uploading an image through form - forms

How should i write my controller class to upload an image through a form?
I have created the form like this
#{form #index(), enctype:'multipart/form-data'}
<br/>
<h4>Image:</h4> <input type="file" name="image" />
<br/>
<br/>
<input type="submit" name="submit" value="Upload Photo" />
#{/form}

There is agoo introduction here : http://www.lunatech-research.fr/playframework-file-upload-blob
If you upload multiple photos, there is some known problems. Here is a workaround that works for me : Multiple file upload in playframework

The enctype="multipart/form-data" is required to have Play framework handle all the upload.
On the controller side you just have to write :
public static void storeImage(File fileUpload) {
if (fileUpload == null) {
//Handle the error case
}
//Store the file in a perenial location.
//For example :
File storeLocation = new File("/relOrAbsPath"), fileUpload.getName());
boolean success = fileUpload.renameTo(storeLocation);
//...
}
By default Play stores the files in a temporary place.
You can also use annotations as you would do for other controller params.

Related

Insert data in H2 using HTML form

I'm using Play! Framework 2.4. I can make a table and insert data via the evolution .sql scripts but how do I setup my Appication.scala, routes etc to make a form insert data?
PS I'm quite new to Play
The best way play fromework recommends is using helper and template system.
https://www.playframework.com/documentation/2.4.x/ScalaForms
But if you want to try with plain html and handler request in the controller by your own so you can try this:
Application.scala
def save = Action { implicit request =>
//here handle params from html form,
//this will return Option[Map[String,ArrayBuffer]]
val optionMap = request.body.asFormUrlEncoded
println(optionMap)
//saving things here
Ok("saved")
}
in plain html form
<form action="/savedata" method="post" accept-charset="UTF-8" enctype="application/x-www-form-urlencoded">
<input type="text" name="name" />
<input type="text" name="lastName" />
<input type="submit" />
</form>
conf/routes
POST /savedata controllers.Application.save
hope this helps.
regards

Rest spring upload file with parameter in form

I am trying to upload a file i can able to upload a file but i need to post one parameter present in the form, now i need to post one parameter and upload file in the form.
I tried Multi part and Form url encoded in the consumes annotation. It is not working i am getting an error.
I used #FormDataParam and #FormParam annotation in the method.
java.lang.NullPointerException
org.glassfish.jersey.media.multipart.internal.FormDataParamValueFactoryProvider$FormDataParamValueFactory.provide(FormDataParamValueFactoryProvider.java:203)
org.glassfish.jersey.server.spi.internal.ParameterValueHelper.getParameterValues(ParameterValueHelper.java:81)
org.glassfish.jersey.server.internal.routing.SubResourceLocatorRouter.getResource(SubResourceLocatorRouter.java:220)
org.glassfish.jersey.server.internal.routing.SubResourceLocatorRouter.apply(SubResourceLocatorRouter.java:133)
org.glassfish.jersey.server.internal.routing.RoutingStage._apply(RoutingStage.java:128)
Server side:
#POST
#Path("import")
#Consumes(MediaType.MULTIPART_FORM_DATA)
#Produces(MediaType.APPLICATION_JSON + "; charset=UTF-8")
public void uploadScenario(#FormDataParam("importScenario") final InputStream is,
#FormDataParam("complectId") final Long complectId) {
// Realisation
}
Client side:
<form id="importForm" target="my_iframe" action="rest/exportimport/import" method="post" enctype="multipart/form-data">
<input id="uploader" type="file" name="importScenario" size="50"><input type="submit">
<input type="text" style="display:none" name="complectId" value="'+id+'">
</form>

Checking which submit button was clicked

I am devoloping a web application using J2EE and Spring Roo as framework.
I want to create a form with two submit buttons:
One for save and continue
Another for save and finish
<form action="mycontroller" method="post">
<input type="submit" value="Save and continue"/>
<input type="submit" value="Save and finish"/>
</form>
So I can choose to either store the data in the database and add more entries or to store the data and finish the process.
How can I check what submit button was clicked in the method of the controller that processes the action?
public class MyController {
void actionMethod(...) {
// check which submit was clicked
}
}
You should add a name field to both buttons:
<input type="submit" name="button" value="Save and continue"/>
<input type="submit" name="button" value="Save and finish"/>
Once in the controller, you can recover the element by this name field and check its value field:
String field = request.getParameter("button");
if ("Save and continue".equals(button)){
// Do stuff
}
else if ("Save and finish".equals(button)){
// Do a different stuff
}
else {
// None of them were pressed
}
Or also you can use a different name value for both buttons:
<input type="submit" name="button1" value="Save and continue"/>
<input type="submit" name="button2" value="Save and finish"/>
In your controller:
String button1 = request.getParameter("button1");
String button2 = request.getParameter("button2");
if (button1 != null){
// Do stuff
}
else if (button2 != null){
// Do a different stuff
}
else {
// None of them were pressed
}
Second solution is preferred because it doesn't depend on the value of the elements
To reduce if-else if you can, and benefit from Spring framework to handle these mapping, try the following solution (if you don't have many meta parameters to send in the form):
<form action="AddToCartListController" method="post">
<input type="submit" value="Save and continue"/>
</form>
<form action="CommitCartListController" method="post">
<input type="submit" value="Save and finish"/>
</form>
...
public class CartListController {
void AddToCartListController(...) {
// business code
}
void CommitCartListController(...) {
// business code
}
}
// ... or ...
public class AddToCartListController {
void actionMethod(...) {
// business code
}
}
public class CommitCartListController {
void actionMethod(...) {
// business code
}
}
and define a proper mapping in Spring configuration.

