how to check file extensions with an array - php

I want to check if an extension is part of an array:
So: if an extension is not part of a forbidden array; do something is allowed
$ext = $path_info['extension'];
$ForbiddenExts = array("php", "html", "htm");
if( $ext != in_array($ForbiddenExts)){
// do something allowed

Change your code to:
$ext = $path_info['extension'];
$ForbiddenExts = array("php", "html", "htm");
if(!in_array($ext, $ForbiddenExts))
{
// do something
}
Check this link for more explanation.

Other than using
if( $ext != in_array($ForbiddenExts)){
You can use
if(!in_array($ext, $ForbiddenExts)){
//your code
}
You have to practice checking the mime type too. Otherwise it may cause errors.
for eg: if someone edit the extension of a ".txt" file to ".pdf"
(assuning pdf is allowed type). Then if you don't check mime type, the
code will accept the file as pdf

Change your code to:
$fileName = 'banner.jpg';
$fileNameParts = explode('.', $fileName);
echo $ext = end($fileNameParts);
$allowed_extensions = array("jpg", "jpeg", "png");
if(in_array($ext, $allowed_extensions))
{
echo 'Allowed Extension';
// do something
}else{
echo 'Not Allowed Extension';
// do something
}
Check this link for more explanation.

Related

how to check uploaded file is video and make ensure really video type

Selected file is not notepad file with extension avi, mp4 etc.
I tried this:
$file_allowed = array('gif','png' ,'jpg','avi','mp4','wmv','mov','flv','webm','ogv','mp3','m4a','ogg','oga');
$uploaded_filename = $_FILES['attachment']['name'];
$file_ext = pathinfo($uploaded_filename, PATHINFO_EXTENSION);
if(!in_array($file_ext,$file_allowed) )
{
exit;
}
Try this variable for validation
$_FILES["attachment"]["type"];
Try changing your code like this:
$file_allowed = array('gif','png' ,'jpg','avi','mp4','wmv','mov','flv','webm','ogv','mp3','m4a','ogg','oga');
$uploaded_filename = $_FILES["attachment"]["type"];
$file_ext = pathinfo($uploaded_filename, PATHINFO_EXTENSION);
if(!in_array($file_ext,$file_allowed) ) {
exit;
}

How to check given word occures at last position of string in php

I want to get an alternate of an sql query WHERE field LIKE '%subStr' in PHP for example
actually I want to conditionally of different file types so I want to check whether the extension is .ppt or .pptx or .pptm .ppm or any thing else.
Make use of array_pop and explode
<?php
$filename='new.newfile.ppt';
$file=explode('.',$filename);
echo $ext = array_pop($file); // ppt
You can easily strip out the last characters of a filename string to achieve this:
<?php
$filename = "myFile.txt";
$pathinfo = pathinfo($filename);
$extension = $pathinfo['extension'];
switch ($extension) {
case "ppt":
echo "Extension is .ppt";
break;
case "txt":
echo "Extension is .txt";
break;
}
?>
You could do it this way.
$file = 'my-file.ppt';
$parts = pathinfo($file);
$extensions = array("ppt", "pptx", "pptm", "ppm");
if (in_array($parts['extension'], $extensions)) {
echo "we have {$parts['extension']}";
}
else {
echo "no matches";
}

Check file PDF or image before upload

I am studying PHP but I don't get the right way by myself. I'd like having Img always required (and I check this in the form input required attribute) but I can decide if upload PDF or not. The script doesn't continue if I don't select both.
I have this:
// image select from form
$img = basename($_FILES['img']['name']);
$allow_img = array('jpg', 'png', 'jpeg');
$ext_img = explode('.', strtolower($_FILES['img']['name']));
$type_img= end($ext_img);
//pdf select from form
$pdf = basename($_FILES['pdf']['name']);
$allow_pdf = array('pdf');
$ext_pdf = explode('.', strtolower($_FILES['pdf']['name']));
$type_pdf= end($ext_pdf);
if ($img || $pdf) {
if(!in_array($type_img, $allow_img) || !in_array($type_pdf, $allow_pdf) ) {
echo "<p><a href='../admin.php'><img style='border:none;' src='../../img/arrow-left.png' /></a>Only jpg, png, jpeg and PDF.</p>";
}
}
Here you go a super fast way to accomplish this:
$filename = $_FILES['img']['name'];
$ext = pathinfo($filename, PATHINFO_EXTENSION);
// allowed extensions
$allowed = array('jpeg', 'png', 'jpeg', 'gif');
if (in_array($ext, $allowed)) {
echo "<p><a href='../admin.php'><img style='border:none;' src='../../img/arrow-left.png' /></a>Only jpg, png, jpeg and PDF.</p>";
}
That's it :)
You need javascript for this.
Before you send return, you have check type of file.
Most likely the second basename() call crashes, since no $_FILES['pdf'] is present when no pdf file is uploaded. But you don't even check for an error here... Take a look into the http servers error log file, most likely you will see the error there.
That said: always look into the log files if something unexpected happens. And always test for a variables existance before you use it. And always do error checking when calling some function which might not return what you expect.
You can use pathinfo() inbuilt php function,
$File = $_FILES['image']['name'];
$Infos = pathinfo($File);
echo $extension = $info[extension];
echo "<pre>"; print_r($Infos); echo "</pre>";
$extension = strtolower( $extension);
if( $extension=='pdf'){
// do your stuff
}

How to check for the correct file extension in PHP?

