanks!

Hi i dont know whether i've overlooked something but im finding it hard to answer this myself..

Background of the problem:

Im keeping the contents of a page in a database, so each page would be a template, and the body text will be stored in the database.

So if i want to call up the contents for my 'About' page, i would use the php command: "<?php getAboutPosts(); ?>"

The code behind that will be:

function getAboutPosts() {
   $query = mysql_query("SELECT * FROM posts WHERE id = '20' ") or die (mysql_error());
   while($subm = mysql_fetch_assoc($query)) {
      echo "<h2>" . $subm['Title'] . " by iVisual Media" . "</h2>";
      echo $subm['Content'];
      }
}

My Actual Problem

The text does echo fine, but it is completely unstructured, I.E. it isn't paragraphed, i cant include any type of tag like <b> because it doesn't seem to echo them, although when i check PHPMyAdmin it has all the tags saved in with the content?

Example:

I have this saved in MySQL Database:

<h4>We are a professional website and graphic design business in Basildon, Essex, providing online marketing strategies for companies throughout the UK.</h4>
<p><br />
Whether you require a brochure website, blog or a full e-commerce solution, Imagine Design Studio can help. Maybe you have a new business and want to create an online presence or an established business looking to pull your company into the 21st century.</p>
<p>

But when it echos, i get this:

We are a professional website and graphic design business in Basildon, Essex, providing online marketing strategies for companies throughout the UK.
Whether you require a brochure website, blog or a full e-commerce solution, Imagine Design Studio can help. Maybe you have a new business and want to create an online presence or an established business looking to pull your company into the 21st century.



Things i've tried:

Adding a class to the DIV which the echo will go into - This works for colouring etc, but i still need to paragraph the text and use italics occasionally.

Please help guys! Thanks in advance

Hi All,

I'm having a major mind blank, and can't find anything in the previous posts resolving what I'm after.

I'm setting

$searchtext = $_POST['searchtext'];

I want to check $seachtext is not null.

I've seen isset($searchtext) but it doesn't solve my problem.

Basiclaly; I want an if statement to say if(isset($searchtext)) {......}

Thoughts?