how pass form from module to component in joomla 3.1

I wana create a front-end joomla component and this is my first experience.
here is important things:
1-component/controller.php
class TestController extends JControllerLegacy
{
public function display($cachable = false, $urlparams = false)
{
$view= JFactory::getApplication()->input->getCmd('view','items');
JFactory::getApplication()->input->set('view', $view);
parent::display($cachable, $urlparams);
}
}
2: com_test/model/items.php
<?php
defined('_JEXEC') or die();
jimport( 'joomla.application.component.modellist' );
class TestModelItems extends JModelList
{
public function __construct($config = array())
{
if (empty($config['filter_fields']))
$config['filter_fields'] = array('id', 'title', 'catid');
parent::__construct($config);
}
function getListQuery()
{
$db = JFactory::getDBO();
$query = $db->getQuery(true);
$query->select(...)
return $query;
}
}
I can print the query result on default.php on view folder!
but I wana another thing.
I have a form like this in the front page of my site in a custom module:
<form action="" method="post">
<input type="text" name="wordsearch" value="search">
.
.
<input type="submit" />
</form>
Now!
I do not know how can I send this form (with post method) to getListQuery() function in model folder...how can do it?
i wana when sb click submit form, the component filter query sql according to values of form and then show new result to user!
i googled for hourse but no chance to solve. thanks for your help.
You can submit Form from module to component as follows.
Suppose your component name is com_helloworld The in your module form should have the following things.
<form action="" method="post">
<input type="text" name="wordsearch" value="search">
.
.
<input type="hidden" name="option" value="com_helloworld" />
<input type="hidden" name="view" value="yourview" />
<input type="hidden" name="task" value="my_controller_fun" />
<input type="hidden" value="your_controller_file_name" name="controller">
<input type="submit" />
</form>
In this example your controller file should have my_controller_fun method from controller to model you can use regular method. This methods will get all the form data in your controller , then you can pass that to model.
Detailed :
In your controller file.
function my_controller_fun(){
$post_array = $_POST;
$model = $this->getModel('Profile', 'UsersModel');//example for including profile model you can specify your model file name
$model->function_inyourmodel($post_array);//this function should be in model
}
Hope its help..

Multiple Form Actions to different pages / different target windows

I have a form on my page that needs to have multiple form actions associated with it. I'm using the following script in conjunction with the following code to achieve this:
<script>
function submitForm(action)
{
document.getElementById('summary').action = action;
document.getElementById('summary').submit();
}
</script>
<form action="go-gold.php" method="post" enctype="multipart/form-data">
<input type="image" id="arrow" name="go_back" onclick="submitForm('go-gold.php')" value="go_back" src="images/arrow_back.png" class="submit_button" /><br>
<input type="image" id="arrow" name="submit_form" onclick="submitForm('registration.php')" value="submit_form" src="images/arrow.png" class="submit_button" />
</form>
The first button needs to "go back" within the same browser window (self), and the second button needs to submit the info to a new window (blank). How do I modify the code to achieve this? Putting "target" functions within the input type doesn't work, and putting the target in the Form tag makes both submit buttons submit to the same window.
Thanks!
Easy with jQuery, also you have to identical ids for two separate form elements. You should have these as distinct ids unless you want to use a class name. Php can submit forms to the same page using the $_SERVER superglobal by using $_SERVER['PHP_SELF'] as the forms action name.
<script>
$(document).ready(function() {
$(".submit_button").click(function() {
clickVal = $(".submit_button").val();
if(clickVal == 'go_back') {
//do go back stuff
}
if(clickVal == 'submit_form') {
// do actions for other page
}
});
});
</script>
<form action="go-gold.php" method="post" enctype="multipart/form-data">
<input type="image" value="go_back" src="images/arrow_back.png" class="submit_button" /><br>
<input type="image" value="submit_form" src="images/arrow.png" class="submit_button" />
</form>