Php – Zend Framework: image upload

file uploadPHPzend-framework

I want to upload an image with Zend Framework version 1.9.6. The uploading itself works fine, but I want a couple of other things as well … and I'm completely stuck.

Error messages for failing to upload an image won't show up.
If a user doesn't enter all the required fields but has uploaded an image then I want to display the uploaded image in my form. Either as an image or as a link to the image. Just some form of feedback to the user.
I want to use Zend_ Validate_ File_ IsImage. But it doesn't seem to do anything.
And lastly; is there some automatic renaming functionality?

All ideas and suggestions are very welcome. I've been struggling for two days now.

These are simplified code snippets:

myform.ini

method = "post"

elements.title.type = "text"
elements.title.options.label = "Title"
elements.title.options.attribs.size = 40
elements.title.options.required = true

elements.image.type = "file"
elements.image.options.label = "Image"
elements.image.options.validators.isimage.validator = "IsImage"

elements.submit.type = "submit"
elements.submit.options.label = "Save"

TestController

<?php
class Admin_TestController extends Zend_Controller_Action
{
  public function testAction ()
  {
    $config = new Zend_Config_Ini(MY_SECRET_PATH . 'myform.ini');
    $f = new Zend_Form($config);

    if ($this->_request->isPost())
    {
      $data = $this->_request->getPost();

      $imageElement = $f->getElement('image');
      $imageElement->receive();

      //$imageElement->getValue();

      if ($f->isValid($data))
      {
        //save data
        $this->_redirect('/admin');
      }

      else
      {
        $f->populate($data);
      }
    }

    $this->view->form = $f;
  }
}
?>

My view just echo's the 'form' variable.

Best Answer

First, put this at the start of your script:

error_reporting(E_ALL);//this should show all php errors

I think the error messages are missing from the form because you re-populate the form before you display it. I think that wipes out any error messages. To fix that, remove this part:

else
{
   $f->populate($data);
}

To show the uploaded image in the form, just add a div to your view template, like this:

<div style="float:right"><?=$this->image?></div>

If the image uploaded ok, then populate $view->image with an img tag.

As for automatic re-naming, no, it's not built in, but it's very easy. I'll show you how below. Here's how I handle my image uploads:

$form = new Zend_Form();
$form->setEnctype(Zend_Form::ENCTYPE_MULTIPART);

$image = new Zend_Form_Element_File('image');
$image->setLabel('Upload an image:')
      ->setDestination($config->paths->upload)
      ->setRequired(true)
      ->setMaxFileSize(10240000) // limits the filesize on the client side
      ->setDescription('Click Browse and click on the image file you would like to upload');
$image->addValidator('Count', false, 1);                // ensure only 1 file
$image->addValidator('Size', false, 10240000);            // limit to 10 meg
$image->addValidator('Extension', false, 'jpg,jpeg,png,gif');// only JPEG, PNG, and GIFs

$form->addElement($image);

$this->view->form = $form;

if($this->getRequest()->isPost())
{
    if(!$form->isValid($this->getRequest()->getParams()))
    {
        return $this->render('add');
    }

    if(!$form->image->receive())
    {
        $this->view->message = '<div class="popup-warning">Errors Receiving File.</div>';
        return $this->render('add');
    }

    if($form->image->isUploaded())
    {
        $values = $form->getValues();
        $source = $form->image->getFileName();

        //to re-name the image, all you need to do is save it with a new name, instead of the name they uploaded it with. Normally, I use the primary key of the database row where I'm storing the name of the image. For example, if it's an image of Person 1, I call it 1.jpg. The important thing is that you make sure the image name will be unique in whatever directory you save it to.

        $new_image_name = 'someNameYouInvent';

        //save image to database and filesystem here
        $image_saved = move_uploaded_file($source, '/www/yoursite/images/'.$new_image_name);
        if($image_saved)
        {
            $this->view->image = '<img src="/images/'.$new_image_name.'" />';
            $form->reset();//only do this if it saved ok and you want to re-display the fresh empty form
        }
    }
}

UPDATE

With XHR2, File upload through AJAX is supported. E.g. through FormData object, but unfortunately it is not supported by all/old browsers.

