I find the documentation and tutorial of Yii 2.0 a bit short.
In a Yii 2.0 Model I would like to add a hidden field with a computed value, let's call it def_id. The model contains fields such as firstname, lastname, email etc. The computed value would be some combination of those three fields. (It is supposed to be some custom type of the logical, unique auto-increment.)
My question: where do I compute def_id with the other given fields so that Create and Update will write def_id into the database-table?
There is no hidden field in the model, there are just fields. If it is calculated field you do not have to even show it on screen so there is no point in putting in a hidden field.
You can however add it to the before save function for the model
public function beforeSave()
{
if ($this->isNewRecord) {
//calculate what you need
} else {
//recalculate if needed
}
return parent::beforeSave();
}
Related
Data table
I'm very new on asp.net core. I have a concern is how we can insert value by default via model. Such as I have a table as picture attached.
I can get model auto-generated and insert by select Transaction_UserID but it's not practical as normally we will get current user logon ID and pass it as value to that field.
My questions are:
How to implement to insert value generated by code not user via input form.
How and on which part we will calculate that value
Such as my code to set is similar like this will not work as it will not allow property with only set.
//- Set user to first user by default
private long _TransactionUserId = 3;
public long TransactionUserId
{
set
{
_TransactionUserId = value;
}
}
Thank for your support in advanced.
I'm trying to show the results from a stored procedure on a view. I have a ASP.NET MVC application with the following code. I used EntityFramework to generate the models.
public class ProjectsController : Controller
{
private DatabaseEntities db = new DatabaseEntities();
// GET: Projects
public ActionResult Index()
{
var projects = db.Projects.Include(p => p.Headquarter);
return View(projects.ToList(), db.CALCULATEBUDGET());
}
}
I get the following errors on this part: db.CALCULATEBUDGET()
Argument 1: cannot convert from 'System.Collections.Generic.List<TestApplication.Models.Project>' to 'string' TestApplication C:\TestApplication\TestApplication\Controllers\ProjectsController.cs 21 Active
Argument 2: cannot convert from 'System.Data.Entity.Core.Objects.ObjectResult<TestApplication.Models.CALCULATEBUDGET_Result>' to 'string' TestApplication C:\TestApplication\TestApplication\Controllers\ProjectsController.cs 21 Active
My stored procedure:
CREATE PROCEDURE dbo.CALCULATEBUDGET
AS
SELECT MonthlyRent, Budget, 100 * H.MonthlyRent/P.Budget AS RentPercentage
FROM Headquarter H, Project P
WHERE H.HeadquarterId = P.Headquarter_HeadquarterId
I'm trying to calculate how much percent the rent is from the budget. And then I want to show the results in a view.
first of all, when you return a View you need to either pass the name or leave it empty and then it works out using the defaults. You can pass a model as well, but what you've done is to pass two data items. The first one needs to be a View name:
return View( "YourViewName", yourDataModel);
The main concept of MVC is this : here is a View and here is the data for a View in the form of a model.
Directly throwing objects from a database at a view is usually a bad idea, I would suggest you decouple your things a little bit.
Have a business layer where you get the data, you map into a model object which matches what the view needs to display. Think of that as a translation layer from what your data looks like and what your View needs to display. Rule of thumb, only send to a View whatever it needs to display and nothing more.
You can combine multiple multiple data items in one data model for the View if that's what you need, but you still pass just one object to your View.
I am looking for the simplest way to get the referenced item value for a droplink field.
#Html.Sitecore().Field("Alignment")
I want to get the value of the choice, what's the best approach?
If you need to have ability to edit fields of alignment item which is chosen in 'Alignment' droplink field of context item or just show values of alignment item's fields for visitors:
#{
Sitecore.Data.Fields.ReferenceField alignmentField = Sitecore.Context.Item.Fields["Alignment"];
Sitecore.Data.Items.Item alignmentItem = alignmentField.TargetItem;
}
<div>
#Html.Sitecore().Field("Text of Alignment", alignmentItem)
</div>
This example assumes that Alignment template contains 'Text of Alignment' field.
