how to avoid duplicate upload in php? - php

i used from function crop_compress() to crop and compress image and then upload them
function crop_compress($source_url,$target_file,$qual) {
$image = imagecreatefromjpeg($source_url);
$width = imagesx($image);
$height = imagesy($image);
$thumb_width = $width;
$thumb_height = (9/16)*$width;
.
.
.
$thumb = imagecreatetruecolor( $thumb_width, $thumb_height );
imagejpeg($thumb , $target_file,$qual);
return $target_file;}
$filename= 'crop_compress($_FILES["fileToUpload"]["tmp_name"][$key],$target_fileL,90)'
and it work fine.
now i want to upload this image with ftp and use ftp_put:
ftp_put($connecti,$target_file, $filename, FTP_BINARY);
and it work fine too, but it upload 2 times that i dont want.
one image in my directory and one with ftp
my question is how to avoid upload in my directory?
i know that ftp_put() and crop_compress() upload separately but i dont know which one should come first and how create a tmp_file to source in another or something else to avoid this problem?!
more info:
$target_dirL = "user"; #in the main host
$tempL=explode(".", $_FILES["fileToUpload"]["name"][$key]);
$target_fileL = $target_dir .$username1. '.' . end($tempL);
$target_dir = "user/$username/"; #another host
$temp=explode(".", $_FILES["fileToUpload"]["name"][$key]);
$target_file = $target_dir .$username1. '.' . end($temp);
$connecti = ftp_connect($anotherHost) or die('Couldn\'t connect to ftp server');

Use unlink( $filename ); after ftp_put($connecti,$target_file, $filename, FTP_BINARY);
This will delete the file after it had been uploaded via FTP.

Related

how to upload image directly on subdomain which is hosted on another server?

I am trying to upload my image directly to my subdomain which is hosted on another server
currently I am using this code to upload image in test.com to demo.test.com
$uploadDirectory = "//demo.test.com/";
i use this code to upload my image
$productname="sdf";
$member_id="123";
$uploadDirectory = "//demo.test.com/";
$randomNumber = rand(0, 99999);
$fn = $_FILES['image']['tmp_name'];
$size = getimagesize($fn);
$ratio = $size[0]/$size[1]; // width/height
$photo_name = $productname.'-'.md5($randomNumber.$member_id);
{
if( $ratio > 1) {
$width = 500;
$height = 500/$ratio;
}
else {
$width = 500*$ratio;
$height = 500;
}
$src = imagecreatefromstring(file_get_contents($fn));
$dst = imagecreatetruecolor($width,$height);
$fileName3 = $uploadDirectory.$photo_name . '-500x500' . ".jpg";
imagecopyresampled($dst,$src,0,0,0,0,$width,$height,$size[0],$size[1]);
imagedestroy($src);
imagepng($dst,$fileName3); // adjust format as needed
imagedestroy($dst);
}
What is the question ?
The subdomain is usually in the domain layer. This layer must not contains infrastructure logic (call to remote data provider). You shouldn't implement the upload in the Domain layer.
To upload your image, you have to call an interface (method : uploadImage) in the domain layer and implement it in the infrastructure (providers) layer.
see https://en.wikipedia.org/wiki/Single-responsibility_principle
and https://en.wikipedia.org/wiki/Dependency_inversion_principle

Laravel show error getimagesize on image update

I am having a strange issue here on image update. After add a new product I need the option of update image and get this error while inserting it everything is ok.
getimagesize(2.png): failed to open stream: No such file or directory
Here is my code
$file = $request->image;
if (isset($file))
{
list($width, $height) = getimagesize($file);
$extension = $file->getClientOriginalExtension();
$final_filename = $file->getFilename() . '.' . $extension;
$new_height_75 = (75 * $height) / $width;
$thumb_img_75 = Image::make($file->getRealPath())->resize(75, $new_height_75);
$thumb_img_75->save(public_path('attachments/product_images/thumbs_75').'/'.$final_filename,80);
$new_height_250 = (250 * $height) / $width;
$thumb_img_250 = Image::make($file->getRealPath())->resize(250, $new_height_250);
$thumb_img_250->save(public_path('attachments/product_images/thumbs_250').'/'.$final_filename,80);
$new_height_150 = (150 * $height) / $width;
$thumb_img_150 = Image::make($file->getRealPath())->resize(150, $new_height_150);
$thumb_img_150->save(public_path('attachments/product_images/thumbs_150').'/'.$final_filename,80);
Storage::disk('product_images')->put($final_filename, File::get($file));
$product->image = $final_filename;
}
If i dd($file ); it gets the file I want to upload.
Thanks
why not you try laravel validator to validate image type as well as image file size https://laravel.com/docs/5.4/validation#rule-image

