I have two wizards ProjectWizard and ModifyEditWizard. I am adding both in the same page like below
projectPage = new ProjectPage(objSelected);
addPage(projectPage);
modifyeditpage=new ModifyEditPage(objSelected);
addPage(modifyeditpage);
also using :
public IWizardPage getNextPage(IWizardPage currentPage) {
currentPage.setPreviousPage(projectPage);
if(currentPage!=null && currentPage instanceof ProjectPage ){
modifyeditpage.setProjectIdValue(projectPage.getProjectIdValue());
return modifyeditpage;
}
return null;
}
The Problem I'm facing is the two page gets loaded at the same time when this wizard. I want to take input from first wizard and according to that populate data in next wizard.
Related
I am using Odoo 10e. I want a simple functionality that whenever i wanted to delete one or more then one item from a list view or from a specific list view only. I want to show all of the items which are selected for deleted to show their name in popup window so that user can have a quick review what's he is going to delete. I know user can see details in list view but i want to give a glimpse to user in shape of model window that this is going to be deleted. Are you sure to delete ?
If user click Confirm then normal delete case should work.
As far i research and worked on it, i have idea that it should be something regarding overriding the do_delete method in the list_view.js in the web module. But i didn't know much about javascript overriding for Odoo.
This is an example how I do it.
I called the name_get for your model and records ids, this names list, I change the text of confirming message with the information of ids selected.
do_delete: function (ids) {
new Model(this.model)
.call('name_get', [ids, this.dataset.get_context()]).done(function (names) {
var text = _t("Do you really want to remove these records?") + ' '+ names.join(' \n')
if (!(ids.length && confirm(text))) {
return;
}
var self = this;
return $.when(this.dataset.unlink(ids)).done(function () {
_(ids).each(function (id) {
self.records.remove(self.records.get(id));
});
// Hide the table if there is no more record in the dataset
if (self.display_nocontent_helper()) {
self.no_result();
} else {
if (self.records.length && self.current_min === 1) {
// Reload the list view if we delete all the records of the first page
self.reload();
} else if (self.records.length && self.dataset.size() > 0) {
// Load previous page if the current one is empty
self.pager.previous();
}
// Reload the list view if we are not on the last page
if (self.current_min + self._limit - 1 < self.dataset.size()) {
self.reload();
}
}
self.update_pager(self.dataset);
self.compute_aggregates();
});
});;
},
I need some tree structure in Aurelia. I got some link for that. It's working fine. But My requirement is like accordion with tree view. Means when I clicked on closed parent all opened parents should close and clicked one should open same as bootstrap accordion. same thing should happen When ever I clicked on child parent element with in parent repeat.
Below is my image for tree structure.
Gist run Link: Gist
The above gist is just tree structure with open and collapse. From that when I clicked on closed tree node , that should open and remaining tree nodes should be closed.
In the above gist "node-model.js" is having events for open and close. So when ever I clicked on icon the clicked event inside this variable will get only clicked node. How can I get other node in that method to hide.
Answer:
Inside your tree-view.js, add the following code (3 methods):
attached() {
window.addEventListener('goCollapseAll', (e) => {
this.closeOtherBranches(e.detail);
}, false);
}
closeOtherBranches(exceptNode) {
// traverse node tree to find current one
var found = null;
for(var i = 0; i < this.nodes.length; i++){
if (this.subSearch(this.nodes[i], exceptNode)) {
found = i;
}
}
if (found !== null) {
for(var i = 0; i < this.nodes.length; i++){
if ((i != found) && (this.nodes[i].expanded)) {
this.nodes[i].toggleNode();
}
}
}
}
subSearch(node, findNode) {
// recursive search of tree for findNode
var match = null;
if (node === findNode) {
match = node;
} else {
for(var i = 0; i < node.children.length; i++){
if (node.children[i] === findNode) {
match = node;
} else {
match = this.subSearch(node.children[i], findNode);
}
}
}
return match;
}
Then, inside your node-model.js, add the following lines at the beginning of toggleNode():
// close other node branches
if (!this.expanded) {
var event = new CustomEvent('goCollapseAll', { 'detail': this });
window.dispatchEvent(event);
}
Explanation:
When a node is expanded, it publishes a custom event to trigger the recursive search to close all nodes that are part of a different branch. It's not the prettiest solution and I think there might be a cleaner way if you adopt a different structure for the tree, but this solution definitely works well and accomplishes your purpose.
GistRun:
I've updated your GistRun to demonstrate the functionality. You can see it working here:
https://gist.run/?id=828c3c79bff0dfbaffec3252ed376c8c
I am new in MVC 4 web development and i am creating a control panel. I have developed a Add user page and submit information in database successfully.
But after submit when i press back button it will show previous form.
i am using redirection the page to same page after submit form.
here is the code to redirect
public ActionResult AdminPanel(RegisterUserModel user)
{
if (ModelState.IsValid) // Check the model state for any validation errors
{
if (user.AddUserToDB(user.username, user.password, user.fullName,user.contactNo,user.COAId)) // Calls the Login class checkUser() for existence of the user in the database.
{
TempData["SuccessMessage"] = "User Added Sucessfully!";
ModelState.Clear();
return Redirect("AdminPanel");
}
else
{
ViewBag.SuccessMessage = "User Not Added";
return View();
}
}
SelectList clientsList = GetClinetList();
ViewBag.clientsList = clientsList;
return View(); // Return the same view with validation errors.
}
I have tried many examples but issue not resolved yet so kindly give my suggesstions
If you don't want the user to be able to see the previous content when clicking back, then you must indicate that content should not be cached by the browser and must revalidate with the origin server
A summary of this behaviour is here - http://blog.55minutes.com/2011/10/how-to-defeat-the-browser-back-button-cache/
You could create a nocache attribute, like this one - https://stackoverflow.com/a/10011896/1538039, and apply it to your controller methods.
