I have a drop down box with multiple options, and based on what value is selected I want it to dynamically update the folder path of the move_uploaded_file() function.
When I echo out the same thing that is in the if($file_name) block it shows the correct path each time I pick from the drop down, but when I actually submit the image upload the variable is basically empty and uploads the image to ../www/html//$file_name
Any help on to why the variable is basically empty on submit would be greatly appreciated. I can share all of the code if needed, but all of the rest of it is working fine, even the echo statement is pulling in the variable $selected just fine, but just not in the if($file_name) block.
$selected = $_POST['countries'];
if(isset($_POST['upload_img'])){
$file_name = $_FILES['image']['name'];
$file_type = $_FILES['image']['type'];
$file_size = $_FILES['image']['size'];
$file_tmp_name = $_FILES['image']['tmp_name'];
if($file_name) {
move_uploaded_file($file_tmp_name, "../www/html/$selected/$file_name");
}
}
echo "../www/html/$selected/$file_name";
Related
I have a table CRUD that display my database.
When I upload an image in my image folder I generate a random name of numbers with that function rand()
Here is what I have coded :
My upload function
function importer_image()
{
if(isset($_FILES["image"]))
{
$extension = explode('.', $_FILES['image']['name']);
$new_name = rand() . '.' . $extension[1];
$destination = './upload/' . $new_name;
move_uploaded_file($_FILES['image']['tmp_name'], $destination);
return $new_name;
}
}
My upload php
if($_POST["operation"] == "Ajouter")
{
$image = '';
if($_FILES["image"]["name"] != '')
{
$image = importer_image();
}
The problem is then when I code the update function, the substitued image stays in my folder and the new one has a new name generated. In order to avoid this, I would like to create a condition that says if $image !='' 1/ erase the old file 2/ upload the new file and keep the same name than the deleted image.
So I'm trying to create an update php process that would 1/ delete unlink() 2/ upload the new image with the name of the previous image.
In order to delete the old image and maintain new name of image as previous, you have to take a hidden input in your form which contain your uploaded image name.
For e.g :
<input type="hidden" name="old-image" value="here is your previous image">
Now when your upload function will hit then you can get previous image name by request and can delete or maintain new image as previous.
if($_FILES['image']['name'] != '') {
$old_image = $_POST['old-image']; // get old image
$new_name = $old_image; // make new image name as previous
unlink('/upload/'.$old_image); // remove old image from folder
$destination = './upload/' . $new_name;
move_uploaded_file($_FILES['image']['tmp_name'], $destination);
return $new_name;
} else {
//if there is no image uploaded in the form then it will maintain old image
return $new_name= $_POST['old-image'];
}
Hope it will help you.
Your code to get extension works only if you have just one dot in the uploaded filename, it won't work for example on image_a.1.jpg. Use instead:
$extension = strrchr($_FILES['image']['name'],'.');
Also, using rand() for naming will eventually give you a headache when it generates the same number the second time and the uploaded image will overwrite the old one or not get saved, possibly producing error showing the user your full server save path. I'd use some hashing, like:
$new_name = md5($_FILES['image']['name'] . time());
If you need to save the file name as int, use crc32() instead of md5(). You could convert md5() result to int, but that would produce a very long int (128bit) that might not fit in your database or even PHP code.
I would like to add user ID to file name in file upload. I tried this:
$date=date("Y-m-d");
$id=mysql_insert_id();
//Abstract File codes
$info = pathinfo($_FILES['abstract']['name']);
$ext = $info['extension']; // get the extension of the file
$newname = $id."_Abstract.".$ext;
$target = "application/abstracts/"; $target = $target .$newname;
if (empty($ext))
{ $abstract='' ;}
else $abstract=$newname;
But I only get 0 for ID.
I can display $name but $id is not working. Is it because I do not have an input for $id on my form? Thank you. I should also mention that the user is filling out an application form (for the first time) and attaching files on the same page. I had the file upload after the application was submitted I don;t think I would have problem call the $id. But since everything is on one page I might be calling an empty $id.
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
I have created a PHP and MySQL script which successfully uploads submitted images via PHP to a folder on my server, and then adds the filename with extension to my MySQL database.
