Javascript and PHP upload system - php

Thanks in advance for any help, I hope my explanation of my request is understandable.
I have a website where I upload various HTML pages with scripts, websites etc. that I have found useful over time... For the purpose of 1) a reference for myself, and 2) to share what I've found with others.
The website consists of 2 sections. A search page to find the script, and an admin page to upload it. The uploaded HTML file gets placed in a "docs/" directory on my server, and the details are added to a MySQL database for the search page.
The form looks like this:
<form name="upload" enctype="multipart/form-data" action="includes/add.php"
method="post" onsubmit="return validateForm();">
<label for="scriptname">Script Name</label><input class="inputarea" type="text"
name="scriptname"><br>
<label for="category">Category</label><input class="inputarea" type="text"
name="category"><br>
<label for="keywords">Keywords</label><input class="inputarea" type="text"
name="keywords"><br>
<label for="content">HTML File</label><input class="inputarea" type="file"
name="content"><br>
<input class="submit" type="submit" value="Add">
</form>
My question is this... Is there any way with JavaScript or PHP to do the following:
generate an automatic file name for the uploaded file (a few random digits would do)
In the "scriptname" input field, add text on submit so that it makes the Script name and file name into a hyperlink that's added to the database as text... eg. When submit button is pressed, the following is added to the database:
"scriptname_input"
Where the bold section is taken from the generated file name and the italic section is from the input field...
The purpose of this is so that in the search results, when the database column with the script name comes up, the script name is a link to the actual file. I have the search feature ready, and it is able to make a link from a database entry, but I just need to simplify the upload process.
If this is not possible, is there a different way to achieve this?
---EDIT---
Thank you all for your help! Much appreciated, I've worked it out using a combination of a few of the suggestions. However, I gave the credit to Ibere as his solution was the closest.
Here is the final code I used for the 'add.php' file that processed the upload and database addition, just in case it ever comes up again (I doubt it) :P
<?php
$filename = md5($_FILES['content']['name']);
$labelForUrl = $_POST['scriptname'];
$url = "$labelForUrl";
$target = "../docs/";
//This gets all the other information from the form
$name=$_POST['scriptname'];
$cat=$_POST['category'];
$key=$_POST['keywords'];
$link=$_POST['link'];
$file=($_FILES['content']['scriptname']);
// Connects to your Database
mysql_connect("localhost", "username", "password") or
die(mysql_error()) ;
mysql_select_db("scripts") or die(mysql_error()) ;
//Writes the information to the database
mysql_query("INSERT INTO scripttable (scriptname,category,keywords,link,content)
VALUES ('$url', '$cat', '$key', '$link', '$file')") ;
if(move_uploaded_file($_FILES['content']['tmp_name'], $target . $filename)) {
echo "The file ". $labelForUrl.
" has been uploaded";
}
else {
echo "There was an error uploading the file, please try again!";
}
?>

You can do something like this for the filename.
$filename = md5($_FILES['content']['name']);
$labelForUrl = $_POST['scriptname'];
md5 is not Random, but is good enough for generating a unreadable string for a filename.
Then you can create a url like this
<a href="docs/<?php echo $filename; ?>" ><?php echo $labelForUrl; ?></a>
Hope this helps.
EDIT: I forgot to add the extension to the filname. So the right code would be something like:
$filename = md5($_FILES['content']['name']).$_FILES['content']['type']

I recommend using uploadify for uploads. But, to do what you asked:
$randomFileName = rand(1000, 9999);
if (file_exists("upload/" . $_FILES["file"]["name"]))
{
echo $_FILES["file"]["name"] . " already exists. ";
}
else
{
move_uploaded_file($_FILES["file"]["tmp_name"],
"upload/" . $randomFileName . $_FILES["file"]["type"]);
// update your db with the location
$loc = "upload/" . $randomFileName . $_FILES["file"]["type"];
mysqli_query("insert into `myTable` (`loc`) values ('$loc')");
echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
}
}
For file uploading help, look at http://www.w3schools.com/php/php_file_upload.asp

This is very easy, if you know codeigniter ( PHP Framework ).
You can use the Upload Class
You can easily create forms and submit them and also display them.
I would do it that way. If you are familiar with MVC you can do that in 10-15 mins.

To generate random file names, I usually find this does the work quite well: md5( rand( 0, 100000 ) );. If you wish to limit the size of the file name, you may use the substr function.
(Assuming a MySQL database), make the connection and then query the database using the INSERT command. This link shows how to do all of this.

