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Is there any way to upload files from the source link without using form? Suppose, an array contains link of 3 file names.

$array = array('C:\test\a.jpg', 'C:\test\b.doc', 'C:\test\c.txt');

Now i want to upload these 3 files in the web server. How can I do it using PHP?

EDIT:

Actually I am currently working on a website (http://eventconnectbd.com/) where admin has to upload event info with images. A csv file has been created with 100 of image links (all the links are admin's pc related). So basically I created an array of those image links and want to upload one by one.

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  • Am I understanding correctly that that array is on the server and the file paths refer to files on the client? You want the server to pull the files out of the user's computer without intervention? ... Have you thought this through? Commented Nov 9, 2012 at 13:07
  • Indeed, as @deceze mentions you have to make more clear what is on the server, what on the client, where you are using php and what upload actually means in your question... Commented Nov 9, 2012 at 13:12
  • Thanks for your important comments. I just updated my question, please check. Commented Nov 9, 2012 at 14:55

4 Answers 4

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No this is impossible.

If you only know the client-paths, your Webserver needs to browse the clients file-system and that's, of course, not allowed. Imagine: if it were allowed, every Server in the Internet could read all your Files, thats really not desirable.

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5 Comments

Who said the files are on the client? Who said client and server are on different systems? And is this really impossible, think for example of an API used by the client for an upload?
I sure wanted to downvote you, but i'am not gonna do it. What if this client is a admin user who wants to upload thousands of important documents from his pc via a csv file?
@Sprits Doesn't matter whether those documents are "important" or not, that doesn't change the fact that Markus is right.
@arkascha: i suggested that because if the client and the server are on the same system it has nothing to do with a upload (rather file-handling)... An upload (where you push data) is something completly different than if there are only client-paths available and the server should pull data...
@Markus - It seems that you are right bro. client file system is not allowed form server. I did it using valid web url which worked great for me.
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<?php
$array = array('/var/www/a.jpg', '/var/www/b.jpg', '/var/www/c.jpg');
$dest_path = '/var/www/myweb/';
foreach($array as $file_path){
$file_name = end(explode('/', $file_path));
copy($file_path, $dest_path . $file_name);
}
?>
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No, it's not possible. The server has no access to a client's harddisk contents, for better or worse (better, really). You'll have to write a browser plugin which can gain such access (say, a Java applet).

Alternatively, you'll have to upload the images in bulk. Either zip them and upload a single zip file which the server can take apart again, or use, again, a plugin which lets you select many files at once to upload (Flash is popular here).

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2 Comments

I've no experience but is it really possible to do that? Why writing spyware if it is that simple and why clients antivirus shouldn't protect him in this situation?
@Markus The client will still have to explicitly allow the plugin. The Java plugin pops up an unmistakable "DO YOU WANT TO ALLOW TO RUN THIS" in recent browsers (if it runs at all) and while I don't know Flash that well, I believe the file selection has to be done through a regular file-open dialog by the user. I never said it was fully automagic. :)
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Certainly that is possible. All you have to do is implement two parts:

  1. a simple API on the server side implemented in a single php script. It accepts a file and stores it somewhere. Maybe you want to add some security or plausibility checks, but that is another story.
  2. a small script on the client side that iterates the array and uploads each file one by one using the API.

That's all, no form required and no manual upload of single objects.

As requested I add a short example. Nothing fancy, just to show the idea:

The server side API, a primitive php script called upload.php here: 

<?php
// this is where uploaded files get dumped to
define('FILES_STORAGE','/some/path/writeable/for/http-server');
// $_FILES holds the meta data of the uploaded file as stored temporarily
syslog(LOG_DEBUG,sprintf("upload API accepted file '%s'",$_FILES['filedata']['name']));
// copy file to its final destination (temp file gets removed automatically)
copy ( $_FILES['filedata']['tmp_name'] , FILES_STORAGE.$_FILES['filedata']['name']);
// some feedback to the client, might be ignored
echo sprintf("file '%s' accepted and stored. ", $_FILES['filedata']['tmp_name']);
?>

The client side script, I made a trivial bash script, though you can use whatever language you want or have available on client side. It uploads only a singel file, but oviously this can be done in a loop processing a list of files or whatever: 

#!/bin/bash
curl -i -F name=test -F [email protected] http://api.myserver.org/upload.php

When you execute the script the file gets uploaded and moved to the folder defined at the top of the API file as FILES_STORAGE. No error, security and plausibility checking is done, since this is a proof of concept.

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12 Comments

nah... i think this is not possible only if you use valid web link instead of file link.
You did not understand my suggestion. Certainly it is possible to implement a trivial script to upload some files using a servers API. How do you think the internet works? Just by people clicking on blue links?
why not you write some code for it and show it to me. Just upload something form your computer without using a form or hitting the submit button. Can you do it? lets see...
Sure, I added a trivial example as a proof of concept to the answer.
where did you get $_FILES['filedata']['name']? Obviously it came from a submitted form where the input type was file and the name was 'filedata'. Can you define a file into $_FILES['filedata']['name'] variable? if yes, then please show me. (Something like $_FILES['filedata'] = 'C:\\a.jpg')
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