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
Related
How to create edit forms. For url edit?id=1121 I want to display pre-filled data
EditForm(twf.Form):
class child(twf.TableLayout):
name= twf.TextField(name="name",value=DBSession.query(student.name).filter(student.id == <id passed in url>).distinct().all())
#expose ('edit')
def edit(self, id)
return dict(page='edit', , form=EditForm(action='/save')
Template:
<div>${form.display()}</div>
There are a few ways to achieve this, but I'd say that the cleanest one is passing the values to the form from the controller action as showcased by http://turbogears.readthedocs.io/en/latest/turbogears/widgets_forms.html#displaying-forms
In the case of your specific example it should result in having a form that only declares the fields that need editing and a reference to the edited object id:
class EditForm(twf.Form):
class child(twf.TableLayout):
student_id = twf.HiddenField()
name = twf.TextField(name="name")
Then within the controller you can fetch the edited object and provide the values to the form:
#expose('edit')
def edit(self, id):
edited_student = DBSession.query(student).filter(student.id==id).first()
return dict(page='edit',
form=EditForm(action='/save',
value=dict(student_id=id,
name=edited_student.name))
Keep in mind that this is just a proof of concept, I haven't tested the code and it lacks proper validation/error handling. But it should pretty much give the idea that you just pass the student name to the form through the value parameter when displaying it.
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);
}
$this->redirect(array('site/success'));
}
So far , all is logical and simple. However , what I end up is that the comment the person wrote is being duplicated to every person_event row in the DB.
I want to put a text box in each row of the grid , and the commnet that will be written there will go to the specific event.
Now , I found this topic in yii about "admin-panel"
which is kind of helpful , BUT:
I already have a foreach in the action , the one that matches the person's id with the event's id , so how can I put another individual comment for each combo?
The default CGridView supports only basic functionality, you would need to extend CGridView or use an extension to make columns editable
Easiest way to do this is use something like TbEditableColumn from Yii-booster library
see http://yiibooster.clevertech.biz/extendedGridView#gridcolumns EditableColumn in the additional column types section
If you do not like or wish to use twitter-bootstrap styling a standalone extension like http://www.yiiframework.com/extension/eeditable will help.
Alternatively you can extend CGridView yourself to extend it to support column level editing
I'm new to Yii. I have a table where I'd like to add a user ID at the time of creation of an entry. I've identified actionCreate() in the associated Controller, which I guess is the appropriate place to make my ammendment. The only thing is I can't find any documentation on what I should now do. Can I ask for help on what I should no next.
thanks
You're probably looking for afterSave(). You can add a method like this to your User model:
protected function afterSave()
{
if($this->isNewRecord) {
// Insert ID into other table here
}
}
Simply set the field to primary and autoincrement in the database. This value will automatically be added after calling ->save() on the new object.
You will be able to access this ID field with $model->primaryKey
A small briefing on what I am trying to do.
I have three tables Content(contentId, body, timeofcreation), ContentAttachmentMap(contentId, attachmentId) and Attachment(attachmentId, resourceLocation).
The reason I adopted to create the mapping table because in future application the attachment can also be shared with different content.
Now I am using HQL to get data. My objectives is as follows:
Get All contents with/without Attachments
I have seen some examples in the internet like you can create an objective specific class (not POJO) and put the attribute name from the select statement within its constructor and the List of that Class object is returned.
For e.g. the HQL will be SELECT new com.mydomain.myclass(cont.id, cont.body) ..... and so on.
In my case I am looking for the following SELECT new com.mydomain.contentClass(cont.id, cont.body, List<Attachment>) FROM ...`. Yes, I want to have the resultList contain contentid, contentbody and List of its Attachments as a single result List item. If there are no attachments then it will return (cont.id, contentbody, null).
Is this possible? Also tell me how to write the SQL statements.
Thanks in advance.
I feel you are using Hibernate in a fundamentally wrong way. You should use Hibernate to view your domain entity, not to use it as exposing the underlying table.
You don't need to have that contentClass special value object for all these. Simply selecting the Content entity serves what you need.
I think it will be easier to have actual example.
In your application, you are not seeing it as "3 tables", you should see it as 2 entities, which is something look like:
#Entity
public class Content {
#Id
Long id;
#Column(...)
String content;
#ManyToMany
#JoinTable(name="ContentAttachmentMap")
List<Attachment> attachments;
}
#Entity
public class Attachment {
#Id
Long id;
#Column(...)
String resourceLocation
}
And, the result you are looking for is simply the result of HQL of something like
from Content where attachments IS EMPTY
I believe you can join fetch too in order to save DB access:
from Content c left join fetch c.attachments where c.attachments IS EMPTY