So basically, I am simply just trying to check for the correct file extension on a file that is being uploaded.
I know, this question has been answered on here a few times before, although I keep getting the same error and there is no solution or suggestions out there to why this is happening.
Here is my code:
$file = fopen($_FILES['upload_csv']['tmp_name'], 'r');
$ext = pathinfo($file, PATHINFO_EXTENSION);
if($ext != "csv")
{
$errors[] = "Sorry, but only CSV files are supported";
}
Here is my error:
Warning: pathinfo() expects parameter 1 to be string
I have tried around 3 other alternatives now, all using pathinfo(). Although, the exact same error is still shown.
Does anyone have any suggestions to why this is happening?
Your problem is here:
$file = fopen($_FILES['upload_csv']['tmp_name'], 'r');
$ext = pathinfo($file, PATHINFO_EXTENSION);
fopen returns a file handle for use reading and writing a file, but pathinfo is expecting a string containing a filename (optionally, with a path), but you're giving it a file handle.
You should, in any case, be looking at $_FILES['upload_csv']['name'], which is the original name of the file, and extracting the file extension from that.
$path_info = pathinfo('/foo/bar/baz.bill');
echo $path_info['extension']; // "bill"
You can simply read the extension from the name of the file. There is no need to fopen the file.
$allowedTypes = 'csv, xls, xlsx';
function getExtension($str) {
$i = strrpos($str,".");
if (!$i) { return ""; }
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}
$filename = stripslashes($_FILES[$fileElementName]['name']);
$extension = getExtension($filename);
$extension = strtolower($extension);
$allowedTypes = explode(',',ltrim(rtrim($allowedTypes,','),','));
array_walk($allowedTypes, create_function('&$val', '$val = ltrim(trim($val),".");'));
if (!in_array($extension, $allowedTypes))
{
$errors[] = "Sorry, but only CSV files are supported";
}
$extension=strtolower(pathinfo($_FILES['upload_csv']['tmp_name'], PATHINFO_EXTENSION));
if($ext != "csv")
{
$errors[] = "Sorry, but only CSV files are supported";
}

PHP image upload - rename without losing extension?

I currently have:
$file_name = $HTTP_POST_FILES['uid']['name'];
$random_digit=rand(0000,9999);
$new_file_name=$random_digit.$file_name;
$path= "uploads/images/users/".$new_file_name;
if($ufile !=none)
{
if(copy($HTTP_POST_FILES['uid']['tmp_name'], $path))
{
echo "Successful<BR/>";
echo "File Name :".$new_file_name."<BR/>";
echo "File Size :".$HTTP_POST_FILES['uid']['size']."<BR/>";
echo "File Type :".$HTTP_POST_FILES['uid']['type']."<BR/>";
}
else
{
echo "Error";
}
}
this generates a random number before the current file name ie 4593example.jpg
but i would just like it to rewrite the whole filename to 4593.jpg removing the ucrrent name (example). When i have tried doing this i lose the extension (.jpg)
any help is appreciated.
If you're dealing with images only, I would consider detecting the image's type using getimagesize() and giving the new file an extension according to that.
That way, you don't have to rely on what extension you get from the user (which could be wrong).
Like so:
$extensions = array(
IMAGETYPE_JPG => "jpg",
IMAGETYPE_GIF => "gif",
IMAGETYPE_PNG => "png",
IMAGETYPE_JPEG2000 => "jpg",
/* ...... several more at http://php.net/manual/en/image.constants.php
I'm too lazy to type them up */
);
$info = getimagesize($_FILES['uid']['tmp_name']);
$type = $info[2];
$extension = $extensions[$type];
if (!$extension) die ("Unknown file type");
move_uploaded_file($_FILES['uid']['tmp_name'], $path.".".$extension);
`
Also:
As #alex says, your method has a considerable risk of collisions of random names. You should add a file_exists() check to prevent those (e.g. a loop that creates a new random number until one is found that doesn't exist yet)
HTTP_POST_FILES is deprecated, use $_FILES instead
I strongly advise you to use move_uploaded_file() instead of copy() which is vulnerable to attacks.
$ext = pathinfo($filename, PATHINFO_EXTENSION);
$newFilename = $random_digit . '.' . $ext;
BTW, what will happen if the random number clashes (which may happen next or in 10 years)?
There are many better ways to name them - for example, hash of the filename and time() should theoretically never clash (though in practice hash collisions do occur).
You can get the original extension by using the following code:
$extension = strrchr($HTTP_POST_FILES['uid']['tmp_name'], '.');
Then you can add the extension to the variable $new_file_name like this:
$new_file_name = $random_digit . $file_name . '.' . $extension;
You can use this code:
$file_name = $HTTP_POST_FILES['uid']['name']; $random_digit=rand(0000,9999);
$extension= end(explode(".", $HTTP_POST_FILES['uid']['name']));
$new_file_name=$random_digit.'.'.$extension; $path= "uploads/images/users/".$new_file_name; if($ufile !=none) { if(copy($HTTP_POST_FILES['uid']['tmp_name'], $path)) { echo "Successful
"; echo "File Name :".$new_file_name."
"; echo "File Size :".$HTTP_POST_FILES['uid']['size']."
"; echo "File Type :".$HTTP_POST_FILES['uid']['type']."
"; } else { echo "Error"; } }
enjoy!!!!!!!!!

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