PHP - Including Files That Are Moved Around

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So far I have managed to create an upload process which uploads a picture, updates the database on file location and then tries to upload the db a 2nd time to update the Thumbnails file location (i tried updating the thumbnails location in one go and for some reason this causes failure)

But the main problem is that it doesn't upload some files

Here is my upload.php

<?php
include 'dbconnect.php';

$statusMsg = '';

$Title = $conn -> real_escape_string($_POST['Title']) ; 
$BodyText = $conn -> real_escape_string($_POST['ThreadBody']) ; 
	 

// File upload path
$targetDir = "upload/";
$fileName = basename($_FILES["file"]["name"]);
$targetFilePath = $targetDir . $fileName;
$fileType = pathinfo($targetFilePath,PATHINFO_EXTENSION);
$Thumbnail = "upload/Thumbnails/'$fileName'";

if(isset($_POST["submit"]) && !empty($_FILES["file"]["name"])){
    // Allow certain file formats
    $allowTypes = array('jpg','png','jpeg','gif','pdf', "webm", "mp4");
    if(in_array($fileType, $allowTypes)){

        // Upload file to server
        if(move_uploaded_file($_FILES["file"]["tmp_name"], $targetFilePath)){

            // Insert image file name into database
            $insert = $conn->query("INSERT into Threads (Title, ThreadBody, filename) VALUES ('$Title', '$BodyText', '$fileName')");
            if($insert){
               $statusMsg = "The file ".$fileName. " has been uploaded successfully."; 

		$targetFilePathArg = escapeshellarg($targetFilePath);
		$output=null;
		$retval=null;
		 //exec("convert $targetFilePathArg -resize 300x200 ./upload/Thumbnails/'$fileName'", $output, $retval);
		 exec("convert $targetFilePathArg -resize 200x200 $Thumbnail", $output, $retval);
		echo "REturned with status $retval and output:\n" ; 
		if ($retval == null) { 
		echo "Retval is null\n" ;
		echo "Thumbnail equals $Thumbnail\n" ; 
		}

            }else{
                $statusMsg = "File upload failed, please try again.";
            } 
        }else{
            $statusMsg = "Sorry, there was an error uploading your file.";
        }
    }else{
        $statusMsg = 'Sorry, only JPG, JPEG, PNG, GIF, mp4, webm & PDF files are allowed to upload.';
    }
}else{
    $statusMsg = 'Please select a file to upload.';
}

//Update SQL db by setting the thumbnail column to equal $Thumbnail
$update = $conn->query("update Threads set thumbnail = '$Thumbnail' where filename = '$fileName'");
            if($update){
               $statusMsg = "Updated the thumbnail to sql correctly.";
		echo $statusMsg ;  } 
else 
{ echo "\n Failed to update Thumbnail. Thumbnail equals $Thumbnail" ; } 


// Display status message
echo $statusMsg;
?>

And this does work on most files however it is not working on a 9.9mb png file  which is named "test.png" I tested on another 3.3 mb gif file and that failed too?

For some reason it returns the following

Updated the thumbnail to sql correctly.Updated the thumbnail to sql correctly.

Whereas on the files it works on it returns

REturned with status 0 and output: Retval is null Thumbnail equals upload/Thumbnails/'rainbow-trh-stache.gif' Failed to update Thumbnail. Thumbnail equals upload/Thumbnails/'rainbow-trh-stache.gif'The file rainbow-trh-stache.gif has been uploaded successfully.

Any idea on why this is?

Hello


I have a simple question about file handling...


Is it possible to list all files in directories / subdirectories, and then read ALL files in those dirs,
and put the content of their file into an array?

Like this:

array:
[SomePath/test.php] = "All In this php file is being read by a new smart function!";
[SomePath/Weird/hello.txt = "Hello world. This is me and im just trying to get some help!";

and so on, until no further files exists in that rootdir.

All my attempts went totally crazy and none of them works... therefore i need to ask you for help.


Do you have any ideas how to do this?
If so, how can I be able to do it?