Related

File upload and display part of the file php

I am so stuck, could someone help, please?
I've been learning PHP for a while and currently, I am creating a file upload form.
What needs to happen is this:
File is uploaded
The name of the file is displayed on the main page as a link
Once the link is clicked on, information about the file displays
(first sentence or so)
The questions I have are:
File directory. I am supposed to create a folder, let's say called "uploads" - that's where the files are going to be downloaded to and have my form+php inside that folder, correct? - noob question, I know
I managed to get the name of the file appears as a link, but I don't know how to display its contents.
Could someone help, please?
Code: https://pastebin.com/rfUgKzSu
//file upload on main page
if (isset($_POST['name'])) {
move_uploaded_file($_FILES['file']['tmp_name'],'add_article_form.php' . $_POST['name'] . '.txt');
echo 'file ' . $_POST['name'] . '.txt' . 'uploaded';
}else{
echo 'There has been a mistake';
}
echo '<br>' ;
echo '<br>' ;
//form on main page
<form action = "add_article_form.php" method = "POST">
<input id = "add" type = "submit" value="add">
</form>
<?php
//display file name
$resource = opendir('../uploads/');
while(($entry = readdir($resource))!== FALSE)
{
if($entry != '.' && $entry != '..'){
echo "$entry" . '<br>';
} else {
"$entry" . '<br>';
}
}
PAGE 2
<!--The upload form, second page -->
<form action = "../uploads/" method="POST" enctype="multipart/form-data" >
Название статьи: <br>
<input type = "text" name = "name" value = "text"><br><br>
Файл:<br>
<input type = "file" name="file"><br><br>
<input id = "add" type = "submit" value="add"><br><br>
</form>
The questions I have are: 1. File directory. I am supposed to create a
folder , let's say called "uploads" - that's where the files are going
to be downloaded to and have my form+php inside that folder, correct?
- noob question, I know
No.
You can keep your upload script and the folder, where the files will be saved, anywhere even on different servers. The only thing that you should be careful that you use the correct links and proper permissions so that they can access each other.
I managed to get the name of the file appears as a link, but I don't
know how to display its contents.
Use PHP header function.
http://php.net/manual/en/function.header.php
(Go through Example 1.)

PHP Image Upload Script Not Uploading But Not Failing

I am trying to enter some data into a database and upload an image to a specific directory. I am using the following script which is a modified version of the top voted answer to this question: How to store file name in database, with other info while uploading image to server using PHP?
require($_SERVER['DOCUMENT_ROOT']."/settings/functions.php");
// This is the directory where images will be saved
$target = $_SERVER['DOCUMENT_ROOT']."/safetyarticles/images/";
$target = $target . basename( $_FILES['article_img']['name']);
date_default_timezone_set("America/Chicago");
// This gets all the other information from the form
$article_name = $_POST['article_title'];
$article_date = date("m/d/Y");
$article_creator = $_POST['article_creator'];
$article_views = "0";
$article_content = $_POST['article_content'];
$article_content_2 = $_POST['article_content_2'];
$article_img = ($_FILES['article_img']['name']);
$article_credit = $_POST['article_credit'];
// Connect to database
$conn = getConnected("safetyArticles");
// moves the image
if(move_uploaded_file($_FILES['article_img']['tmp_name'], $target))
{
// if upload is a success query data into db
mysqli_query($conn, "INSERT INTO currentArticles (article_name, article_date, article_creator, article_content, article_content_2, article_img, article_credit)
VALUES ('$article_name', '$article_date', '$article_creator', '$article_views', '$article_content', '$article_content_2', '$article_img', '$article_credit')") ;
echo "The file ". basename( $_FILES['article_img']['name']). " has been uploaded, and your information has been added to the directory";
}
else {
echo "Sorry, there was a problem uploading your file.";
}
I have successfully connected to my database since my getConnected() function contains error handling for if the connection fails.
For some reason I keep getting the Sorry, there was a problem uploading your file. error from the bottom of the script.
Am I missing something? All I did was change some minor lines here and there such as how the database connects, and the variables. I also moved the query to only happen if the file uploads.
I'm not sure what I'm missing.
Is it also possible to modify this current script to rename the image to whatever the value of $article_name is? For example if the article name is "This Is The First Article" then the image would be this-is-the-first-article.jpg?
My HTML form is:
<form method="post" action="http://example.com/admin/articleCreate.php" enctype='multipart/form-data'>
<input type="text" name="article_title" placeholder="What Is The Name Of This Article?" id="article_title_input">
<textarea name="article_content" placeholder="Write The Top Half Of Your Article Here." id="article_content_input"></textarea>
<input type="file" name="article_img" placeholder="If The Article Has An Image, Upload It Here." id="article_img_input">
<textarea name="article_content_2" placeholder="Write The Bottom Half Of Your Article Here." id="article_content_2_input"></textarea>
<input type="text" name="article_creator" placeholder="Who Is Writing This Article?" id="article_creator_input">
<input type="text" name="article_credit" placeholder="If This Article Is Copied, What Website Was It Taken From?" id="article_credit_input">
<input type="submit" value="Submit">
</form>
And I did var_dump(is_uploaded_file($_FILES['article_img']['tmp_name'])); and it's returnign true.
Sidenote edit: This being before you edited your question with only one of them being renamed. https://stackoverflow.com/revisions/36367407/4
$_FILES['photo']
$_FILES['uploadedfile']
are two different file arrays and you're using name="article_img" as the name attribute.
You need to use the same one for all of them.
Error reporting http://php.net/manual/en/function.error-reporting.php would have told you about it.
Add error reporting to the top of your file(s) which will help find errors.
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// Then the rest of your code
Sidenote: Displaying errors should only be done in staging, and never production.
Additional edit:
$target = $target . basename( $_FILES['photo']['name']);
if that's your real edit, still the wrong array name.
I think the problem is in this line:
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
Change it to this:
if(move_uploaded_file($_FILES['article_img']['tmp_name'], $target))
Your html has:
<input type="file" name="article_img" placeholder="If The Article Has An Image, Upload It Here." id="article_img_input">
And your php is waiting for $_FILES['photo']['tmp_name']
Change your html file input to:
<input type="file" name="photo" placeholder="If The Article Has An Image, Upload It Here." id="article_img_input">

