How to get a link from an uploaded file?

I use this simple script to build a simple image upload form - http://www.w3schools.com/PHP/php_file_upload.asp

The script is working 开发者_运维百科well and the image is stored in a folder called 'upload'.

The question is, how to echo back the uploaded file in my php script so that I can display the uploaded image in my website.

Thanks for helping! :D

---

The last script in your linked code (section entitled Saving the Uploaded File) states where it has been saved:

echo "Stored in: " . "upload/" . $_FILES["file"]["name"];

So, assuming upload is right below the webroot, your src would be:

echo '<img src="/upload/'.$_FILES["file"]["name"].'" />';

---

Typically, depending on your server config, the image (or any upload) is uploaded in /tmp. You'll need to MOVE the copied file to your web directory (possibly in a subdirectory like "uploaded images". You can rename the file in this process as well, because it's name will be adlkfa23ur20dasjsdablah.jpg

---

The uploaded file will be deleted if you do not copy it so. So copy it to either public or private directory, as in link w3school...:

 move_uploaded_file($_FILES["file"]["tmp_name"],
  "upload/" . $_FILES["file"]["name"]);
  echo "Stored in: " . "upload/" . $_FILES["file"]["name"];

Public, point to the file as in

<img src="path-to-img.jpg" />

and then the file needs to reside in a readable directory by the webserver. The default fileupload-directory is not readable, so copy the file.

Private, you use php to get the file for you:

<img src="getFile.php?filename=myfile.jpg">

//getFile.php
header('content-type: image/jpeg');
echo file_get_contents('myjpg.jpg');

regards, /t