I have a JTable with several columns and rows. Once the user has moved the column, I would like to be informed via an event of some sort, so that I can do some calculations. However, since all I needed to do was
table.getTableHeader().setReorderingAllowed(true);
to get the columns to be movable, I am somewhat unsure what I can use to find out once they have been moved.
Any suggestions?
-F
To answer my own question..
You need to implementTableModelListener, and the tableChanged method in order to do this.
This tutorial looks at adding a Column Model Listener to a table and detecting when a column is moved. It might be what you are looking for.
Related
I use Backpack for Laravel.
I know that to add a field, we can use:
CRUD::addField(['name'=>..])
But, I don't know how to add a field that lookup another field input from different table to show a lookup value.
Please look at the illustration below.
Thanks.
Edit:
I found a temporary solution to this, but not effective.
Reference: https://backpackforlaravel.com/docs/4.1/crud-how-to#add-a-select2-field-that-depends-on-another-field
With select2_from_ajax you still have to select the option (even its just one option left), meanwhile what I really want is its automatically select that one option.
It's depends on what you need to do. You can use relationship field and show a select with the list of user. If you really wanna do it like this. You can probably make an ajax request to show it. In that case you ll need to use a custom view for the edit or the create view.
I have a datepicker calendar in my views, and basically I need to synchronize the selected dates (that I send in ajax) with a BoxDeliveryDate model (which only has one column named "date").
So in my Controller I was able to write a pretty nice method to only create a new record if one the selected dates is not yet stored in the database, like this :
foreach (Request::get('dates') as $date) {
$date_formated = date('Y-m-d', strtotime($date));
BoxDeliveryDate::firstOrCreate(['date'=>$date_formated]);
}
Now, if the user de-select one the dates in the datepicker, later, and synchronize, I need to delete it from the database.
Is there a nice way to do that in Laravel ? In other words, to delete every record of the table that are NOT in my Request::get('dates') ?
Also, I searched for a simple way to synchronize everything with only one method, but couldn't find anything.
Thanks for helping !
You can use whereNotIn() for that:
BoxDeliveryDate::whereNotIn('date', Request::get('dates'))->delete();
Note that this will not trigger any model events nor will it work with soft delete. Also, depending on the format of dates you might have to format the array before passing it to whereNotIn().
Also I believe it should be Request::input('dates') but it's possible that both actually works...
I would highly recommend using the soft delete trait (built into Laravel) if your doing mass deletes!
http://laravel.com/docs/4.2/eloquent#soft-deleting
$this->model->whereNotIn('id', $ids)->delete();
I created an extension for joomla using:
$id=$database->insertid();
I just covered that if two users are logged on to the site will fit together perform two records in the database and then this statement will return in both cases the same value.
in php you can solve this problem with the transactions.
In joomla how do I solve this problem?
If you have a table you are working with that extends JTable then make sure that you included the check out functionality that is optionally a part of that. THis must means adding a couple of fields like what is in the content table. This will prevent two people from editing the same row at the same time which creates a race condition in which one of the other will lose their data.
Please note that both php and joomla functions to return the last insert id rely on the mysql implementation, and mysql returns the last id inserted on the currently open connection so concurrency is not an issue
#iacoposk8 Your are right it might possible that in very rear case. Such time try to add current logged in user id in your sql query or any where so that it doesn't make any confict. I hope you get it what i want to say. Thanks
I got a silly general question...
If I generate a strongly typed view of an entity and chose "edit" as scaffolding, then the view does contain every column for that table. Changing and saving the values via setting it modifierd and call db.SaveChanges() does work in the controller. So far, so good.
But if I remove just one of that columns inside the view, then saving doesn't work anymore.
Is there a rule describing this? Is it only possible to make view with every column when wanting to save the model later on? I don't want to make 90 of 100 columns "hidden"...
PS: When editing a value in another table which is connected via Foreign Key (like customer.address.STREET) saving also does not work. Does everything of the entity ADDRESS has to be inside the view? I really don't get that.
Besides that: If I create my own ViewModel containing two entities: Do they also have to hold every column of both entities? This would be a whole bunch of traffic...
Answer is: You should not use the .Modified state. Instead using the UpdateModel method works fine without every field.
Recently I've been requested to add on something for the administrator of a site where he can 'feature' something.
For this discussion let's say it's a 'featured article'.
So naturally we already have a database model of 'articles' and it has ~20 columns as it is so I really do not feel like bloating it anymore than it already is.
My options:
Tack on a 'featured' bool (or int) and realize that only one thing will be featured at any given time
Create a new model to hold this and any other feature-creep items that might pop up.
I take your suggestions! ;)
What do you do in this instance? I come across this every now and then and I just hate having to tack on one more column to something. This information DOES need to be persisted.
I'd probably just add a simple two-column table that's basically a key-value store. Then add a new column with values like (featured_article_id, 45) or whatever the first featured ID is.
Edit: as pointed out in the comments by rmeador, it should be noted that this is only a good solution as long as things stay relatively simple. If you need to store more complex data, consider figuring out a more flexible solution.
If only one article can be featured at a time it is a waste to add a bool column. You should go up a level and add a column for the FeaturedArticleID. Do you have a Site_Settings table?
You could use an extensible model like having a table of attributes, and then a linking table to form a many-to-many relationship between articles and attributes. This way, these sorts of features do not require the schema to be modified.
Have some kind of global_settings table with a parameter_name and parameter_value columns. Put featured article id here.
For quick-and-dirty stuff like this, I like to include some sort of Settings table:
CREATE TABLE Settings (
SettingName NVARCHAR(250) NOT NULL,
SettingValue NVARCHAR(250)
)
If you need per-user or per-customer settings, instead of global ones, you could add a column to identify it to that specific user/customer. Then, you could just add a row for "FeaturedArticle" and parse the ID from a string. It's not super optimized, but plaintext is very flexible, which sounds like exactly what you need.