how to use filemtime function in PHP?

i have a database like picture bellow :
database
i want to insert date and time based on last modified date from a file on my computer in uploaddate column automtically using filemtime() function in php.
i have tried to use this code :
$namefile= $_FILES['filename']['name']; //from file i have uploaded
if (file_exists($namefile))
{
$uploaddate = date ("Y-m-d H:i:s", filemtime($namefile));
}
echo $uploaddate;
and this is my SQL Query :
$import="INSERT into scan (UploadDate, ScanDate, FileName)
values('$uploaddate', '$date', '$namefile')
ON DUPLICATE KEY UPDATE UploadDate='$uploaddate', ScanDate='$date',
FileName='$namefile'";
echo function is running and true but i still can't insert into database.
May you know where is the problem? Thank you so much for your help.
The problem is - you are trying to get modified time of name(!) not a path of the file.
This sample will work, if you will create and upload_from directory and put it near your form-processor file. You should then upload your files from there. Or you could specify another one (the path should be absolute). Unfortunately, this decision will work only in those specific cases.
However, the manipulations with modify-time of file - are tricky, because it could be set by users manually, it could come from users with different time-settings, so, you can only believe to your machine with that local decision, that I provided.
PS: I did not find the ways to access the absolute file name on uploaded file.
index.php contents:
<form method="POST" action="index.php" enctype="multipart/form-data">
<input type="file" name="filename">
<input type="submit">
</form>
<?php
$upload_from_dir = 'upload_from';
if (!empty($_FILES)) {
$namefile = $upload_from_dir . DIRECTORY_SEPARATOR . $_FILES['filename']['name'];
if (file_exists($namefile)) {
$uploaddate = date("Y-m-d H:i:s", filemtime($namefile));
echo $uploaddate;
}
}
?>