With an FTP program I can see the submitted image inside the correct folder on my server with its correct file size. However, when I type the file path of the newly uploaded image (http://xxxxxx.com/images/image.jpg) into my browser, I get a blank page. Also when I try to import the image onto a website, nothing shows up.
However, when I re-download the image via the FTP program onto my computer, I can see that the image is TOTALLY OK. What am I missing?
Excerpts of my code are below:
<?php
// getting current post id and slug
$pid = $_POST['pid'];
$slug = $_POST['slug'];
//This is the directory where images will be saved
$target = '../company/'.$slug.'/images/';
$target = $target . basename( $_FILES['image']['name']);
//This gets all the other information from the form
$pic = ($_FILES['image']['name']);
$fileTmpLoc = ($_FILES["image"]["tmp_name"]);
$extract = explode(".", $pic);
$fileExt = end($extract);
list($width, $height) = getimagesize($fileTmpLoc);
if($width < 10 || $height < 10){
header("location: ../message.php?msg=ERROR: That image has no dimensions");
exit();
}
$rename = rand(100000000000,999999999999).".".$fileExt;
// check for correct filetype
if (!preg_match("/\.(gif|jpg|png)$/i", $pic) ) {
header("location: ../message.php?msg=ERROR: incorrect filetype");
exit();
}
include_once "../database-connect.php";
//Writes the information to the database
mysqli_query($dbconnection,"UPDATE companies SET picture='$rename' WHERE ID='$pid'") ;
//Writes the photo to the server
if(move_uploaded_file($fileTmpLoc, "../company/'.$slug.'/images/$rename"))
{
.... etc
What am I missing that it does not show up in the browser?
Maybe the path is not what you think it is when you try to link to the image or when you try to open it.
Note that this looks very wrong:
if(move_uploaded_file($fileTmpLoc, "../company/'.$slug.'/images/$rename"))
This will add two quotes and two dots to your path, so if $slug is some_company, the path will be:
/company/'.some_company.'/images/123456789.jpg
Perhaps you don't see or didn't notice that in your ftp program.
Also note that you have an sql injection problem, you should switch to prepared statements with bound variables.
Problem was indeed the URL output structure. Have changed it, like jeroen suggested:
if(move_uploaded_file($fileTmpLoc, "../images/$rename"))
Works fine now
I am using the code below to handle image upload. Every thing is going fine. I have allowed the option to upload two images, one will be the logo and another the screenshot.
Now when I am providing the option to edit the uploads, say choose new files to upload and replace the old ones, I am having this problem.
In my images[] array, I will get the image names one by one with the help of loop. However when the user wants to change the second image(screenshot) and selects nothing in the logo field, only screenshot should be changed.
But in the images[] array, the first element will be the first from the uploads. If i select both the images it is all right, I will get logo.jpg in images[0] and screenshot.jpg in images[1]. But when I select only the second image, i will get screenshot.jpg in images[0]. However, the first element of the array should be blank and the second element of the array should have the data of the second image upload option so the the changes to take into effect. How can i acheive that?
$config['upload_path'] = '../uploads/';
$config['allowed_types'] = '*';
$config['max_size'] = '0';
$config['max_width'] = '0';
$config['max_height'] = '0';
$this->load->library('upload', $config);
$image = array();
foreach ($_FILES as $key => $value) {
if (!empty($value['tmp_name'])) {
if ( ! $this->upload->do_upload($key)) {
$error = array('error' => $this->upload->display_errors());
//failed display the errors
} else {
//success
$data = $this->upload->data();
$image[]= $data['file_name'];
}
}
}
Remember, You must to initialize the upload library with each upload.
$this->load->library('upload');
$this->upload->initialize($config);
User guide: http://ellislab.com/codeigniter/user-guide/libraries/file_uploading.html
Edit:
You can check the $_FILES array and check if there are the name of the screenshot input and the logo input, if they aren´t, you can create the array you need
I´ve found this:
A simpler way to have create that data structure is to name your HTML file inputs different names. >If you want to upload multiple files, use:
<input type=file name=file1>
<input type=file name=file2>
<input type=file name=file3>
Each field name will be a key in the $_FILES array.
http://www.php.net/manual/es/features.file-upload.multiple.php#53933