Create a unique name for the uploaded image

I have a certain code for image upload .I found this on the internet and there was no explanation of the code either.What i can understand from the code is that php upload a certain file makes it a temporary file and then moves the temporary file to the original location
Code Looks something like this
$filename = $_FILES["img"]["tmp_name"];
list($width, $height) = getimagesize( $filename );
move_uploaded_file($filename, $imagePath . $_FILES["img"]["name"]);
What happens now is that when i try to provide an unique name to the image when it is being moved using the move_uploaded_file then a file does come up inside the folder but it says an invalid file and with the extension type of file.
My code for trying to achieve the same but with an unique name/id for the uploaded image.
$uniquesavename=time().uniqid(rand());
$filename = $_FILES["img"]["tmp_name"];
list($width, $height) = getimagesize( $filename );
move_uploaded_file($filename, $imagePath . $uniquesavename);
How to achieve the same as before and could you please explain me the previous code as well?
Sample code:
// Get file path from post data by using $_FILES
$filename = $_FILES["img"]["tmp_name"];
// Make sure that it's a valid image which can get width and height
list($width, $height) = getimagesize( $filename );
// Call php function move_uploaded_file to move uploaded file
move_uploaded_file($filename, $imagePath . $_FILES["img"]["name"]);
Please try this one:
// Make sure this imagePath is end with slash
$imagePath = '/root/path/to/image/folder/';
$uniquesavename=time().uniqid(rand());
$destFile = $imagePath . $uniquesavename . '.jpg';
$filename = $_FILES["img"]["tmp_name"];
list($width, $height) = getimagesize( $filename );
move_uploaded_file($filename, $destFile);
Edit 1:
To get image type in two ways:
Get the file type from upload file name.
Use php function as below
CODE
// Get details of image
list($width, $height, $typeCode) = getimagesize($filename);
$imageType = ($typeCode == 1 ? "gif" : ($typeCode == 2 ? "jpeg" : ($typeCode == 3 ? "png" : FALSE)));
$name = $_FILES['file']['name'];
$tmp_name = $_FILES['file']['tmp_name'];
$location = "uploads/";
$new_name = $location.time()."-".rand(1000, 9999)."-".$name;
if (move_uploaded_file($tmp_name, $new_name)){
echo "uploaded";
}
else{
sleep(rand(1,5));
$new_name = $location.time()."-".rand(1000, 9999)."-".$name;
if (move_uploaded_file($tmp_name, $new_name)){
echo "uploaded";
}
else{
echo"failed, better luck next time";
}
}
here, location is folder inside directory, i mainly create folder "uploads"
time() adds timestamp , which is always unique, until two person upload at same time, which is rare.
moreover, adding 4 digit random number to it , making combination rarest
after that adding actual file name , to making combination unique.
why i use it :
u can extract timestamp later if u need to know when image was uploaded.
u can extract actual filename too.
Lets, say our so unique combination somehow fails,
then, php instance will wait for 1 to 5 second whatever random number is generated. and rename with latest timestamp and regenerated random number.
It's the best u can think of without being resource hog.
you can use
$strtotime = strtotime("now");
$filename = $strtotime.'_'.$_FILES['file']['name'];

