Created a form with button actions 'Approved' and 'Rejected' and selected workflow 'One step approval'.But after approving/rejecting the form status is still shown as 'New'.Is there anything missing from my side
You missed adding custom code for updating the action Type at the form Design. just adding the button from form design for update the Application Status . You need to define when to emit the customEvent with actionComplete Type example use the below for approved.
form.emit('customEvent', {
type: "actionComplete",
component: component,
actionType: "Approved" // action Type you need to pass
});
Click here to know more custom Events Available
Find a sample code from the Freedom of Information form given as a sample below which does a similar behaviour :
const submissionId = form._submission._id;
const formDataReqUrl = form.formio.formUrl+'/submission/'+submissionId;const formDataReqObj1 = { "_id": submissionId, "data": data};
const formio = new Formio(formDataReqUrl);
formio.saveSubmission(formDataReqObj1).then( result => {
form.emit('customEvent', {
type: "actionComplete",
component: component,
actionType: "data.actionType" // action Type you need to pass
});
}).catch((error)=>{
//Error callback on not Save
});
You need to choose custom Action on the button as shown:
Related
I'm building a angular 2 (2.0.0-rc.4) application with the new form API (2.0.2) and i've got a problem when i'm trying to update checkbox form controls with updateValue().
This is what i've done:
I've built a dynamic form with the new form API (based on the section in the cookbook: https://angular.io/docs/ts/latest/cookbook/dynamic-form.html). I've extended the form class to also handle checkboxes:
export class FormControlCheckbox extends FormBase <string> {
controlType: string;
options: { key: string, value: string, checked: boolean } [] = [];
checked: boolean = false;
constructor(options: {} = {}) {
super( options );
this.controlType = "checkbox";
this.options = options['options'] || [];
this.checked = this.options[0].checked;
}
}
This is what it looks like when it's created:
new FormControlCheckbox({
type: "checkbox",
label: 'A label',
name: "a-name",
value: "",
description: "a description",
options: [
{
label: 'A label',
name: "a-name",
checked: false
}
],
})
When the application are loaded the form controls are created and grouped together, everything works fine and the form get submitted as intended. I only had to do a workaround to make the checkbox update on change and the markup are as followed:
<input [formControlName]="control.key" [(ngModel)]="control.checked" [id]="control.key" [type]="control.controlType" [attr.checked]="control.checked ? true : null" [value] = "control.checked" (change)="control.checked = $event.target.checked">
I've also tried this markup (it also works fine):
<input [formControlName]="control.key" [(ngModel)]="control.checked" [id]="control.key" [type]="control.controlType" [attr.checked]="control.checked ? true : null" [value] = "control.checked" (change)="control.checked = check.checked" #check>
This is where my problem occurs
I'm adding a feature that updates the control values when the form just have been loaded (the user should be able to revisit the page and update previous values). The code below update all the control values.
for (var key in new_values) {
//the form control keys are the same as the key in the list of new_values
this.form.controls[key].updateValue(new_values[key]); //works for text, select and option
this.form.controls[key].updateValueAndValidity(); //not sure if needed?
this.form.controls[key].markAsDirty();
}
Text, option and select inputs gets updated but the checkboxes are unchanged. No errors or warnings. I've searched a lot for similar problems but have not found a similar problem or a solution.
Am I missing something obvious? Or have somebody had the same problem and want to share their solution?
Thanks!
Solved the problem by changing the values before creating the control-groups (before this it's possible to change the values (ex. x.value). It solved my problem but do not solve the fact that dynamically changes to checkbox form controls are not reflected in the DOM element.
Is there any good solutions when editing a form to save the model only if the user clicked on the save button and retrieve the old datas if the user canceled the action ?
I've seen some solutions like duplicating the object that is data-binded with each form fields and set the the initial object with the duplicated one when it is saved.
If you could give answers without using ember data could be great.
I understand you would prefer a solution that doesn't use ember-data, but I would argue that using ember-data is best practices. Here is a solution using ember-data because I imagine a lot of people may come across this question...
If you set up your route as follows, it will do exactly that.
App.CommentEditRoute = Em.Route.extend({
model: function(params) {
return this.store.find('comment', params.comment_id);
},
actions: {
willTransition: function(transition) {
var model = this.get('controller.content');
if (model.get('isDirty')) {
model.rollback();
}
}
},
});
If you call this.get('content').save() in the controller (because the user clicked the save button) it will persist the changes through the adapter and isDirty will be set to false. Thus, the model will not rollback. Otherwise, if you did not call this.get('content').save() in the controller, the isDirty property will be true and the unsaved changes will be discarded. See the DS.Model docs for more info.
willTransition is an event automatically called when the route is about to change - you don't have to call it directly.
Your controller might look like this:
App.CommentEditController = Em.ObjectController.extend({
save: function() {
var _this = this;
_this.get('content').save().then(function() {
// Success
_this.transitionToRoute('comments');
}, function() {
// Failure
Em.assert('Uh oh!');
});
},
cancel: function() {
this.transitionToRoute('comments');
},
});
Also, be sure to utilize the default HTML form submission using a proper HTML button or input for submission so you can capture the submission event in your view as follows:
App.CommentEditView = Em.View.extend({
submit: function() {
this.get('controller').save();
},
});
I am on my way building a Fiori like app using SAPUI5. I have successfully built the Master page, and on item click, I pass the context and navigate to Detail page.
The context path from Master page is something like /SUPPLIER("NAME"). The function in App.controoler.js is as follows:
handleListItemPress: function(evt) {
var context = evt.getSource().getBindingContext();
this.myNavContainer.to("Detail", context);
// ...
},
But I would like to know how I can access this context in the Detail page. I need this because I need to use $expand to build the URL and bind the items to a table.