FormData support starts from following desktop browsers versions.

IE 10+
Firefox 4.0+
Chrome 7+
Safari 5+
Opera 12+

For more detail, see MDN link.

Java – How to upload files to a server using JSP/Servlet

Introduction

To browse and select a file for upload you need a HTML <input type="file"> field in the form. As stated in the HTML specification you have to use the POST method and the enctype attribute of the form has to be set to "multipart/form-data".

<form action="upload" method="post" enctype="multipart/form-data">
    <input type="text" name="description" />
    <input type="file" name="file" />
    <input type="submit" />
</form>

After submitting such a form, the binary multipart form data is available in the request body in a different format than when the enctype isn't set.

Before Servlet 3.0, the Servlet API didn't natively support multipart/form-data. It supports only the default form enctype of application/x-www-form-urlencoded. The request.getParameter() and consorts would all return null when using multipart form data. This is where the well known Apache Commons FileUpload came into the picture.

Don't manually parse it!

You can in theory parse the request body yourself based on ServletRequest#getInputStream(). However, this is a precise and tedious work which requires precise knowledge of RFC2388. You shouldn't try to do this on your own or copypaste some homegrown library-less code found elsewhere on the Internet. Many online sources have failed hard in this, such as roseindia.net. See also uploading of pdf file. You should rather use a real library which is used (and implicitly tested!) by millions of users for years. Such a library has proven its robustness.

When you're already on Servlet 3.0 or newer, use native API

If you're using at least Servlet 3.0 (Tomcat 7, Jetty 9, JBoss AS 6, GlassFish 3, etc), then you can just use standard API provided HttpServletRequest#getPart() to collect the individual multipart form data items (most Servlet 3.0 implementations actually use Apache Commons FileUpload under the covers for this!). Also, normal form fields are available by getParameter() the usual way.

First annotate your servlet with @MultipartConfig in order to let it recognize and support multipart/form-data requests and thus get getPart() to work:

@WebServlet("/upload")
@MultipartConfig
public class UploadServlet extends HttpServlet {
    // ...
}

Then, implement its doPost() as follows:

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    String description = request.getParameter("description"); // Retrieves <input type="text" name="description">
    Part filePart = request.getPart("file"); // Retrieves <input type="file" name="file">
    String fileName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString(); // MSIE fix.
    InputStream fileContent = filePart.getInputStream();
    // ... (do your job here)
}

Note the Path#getFileName(). This is a MSIE fix as to obtaining the file name. This browser incorrectly sends the full file path along the name instead of only the file name.

In case you want to upload multiple files via either multiple="true",

<input type="file" name="files" multiple="true" />

or the old-fashioned way with multiple inputs,

<input type="file" name="files" />
<input type="file" name="files" />
<input type="file" name="files" />
...

then you can collect them as below (unfortunately there is no such method as request.getParts("files")):

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    // ...
    List<Part> fileParts = request.getParts().stream().filter(part -> "files".equals(part.getName()) && part.getSize() > 0).collect(Collectors.toList()); // Retrieves <input type="file" name="files" multiple="true">

    for (Part filePart : fileParts) {
        String fileName = Paths.get(filePart.getSubmittedFileName()).getFileName().toString(); // MSIE fix.
        InputStream fileContent = filePart.getInputStream();
        // ... (do your job here)
    }
}

When you're not on Servlet 3.1 yet, manually get submitted file name

Note that Part#getSubmittedFileName() was introduced in Servlet 3.1 (Tomcat 8, Jetty 9, WildFly 8, GlassFish 4, etc). If you're not on Servlet 3.1 yet, then you need an additional utility method to obtain the submitted file name.

private static String getSubmittedFileName(Part part) {
    for (String cd : part.getHeader("content-disposition").split(";")) {
        if (cd.trim().startsWith("filename")) {
            String fileName = cd.substring(cd.indexOf('=') + 1).trim().replace("\"", "");
            return fileName.substring(fileName.lastIndexOf('/') + 1).substring(fileName.lastIndexOf('\\') + 1); // MSIE fix.
        }
    }
    return null;
}

String fileName = getSubmittedFileName(filePart);

Note the MSIE fix as to obtaining the file name. This browser incorrectly sends the full file path along the name instead of only the file name.