Trying to display images from a filesystem in php

Today i am looking for help. This is my first time asking so sorry in advance if I make a few mistakes
I am trying to code a small web application that will display images.Originally I used the blob format to store my images in a database, however from researching on here People suggest to use a file system. My issue is I cannot display an image. It could be a very small error or even a bad reference to a files location however I cannot make it work.
This is a small project that I hope to be able to improve on and hopefully create into a sort of photo gallery. I am running this application on a localhost.
I am having an issue with displaying images from a filesystem.
// index.php
<form action="process.php" method="post" enctype="multipart/form-data">
<input type="file" name="image" />
<input type="submit" name="submit" value="Upload" />
</form>
My form then leads to a process page where the request is dealt with.
<?php
// process.php
// connect to the database
include 'connection.php';
// take in some file data
$filename =$_FILES['image']['name'];
// get the file extension
$extension = strtolower(substr($filename, strpos($filename, '.')+1));
// if the file name is set
if(isset($filename)){
// set save destination
$saved ='images/';
// rename file
$filename = time().rand().".".$extension;
$tmp_name=$_FILES['image']['tmp_name'];
// move image to the desired folder
if(move_uploaded_file($tmp_name, $saved.$filename)){
echo "Success!";
// if success insert location into database
$insert="INSERT INTO stored (folder_name,file_name) VALUES('$saved', '$filename')";
// if the query is correct
if($result=mysqli_query($con,$insert)){
echo "DONE";
echo"</br>";
// attempt to print image
echo "<img src=getimage.php?file_name=$filename>";
}
}
}
else{
echo "Please select a photo!!";
}
?>
Now as you can see I have an < img > tag. To try and learn, I was trying to just display the recently uploaded image. To try and do this I created a getimage file.
<?php
//getimage.php
// set the page to display images
header("Content-Type: image/jpeg");
include "connection.php";
// get requested filename
$name = ($_GET['file_name']);
$query = "SELECT * FROM stored WHERE file_name=$name";
$image = mysqli_query($con,$query);
$row = mysqli_fetch_array($image,MYSQLI_ASSOC);
$img = $row['file_name'];
echo $img;
?>
My database structure is as follows:
database name = db_file.
table name = stored.
columns = folder_name, file_name
Again, this is just a small project so I know I will have to alter the database if I wish to create a larger more efficient application.
It seems you use the database lookup to get just the file name, but you already have the file name. Try adding the folder name, create a valid path.
change
$img = $row['file_name'];
to
$img = $row['folder_name'] . '/' . $row['file_name'];
check your <img>tag to see if the correct url is present. You may or may not need the '/', it depends on how you stored the folder name. You may need to add the domain name. There is just not enough information know what is needed.
Your <img> should look like this
<img href="http://www.yourdomain.com/folder name/file name">
in the end

PHP File Upload Failing

For some reason my PDF upload form is failing consistently, I have this code:
<?php
if($_POST["submit"] == "Add PDF to Comm and Special Projects")
{
$addsubp = $_POST["addsubp"];
$addsubp_name = $_POST["addsubp_name"];
$commuploadedfile = $_FILES['uploadedfile']['name'];
$sqldoc = "INSERT INTO projects_links (pid, display_name, link) VALUES ('".$addsubp."','".$addsubp_name."','".$commuploadedfile."')";
mysql_query($sqldoc) or die(mysql_error());
echo "<BR>";
$target_path = "D:\\Hosting\\69903\\html\\pdfs\\comm\\";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "<br>The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded<br>";
} else{
echo "<br>There was an error uploading the file, please try again.<br>";
}
}
?>
<form method="post">
Add PDF to Project for Committees and Special Projects <br>Choose Project<select name="addsubp"><?php
$query = "SELECT
projects.*
FROM
projects";
$showresult = mysql_query($query);
$csp_c = 1;
while($buyarray = mysql_fetch_assoc($showresult))
{
echo "<option value=".$buyarray['id'].">".$buyarray["pname"]."</option>";
}
?></select><br>
Choose Display Name for PDF <input type="text" name="addsubp_name" /> <Br>
Choose PDF: <input name="uploadedfile" type="file" /> <Br>
<input type="submit" value="Add PDF to Comm and Special Projects" name="submit" />
</form>
I have made sure that the application has write privileges to the "comm" directory. I have godaddy and used the file manager to make sure of that. I have had problems with permissions in this project before, so I know this isn't case. It keeps printing
There was an error uploading the file, please try again.
It doesn't attempt to upload any PDF at all, what am I doing wrong?
thanks!
You may have permissions issues, but for file uploads your form tag should contain the proper enctype attribute.
<form enctype="multipart/form-data" method="POST">
and defining a file size limit is also a good idea:
<input type="hidden" name="MAX_FILE_SIZE" value="1000000" />
try checking the Upload error message: http://php.net/manual/en/features.file-upload.errors.php
Your code is blindly assuming the file upload succeeded. At bare minimum you should have something like
if ($_FILES['uploadedfile']['error'] === UPLOAD_ERR_OK) {
... handle the upload
}
Your code is vulnerable to SQL injection. You do not escape any of the 3 values you're inserting into the database
You're creating the database record before making sure the file was successfully moved into the target directory. What happens if the file can't be written for any reason (as it is now with your problem)? The database will say it's there, file system will say it isn't
You're not checking for file collisions. If two seperate uploads send "file.txt", the second upload will overwrite the first one.
You're storing the files with the user-supplied name, which is under user control. If this file is web-accessible, anyone with access to your upload form can upload anything they want (e.g. a php file) and the server will happily execute it for them.

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