Im creating a site who works with ajaxRequest, when I click a link, it will load using ajaxRequest. When I load for example user/login UserController actionLogin, I renderPartial the view with processOUtput to true so the js needed inside that view will be generated, however if I have clientScriptRegister inside that view with events, how can I avoid to generate the scriptRegistered twice or multiple depending on the ajaxRequest? I have tried Yii::app()->clientScript->isSCriptRegistered('scriptId') to check if the script is already registered but it seems that if you used ajaxRequest, the result is always false because it will only be true after the render is finished.
Controller code
if (Yii::app()->request->isAjaxRequest)
{
$this->renderPartial('view',array('model'=>$model),false,true);
}
View Code
if (!Yii::app()->clientScript->isScriptregistered("view-script"))
Yii::app()->clientScript->registerScript("view-script","
$('.link').live('click',function(){
alert('test');
})
");
If I request for the controller for the first time, it works perfectly (alert 1 time) but if I request again for that same controller without refreshing my page and just using ajaxRequest, the alert will output twice if you click it (because it keeps on generating eventhough you already registered it once)
This is the same if you have CActiveForm inside the view with jquery functionality.. the corescript yiiactiveform will be called everytime you renderPartial.
To avoid including core scripts twice
If your scripts have already been included through an earlier request, use this to avoid including them again:
// For jQuery core, Yii switches between the human-readable and minified
// versions based on DEBUG status; so make sure to catch both of them
Yii::app()->clientScript->scriptMap['jquery.js'] = false;
Yii::app()->clientScript->scriptMap['jquery.min.js'] = false;
If you have views that are being rendered both independently and as HTML fragments to be included with AJAX, you can wrap this inside if (Yii::app()->request->isAjaxRequest) to cover all bases.
To avoid including jQuery scripts twice (JS solution)
There's also the possibility of preventing scripts from being included twice on the client side. This is not directly supported and slightly more cumbersome, but in practice it works fine and it does not require you to know on the server side what's going on at the client side (i.e. which scripts have been already included).
The idea is to get the HTML from the server and simply strip out the <script> tags with regular expression replace. The important point is you can detect if jQuery core scripts and plugins have already been loaded (because they create $ or properties on it) and do this conditionally:
function stripExistingScripts(html) {
var map = {
"jquery.js": "$",
"jquery.min.js": "$",
"jquery-ui.min.js": "$.ui",
"jquery.yiiactiveform.js": "$.fn.yiiactiveform",
"jquery.yiigridview.js": "$.fn.yiiGridView",
"jquery.ba-bbq.js": "$.bbq"
};
for (var scriptName in map) {
var target = map[scriptName];
if (isDefined(target)) {
var regexp = new RegExp('<script.*src=".*' +
scriptName.replace('.', '\\.') +
'".*</script>', 'i');
html = html.replace(regexp, '');
}
}
return html;
}
There's a map of filenames and objects that will have been defined if the corresponding script has already been included; pass your incoming HTML through this function and it will check for and remove <script> tags that correspond to previously loaded scripts.
The helper function isDefined is this:
function isDefined(path) {
var target = window;
var parts = path.split('.');
while(parts.length) {
var branch = parts.shift();
if (typeof target[branch] === 'undefined') {
return false;
}
target = target[branch];
}
return true;
}
To avoid attaching event handlers twice
You can simply use a Javascript object to remember if you have already attached the handler; if yes, do not attach it again. For example (view code):
Yii::app()->clientScript->registerScript("view-script","
window.myCustomState = window.myCustomState || {}; // initialize if not exists
if (!window.myCustomState.liveClickHandlerAttached) {
window.myCustomState.liveClickHandlerAttached = true;
$('.link').live('click',function(){
alert('test');
})
}
");
The cleanest way is to override beforeAction(), to avoid any duplicated core script:
class Controller extends CController {
protected function beforeAction($action) {
if( Yii::app()->request->isAjaxRequest ) {
Yii::app()->clientScript->scriptMap['jquery.js'] = false;
Yii::app()->clientScript->scriptMap['jquery-2.0.0.js'] = false;
Yii::app()->clientScript->scriptMap['anything.js'] = false;
}
return parent::beforeAction($action);
}
}
Note that you have to put the exact js file name, without the path.
To avoid including script files twice, try this extension: http://www.yiiframework.com/extension/nlsclientscript/
To avoid attaching event handlers twice, see Jons answer: https://stackoverflow.com/a/10188538/729324
I have created a silverlight app and have a listbox that is populated using a domaindatasource. I have created a button that takes the selected list item and updates a field (in my case its called IsDeleted)
my domain service looks like this
public IQueryable<Employee> GetEmployees(int storeID)
{
return this.ObjectContext.Employees.Where(e=>(e.StoreID==storeID)&&(e.IsDeleted==false));
}
In my button event I am doing this
EmployeeRecord.IsDeleted = true;
dsEmployee.SubmitChanges();
The database is been updated as expected until the application is reloaded or I F5 than I don't see the update.
What else should I be doing to see the update straight away?
You need to reload your domain context. Call the following:
myDomainDataSource.SubmitChanges((submitArgs) =>
{
if (submitArgs.IsComplete)
{
myDomainDataSource.Load<MyType>(myDomainDataSource.MyQuery(filterTextBox.Text), System.ServiceModel.DomainServices.Client.LoadBehavior.RefreshCurrent, true);
if (myDataSource.CanLoad)
myDataSource.Load();
}
else if (submitArgs.HasError)
{
throw submitArgs.Error;
}
}, null);