When you're not on Servlet 3.0 yet, use Apache Commons FileUpload

If you're not on Servlet 3.0 yet (isn't it about time to upgrade?), the common practice is to make use of Apache Commons FileUpload to parse the multpart form data requests. It has an excellent User Guide and FAQ (carefully go through both). There's also the O'Reilly ("cos") MultipartRequest, but it has some (minor) bugs and isn't actively maintained anymore for years. I wouldn't recommend using it. Apache Commons FileUpload is still actively maintained and currently very mature.

In order to use Apache Commons FileUpload, you need to have at least the following files in your webapp's /WEB-INF/lib:

Your initial attempt failed most likely because you forgot the commons IO.

Here's a kickoff example how the doPost() of your UploadServlet may look like when using Apache Commons FileUpload:

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    try {
        List<FileItem> items = new ServletFileUpload(new DiskFileItemFactory()).parseRequest(request);
        for (FileItem item : items) {
            if (item.isFormField()) {
                // Process regular form field (input type="text|radio|checkbox|etc", select, etc).
                String fieldName = item.getFieldName();
                String fieldValue = item.getString();
                // ... (do your job here)
            } else {
                // Process form file field (input type="file").
                String fieldName = item.getFieldName();
                String fileName = FilenameUtils.getName(item.getName());
                InputStream fileContent = item.getInputStream();
                // ... (do your job here)
            }
        }
    } catch (FileUploadException e) {
        throw new ServletException("Cannot parse multipart request.", e);
    }

    // ...
}

It's very important that you don't call getParameter(), getParameterMap(), getParameterValues(), getInputStream(), getReader(), etc on the same request beforehand. Otherwise the servlet container will read and parse the request body and thus Apache Commons FileUpload will get an empty request body. See also a.o. ServletFileUpload#parseRequest(request) returns an empty list.

Note the FilenameUtils#getName(). This is a MSIE fix as to obtaining the file name. This browser incorrectly sends the full file path along the name instead of only the file name.

Alternatively you can also wrap this all in a Filter which parses it all automagically and put the stuff back in the parametermap of the request so that you can continue using request.getParameter() the usual way and retrieve the uploaded file by request.getAttribute(). You can find an example in this blog article.

Workaround for GlassFish3 bug of `getParameter()` still returning `null`

Note that Glassfish versions older than 3.1.2 had a bug wherein the getParameter() still returns null. If you are targeting such a container and can't upgrade it, then you need to extract the value from getPart() with help of this utility method:

private static String getValue(Part part) throws IOException {
    BufferedReader reader = new BufferedReader(new InputStreamReader(part.getInputStream(), "UTF-8"));
    StringBuilder value = new StringBuilder();
    char[] buffer = new char[1024];
    for (int length = 0; (length = reader.read(buffer)) > 0;) {
        value.append(buffer, 0, length);
    }
    return value.toString();
}

String description = getValue(request.getPart("description")); // Retrieves <input type="text" name="description">

Saving uploaded file (don't use `getRealPath()` nor `part.write()`!)

Head to the following answers for detail on properly saving the obtained InputStream (the fileContent variable as shown in the above code snippets) to disk or database:

Serving uploaded file

Head to the following answers for detail on properly serving the saved file from disk or database back to the client:

Ajaxifying the form

Head to the following answers how to upload using Ajax (and jQuery). Do note that the servlet code to collect the form data does not need to be changed for this! Only the way how you respond may be changed, but this is rather trivial (i.e. instead of forwarding to JSP, just print some JSON or XML or even plain text depending on whatever the script responsible for the Ajax call is expecting).

Hope this all helps :)