Thanks in Advance, pros

Hey all,

I've a problem with including of my pages.
When I do choose lotto on index.php , I do get the page and also can choose how many roster I want.
But I can't see my lotto results, because it does refresh to the index.php again.
How can I fix this problem ?

this is the code of index.php:
Code: [Select]
<?php
        echo '<table class="achtergrond" border="1">';
        echo '<tr><td colspan="3">';
        /* Keuzelijst tonen */
        echo '<center><form action="'.$_SERVER['php_self'].'" method="POST"><select name="test">';
        echo '<option selected>Kies een formulier</option></center>';
        echo '<option value="1">Lotto</option>';
        echo '<option value="2">Euromillions</option>';
        echo '</select>';
        echo '<button type="submit" name="knop">OK</button>';
        echo '</form>';
        echo '</td></tr>';
            /* Als er op de knop gedrukt is , voer het lottoformulier uit*/
           if(isset($_POST['knop']))
           {
                  echo '<tr><td>';
                  echo $waarde;
                  /* variabel $waarde een waarde insteken via keuzelijst  */
                  switch($_POST['test'])
                  {
                      case 1;
                          include('formulieren/lotto.php');
                      case 2;
                          include('formulieren/euromillions.php');
                  }
                  echo '</td></tr>';
                  echo '</table>';

           }          
        ?>

and this is the code of lotto.php:
Code: [Select]
<?php
    echo '<table class="achtergrond">';
    echo '<tr><td colspan="3">';
    /* Keuzelijst tonen */
        echo '<center><form action="'.$_SERVER['php_self'].'" method="POST"><select name="lotto">';
        echo '<option selected>Kies aantal rooster</option></center>';
        echo '<option value="1">2 roosters</option>';
        echo '<option value="2">4 roosters</option>';
        echo '<option value="3">6 roosters</option>';
        echo '<option value="4">8 roosters</option>';
        echo '<option value="5">10 roosters</option>';
        echo '<option value="6">12 roosters</option>';
        echo '</select>';
        echo '<button type="submit" name="knop1">OK</button>';
        echo '</form>';
    echo '</td></tr>';
    /* Als er op de knop gedrukt is , voer het lottoformulier uit*/
   if(isset($_POST['knop1']))
   {
          /* variabel $waarde een waarde insteken via keuzelijst  */
          switch($_POST['lotto'])
          {
                case 0;
                   $waarde2 = 0;
                   break;
                case 1;
                   $waarde2 = 1;
                   break;
                case 2;
                   $waarde2 = 2;
                   break;
                case 3;
                   $waarde2 = 3;
                   break;
                case 4;
                   $waarde2 = 4;
                   break;
                case 5;
                   $waarde2 = 5;
                   break;
                case 6;
                   $waarde2 = 6;
                   break;
          }...

Okay so we all know

Code: [Select]
<?php

include "theme.html";

print: "Hello";

?>
Would output the theme and output hello at the bottom of the page, my question is, how do you position hello to go on the current theme? Thanks.

Hi,
I have a class:
Code: [Select]
class Toplist {
    public function get_article($id) {
        $id = (int) $id;
       
        $articles = array();
       
        $query = $db->sql_query("SELECT * FROM `toplist` WHERE `id` = '$id'");
        $result = $db->sql_fetchrowset($query);

        return $articles;
    }
}
and I get
Quote

Notice: Undefined variable: db in classToplist.php on line 11

Fatal error: Call to a member function sql_query() on a non-object in classToplist.php on line 11


the variable db is in ./models/config.php and it isn't working on the classToplist.
I have a feeling that this happens because the config file is not a class.

How can I fix this?

Hello:

I wanted to see if someone can help me figure out how to include data from my included navigation file.

Meaning:

I have this file called myNav.php (the database table storing this is "myUpcomingEvents"):
Code: [Select]
function spLeftMenu() {
$spLeftMenu =
"

<div id=\"sideEvents\">
echo \" $mySideBarData; \";
</div>

";
return $spLeftMenu;
}

I am trying to pull it into the Index.php page like this (the database table storing this is "myHome"):
Code: [Select]
<?php
include('include/myConn.php');
include('include/myNav.php');

$query=mysql_query("SELECT * FROM myHome,myUpcomingEvents") or die("Could not get data from db: ".mysql_error());
while($result=mysql_fetch_array($query))
{

  $myPageData=$result['myPageData'];
  
  $mySideBarPageData=$result['mySideBarPageData'];
}

?>

<!DOCTYPE HTML>

<html>

<head></head>

<body>

<div id="siteContainer">


<div id="leftColumn">
<?php echo spLeftMenu(); ?>
</div>

<div id="mainContent">
<?php echo $myPageData; ?>
</div>


</div>

</body>

</html>

Can't get it to work. It just writes out:
Code: [Select]
echo " ; ";

Anyone know how I can get this to work?

Thanks.

Hi everyone.... I have a simple search.php And I want to include that in my html output page.  I tried using

<?php
   include ('search.php');
?> 

Below is my HTML code which is part of a php file...what am I doing wrong here?

Code: [Select]
// Final Output


echo <<<__HTML_END
 
<html>

<head>

<title>Photography</title>

</head>

<body>

<table width='100%' border='0' align='center' style='width: 100%;'>

$result_final 

<?php include("search.php");?>

</table>

</body>

</html>

__HTML_END;