The Droplink field stores the referenced item's ID. To retrieve this ID (providing the field is present in your current item/model):
((LinkField)Model.Item.Fields["Alignment"]).Value
To output the referenced item's name, you could do something like this:
#(Model.Item.Database.GetItem(((LinkField)Model.Item.Fields["Alignment"]).Value).Name)
But that's really ugly. The preferred approach would be to create an extension method encapsulating some of the above so you're not having to re-type that out :D
The article Extending the SitecoreHelper Class by John West shows how to extend the SitecoreHelper class to add custom field renderers, so you could end up creating a neat re-usable snippet like:
#(Html.Sitecore().ReferenceField("Alignment","Name"))
If this is in a partial view i.e. .cshtml file you can also use something like below:
Sitecore.Data.Fields.TextField alignment= Model.Item.Fields["Alignment"];
This will give you the id of the set item in the drop link , then from that id can retrieve it from the database like:
#if (!string.IsNullOrWhiteSpace(alignment.Value))
{
var setAlignment = Sitecore.Context.Database.GetItem(alignment.Value);
if (setAlignment != null && !string.IsNullOrWhiteSpace(setAlignment.Name))
{
setAlignment.Name
}
}
Personally i prefer this way as i can check if the droplink is set before trying to use the value.
I'm not sure what is the way to do this , so here I ask:
I have a Person model and Event model, and a connection table Person_Event.
The interface that I got now works in the following way:
A person is logging in and his id is being send via URL
The person is selecting events he is interested in from the cGridView (checkbox column)
Writing some comment
4.Pressing send button , and the following create action is being triggered:
public function actionXcreate()
{
$model=new Person_Event;
if(isset($_POST['Person_Event']))
{
foreach ($_POST['selectedIds'] as $eventId)
{
$pmodel=new Person_Event;
$pmodel->person_id=$this->_person->id; //the id of the person who is logged in
$pmodel->attributes=$_POST['Person_Event']; //the comment
$pmodel->event_id = $eventId; //all the events he checked in the grid
if (!$pmodel->save()) print_r($pmodel->errors);
}
$this->redirect(array('site/success'));
}
So far , all is logical and simple. However , what I end up is that the comment the person wrote is being duplicated to every person_event row in the DB.
I want to put a text box in each row of the grid , and the commnet that will be written there will go to the specific event.
Now , I found this topic in yii about "admin-panel"
which is kind of helpful , BUT:
I already have a foreach in the action , the one that matches the person's id with the event's id , so how can I put another individual comment for each combo?
The default CGridView supports only basic functionality, you would need to extend CGridView or use an extension to make columns editable
Easiest way to do this is use something like TbEditableColumn from Yii-booster library
see http://yiibooster.clevertech.biz/extendedGridView#gridcolumns EditableColumn in the additional column types section
If you do not like or wish to use twitter-bootstrap styling a standalone extension like http://www.yiiframework.com/extension/eeditable will help.
Alternatively you can extend CGridView yourself to extend it to support column level editing
I am working on create a custom field type and for implementation issue I need to retrieve the ID of the SPListItem in the SPField class which belong to these field type but I can't retrieve it.
For example:
public class myField:SPFieldText
{
// I need ListItemID in this class
}
Can anyone help me please?
SPFieldText is an SPField, which is the schema definition for a field. Its like saying, given an SQL create table statement, give me the id of row x. Can't be done.
I think the logic you are trying to perform should be done in an event receiver, so say when an item is saved, you take the ID and add it to the text field.
I didn't find a solution but I've tried another solution which is
I can get the ID of the item in the New and Edit forms so I saved the ID and the field value as a one value separated by '/' and in the SPFieldText class I've been able to retrieve the ID value from the value of the field