There is an example in the UI5 Documentation on how to deal with this problem using an EventDelegate for the onBeforeShow function which is called by the framework automatically. I adapted it to your use case:
this.myNavContainer.to("Detail", context); // trigger navigation and hand over a data object
// and where the detail page is implemented:
myDetailPage.addEventDelegate({
onBeforeShow: function(evt) {
var context = evt.data.context;
}
});
The evt.data object contains all data you put in to(<pageId>, <data>). You could log it to the console to see the structure of the evt object.
Please refer the "shopping cart" example in SAP UI5 Demo Kit.
https://sapui5.hana.ondemand.com/sdk/test-resources/sap/m/demokit/cart/index.html?responderOn=true
Generally, in 'Component.js', the routes shall be configured for the different views.
And in the views, the route has to be listened to. Please see below.
In Component.js:
routes: [
{ pattern: "cart",
name: "cart",
view: "Cart",
targetAggregation: "masterPages"
}
]
And in Cart.controller.js, the route has to be listened. In this example, cart is a detail
onInit : function () {
this._router = sap.ui.core.UIComponent.getRouterFor(this);
this._router.attachRoutePatternMatched(this._routePatternMatched, this);
},
_routePatternMatched : function(oEvent) {
if (oEvent.getParameter("name") === "cart") {
//set selection of list back
var oEntryList = this.getView().byId("entryList");
oEntryList.removeSelections();
}
}
Hope this helps.
Presently, a messagebox appears with the failing class name:
Is it possible to override the default behavior in Alfresco? Could we use forms service to present a different message ?
Additional to zladuric answer,
you can use failureCallback method to show message what you want.
But it is difficult to search failureCallback method of workflow forms for a new one because workflow forms such as "Start Workflow", "Task Edit", "Task Detail" are used form engine.
For example, in "Start Workflow" form, you can add our own successCallBack and failureCallBack by writing onBeforeFormRuntimeInit event handler in start-workflow.js like this.
onBeforeFormRuntimeInit: function StartWorkflow_onBeforeFormRuntimeInit(layer, args)
{
var startWorkflowForm = Dom.get(this.generateId + "-form");
Event.addListener(startWorkflowForm, "submit", this._submitInvoked, this);
args[1].runtime.setAJAXSubmit(true,
{
successCallback:
{
fn: this.onFormSubmitSuccess,
scope: this
},
failureCallback:
{
fn: this.onFormSubmitFailure,
scope: this
}
});
}
onFormSubmitSuccess: function StartWorkflow_onFormSubmitSuccess(response)
{
this.navigateForward(true);
// Show your success message or do something.
}
onFormSubmitFailure: function StartWorkflow_onFormSubmitFailure(response)
{
var msgTitle = this.msg(this.options.failureMessageKey);
var msgBody = this.msg(this.options.failureMessageKey);
// example of showing processing response message
// you can write your own logic
if (response.json && response.json.message)
{
if(response.json.message.indexOf("ConcurrencyFailureException") != -1)
{
msgTitle = this.msg("message.concurrencyFailure");
msgBody = this.msg("message.startedAgain");
}
else
msgBody = response.json.message;
}
Alfresco.util.PopupManager.displayPrompt(
{
title: msgTitle,
text: msgBody
});
}
Since Alfresco.component.StartWorkflow(in start-workflow.js) extends Alfresco.component.ShareFormManager(in alfresco.js). You can override onBeforeFormRuntimeInit event in start-workflow.js. I hope this your help you.
I'm not looking at the code right now, but this looks like a regular YUI dialog. So it's fired by YUI. So this YUI is client side, probably in My-tasks dashlet or my tasks page.
Furthermore, the error message looks like it is a status.message from the failed backend message/service.
You could probably locate that client-side javascript file, find the method that starts the task and see what its' failureCallback handler is. Then edit that failureCallback method and make it show something different then the response.status.message or whatever it is. Perhaps something like this.msg("message.my-custom-error-message"); which you then customize on your own.
Modifying YUI dialog scripts will might affect the other functionalities as well.
If we customize start-workflow. js, its only going to be achieved in start workflow form.
So as generic solution, below is the suggestion.
When alfresco is rendering the workflow form , it is rendering the transition button using the activiti-transition.js file.Basically this buttons are doing nothing more but submitting the workflow form.
So the best way would be , customizing this activiti-transition.ftl and activiti-transition.js file , to make an ajax call and handle the response as we want.
I just had a look on full flow of how this front end error is shown.
activiti-transition is submiting the workflow form.
Using a function named as submitForm which resides inside alfresco.js, it is invoking an submit event of form
Inside the forms-runtime.js file there is one function named as _submitInvoked(handles the submit event of form), which is responsible for making an ajax call and submitting the workflow form.If there is error while submitting , it will display the error which is from backend.
I am new for extjs.
I am having one store in which i am having some data coming from json file.
I have created a form.
i want to save new data in that store through form.
How can I create button and submit form on that button's click?
please let me know.
Thanks in advance
If you're using ExtJS 4 and your form is bound to a 'model' instance, then you get get a reference to your model in your buttons click handler and call:
model.save();
Which will send a post request to the url defined as the proxy on the model class.
Post your code and perhaps we can be some more direct help.
In the API of ExtJS is a good example on how to create a Ext.form.Panel with a button. When you click on this button the form gets submitted.
This example doesn't work because it doesn't submit to a page but it is very configurable.
Snippet of the button+handler:
buttons: [{
text: 'Submit',
handler: function() {
var form = this.up('form').getForm();
if (form.isValid()) {
form.submit({
url: '', //this is the url where the form gets submitted
success: function(form, action) {
Ext.Msg.alert('Success', action.result.msg);
},
failure: function(form, action) {
Ext.Msg.alert('Failed', action.result.msg);
}
});
}
}
}]