PHP upload image different directory

i have a problem with uploading images into different directory.
$path = "../uploads/";
$path2 = "../uploads2/";
$imagename = $_FILES['photoimg']['name'];
$actual_image_name = $imagename;
$uploadedfile = $_FILES['photoimg']['tmp_name'];
$widthArray = array(600,240); //resize width.
foreach($widthArray as $newwidth)
{
$filename = $uploadedfile,$path,$actual_image_name,$newwidth;
//Original Image
if(move_uploaded_file($uploadedfile, $path.$actual_image_name))
{}
if(move_uploaded_file($uploadedfile, $path2.$actual_image_name))
{}
i want to upload image into uploads and uploads2 folders also?
for example width width = 600px into uploads, width = 240px into folder upload2.
what's wrong with my code?
After moving the file with move_uploaded_file it isn't available in the location stored in $uploadedfile anymore. For the second file you have to use copy function.
Please try the following:
if(move_uploaded_file($uploadedfile, $path.$actual_image_name))
{}
if(copy($path.$actual_image_name, $path2.$actual_image_name))
{}
Remove this line. I don't know for what purpose it's there.
$filename = $uploadedfile,$path,$actual_image_name,$newwidth;
To resize the uploaded image use any library to resize it then pass it.
But here is the full code to upload in different directory. But you will have to resize these two $file1 & $file2 to your expected resize file and replace the same in $file1 & $file2
To resize you can use any code suggested here
$path1 = "../uploads/";
$path2 = "../uploads2/";
$file1= $_FILES['photoimg'];
$file2= $_FILES['photoimg'];
$file1_imagename = $file1['name'];
$file2_imagename = $file2['name'];
$file1_actual_image_name = $file1_imagename;
$file2_actual_image_name = $file2_imagename;
$file1_uploadedfile = $file1['tmp_name'];
$file2_uploadedfile = $file2['tmp_name'];
$widthArray = array(600, 240); //resize width.
if (move_uploaded_file($file1_uploadedfile, $path1 . $file1_actual_image_name)) {
echo "Uploaded Successfully!";
}
if (move_uploaded_file($file2_uploadedfile, $path2 . $file2_actual_image_name)) {
echo "Uploaded Successfully!";
}

Creating thumbnails PHP problem

I have this source code where I got it from net tutsplus. I have configured it and made it work in one PHP file. It does work by transferring the original image, but it does not generate to the thumbnails folder.
<?php
$final_width_of_image = 100;
$path_to_image_directory = "../../img/events/" . urldecode($_GET['name']) . "/";
$path_to_thumbs_directory = "../../img/events/" . urldecode($_GET['name']) . "/thumbnails/";
function createThumbnail($filename)
{
if(preg_match('/[.](jpg)$/', $filename))
{
$im = imagecreatefromjpeg($path_to_image_directory . $filename);
}
elseif(preg_match('/[.](gif)$/', $filename))
{
$im = imagecreatefromgif($path_to_image_directory . $filename);
}
elseif(preg_match('/[.](png)$/', $filename))
{
$im = imagecreatefrompng($path_to_image_directory . $filename);
}
$ox = imagesx($im);
$oy = imagesy($im);
$nx = $final_width_of_image;
$ny = floor($oy * ($final_width_of_image / $ox));
$nm = imagecreatetruecolor($nx, $ny);
imagecopyresized($nm, $im, 0,0,0,0,$nx,$ny,$ox,$oy);
imagejpeg($nm, $path_to_thumbs_directory . $filename);
$tn = '<img src="' . $path_to_thumbs_directory . $filename . '" alt="image" />';
echo $tn;
}
if(isset($_FILES['fupload'])) {
if(preg_match('/[.](jpg)|(gif)|(png)$/', $_FILES['fupload']['name'])) {
$filename = $_FILES['fupload']['name'];
$source = $_FILES['fupload']['tmp_name'];
$target = $path_to_image_directory . $filename;
move_uploaded_file($source, $target);
createThumbnail($filename);
}
}
?>
Basically it is supposed to generate a thumbnail of the uploaded image and store the original image into a different folder.
The paths are correct, it works by getting the folder name in the URL, it does work, but nothing works for the thumbnails folder.
BEFORE you ask this related question, yes, thumbnails generation does work on my server by the PHP GD, I have tested it separately. So this is not the problem. :)
How do I get this to work? :(
Well, first off, use imagecopyresampled(), as it will generate a better thumbnail.
Secondly, you shouldn't use the same variable for filesystem directory and for url directory. You should have $filesystem_path_to_thumbs and $url_path_to_thumbs. So you can set them differently.
Third, you may want to do a size check for both width and height. What happens if someone uploads a tall image? Your thumbnail will be outside the target box (but this may not be a big issue).
Fourth, you should prob do a check to see if the thumbnail file exists in the thumbs directory before generating the thumbnail (for performance reasons)...
Finally, the reason it's actually failing, is $final_width_of_image, $path_to_image_directory and $path_to_thumbs_directory are not defined within the function. Either:
Make them global at the start of the function, so you can access them inside of the function global $final_width_of_image, $path_to_image_directory, $path_to_thumbs_directory;
Make them arguments to the function: function createTumbnail($image, $width, $path_to_images, $path_to_thumbs) {
Hard code them inside of the function (Move their declaration from outside the function to inside the function).
Personally, I'd do #2, but it's up to what your requirements and needs are...

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