I have two Classes. "Product" and Product can have a number of objects of class "Attribute"
When someone edits the "Product" I would like to delete all Attributes associated with that product and recreate the new ones (second part works fine) but I'm getting a error which stops me removing the attributes.
public ProductMapping()
{
Id(x => x.Id);
Map(x => x.ProductName);
Map(x => x.Description);
Map(x => x.Quantity);
Map(x => x.Price);
Map(x => x.Live);
HasMany(x => x.Images)
.KeyColumn("ProductId")
.Cascade.All()
.Inverse();
HasMany(x => x.Attribute)
.KeyColumn("ProductId")
.Cascade.All();
}
And my Attribute mapping
public AttributeMapping()
{
Id(x => x.Id);
Map(x => x.Name);
Map(x => x.Value);
Map(x => x.Live);
References(x => x.Product)
.ForeignKey()
.Column("ProductId")
.Cascade.SaveUpdate();
}
I'm not sure if my cascading is correct but I tried them all!
I've tried to get all the attributes and do
var AttRepository = GetRepository<ForSale.Domain.Attribute>();
foreach (var a in dbProduct.Attribute)
{
AttRepository.Delete(a);
}
removing each one individually, also tried doing a:
dbProduct.Attributes.Clear();
Again it doesn't succeed!
The error I get is
"Cannot insert the value NULL into column 'ProductId', table '{LocalFilePath}/PRODUCTS.MDF.dbo.Attribute'; column does not allow nulls. UPDATE fails.
The statement has been terminated."
dbProduct
You should map Product.Attribute as Inverse.
Related
I have the following tables/models: A, B, C, BC, D, BCD
(A:B 1:N) Connecting table D to BCD would be no problem. However I would like to filter key attributes as dropdown from A, B, C and D to find results in BCD (because in the end I need BCDid). In BCD next to BCid and Did I can store of course Aid, Bid and Cid, and it would seem to me quite an easy workaround, however I know it's totally against db normalisation. Is there another, better way (with eager loading of course)?
I've now this in BCD:
public function getB() {
return $this->hasOne(\app\models\B::className(), ['id' => 'Bid'])
->via('BC');
}
and it seems to work, but it's not eager loading.
And how do I get to model A? can I define it like this in BCD?:
public function getA() {
return $this->hasOne(\app\models\A::className(), ['id' => 'Aid'])
->via('BC')
->via(B);
}
It doesn't really work yet.
This way it works (BCSearch):
public function search($params) {
$query = BC::find()->joinWith('A', true)->joinWith('C', true);
Relation A in BC defined with "via". Dropdown filter also works.
But I still don't know how to achieve one more level deep into db structure.
This way it seems to work fully:
models/BCDSearch.php
public function search($params) {
$query = BCD::find()
->select([
'BCD.id',
'BCD.amount',
'A.id AS A_Id',
'A.name AS A_name',
'B.name AS B_name',
'B.name2 AS B_name2',
'C.name AS C_name',
'D.name AS D_name',
])
->leftJoin('BC', 'BC.id = BC_Id')
->leftJoin('B', 'B.id = B_Id')
->leftJoin('A', 'A.id = A_Id')
->leftJoin('C', 'C.id = C_Id')
->leftJoin('D', 'D.id = D_Id');
$query->andFilterWhere([
...
'A.id' => $this->A_Id,
'B.name' => $this->B_name,
'B.name2' => $this->B_name2,
'C.name' => $this->C_name,
'D.name' => $this->D_name,
]);
public function rules() {
return [
...
[[... 'A_Id', 'B_name', 'B_name2', 'C_name', 'D_name'], 'safe'],
];
}
models/base/BCD.php:
class BCD extends Whatever {
public $A_Id;
public $A_name;
public $B_name;
public $B_name2;
public $C_name;
public $D_name;
views/BCD/index.php:
GridView::widget([
'layout' => '{summary}{pager}{items}{pager}',
'dataProvider' => $dataProvider,
'pager' => [
'class' => yii\widgets\LinkPager::className(),
'firstPageLabel' => Yii::t('app', 'First'),
'lastPageLabel' => Yii::t('app', 'Last')],
'filterModel' => $searchModel,
'columns' => [
[
'attribute' => 'A_Id',
'value' => 'A_name',
'filter' => yii\helpers\ArrayHelper::map(app\models\A::find()->all(), 'id', 'name')
],
[
'attribute' => 'B_name',
'value' => 'B_name',
'filter' => yii\helpers\ArrayHelper::map([['id' => 'name1', 'name' => 'name1'], ['id' => 'name2', 'name' => 'name2']], 'id', 'name')
],
[
'attribute' => 'B_name2',
'value' => 'B_name2',
'filter' => yii\helpers\ArrayHelper::map([['id' => 'name2_1', 'name' => 'name2_1'], ['id' => 'name2_2', 'name' => 'name2_2']], 'id', 'name')
],
[
'attribute' => 'C_name',
'value' => 'C_name',
'filter' => yii\helpers\ArrayHelper::map(app\models\C::find()->all(), 'C_name', 'C_name')
],
[
'attribute' => 'D_name',
'value' => 'D_name',
'filter' => yii\helpers\ArrayHelper::map(app\models\D::find()->all(), 'D_name', 'D_name')
],
'amount',
...
hope this helps others. It was not easy to figure out (at least for me) this basically easy solution. I don't know why but so far I couldn't find any relevant info on the web like this.
EF 6.1.3
Mapping
var b = mb.Entity<PAYMENT_INVOICE>();
b.ToTable("T_PAYMENT_INVOICE");
b.HasPrimaryKey(x => x.Id);
b.Property(x => x.OrdersSumm).IsRequired();
b.Property(x => x.MargaSumm).IsRequired();
b.Property(x => x.DeliverySumm).IsRequired();
b.Property(x => x.PaySumm).IsRequired();
b.Property(x => x.CreateDate).IsRequired();
b.Property(x => x.Sended).IsRequired();
b.Property(x => x.LastError).IsOptional();
b.HasOptional(x => x.Payment).WithMany(x => x.Invoices).HasForeignKey(x => x.PaymentId);
b.HasRequired(x => x.Client).WithMany().HasForeignKey(x => x.ClientId);
b.HasRequired(x => x.Purchase).WithMany(x => x.Invoices).HasForeignKey(x => x.PurchaseId).WillCascadeOnDelete(false);
Why generate this key and column? Both are equal.
I remove column PURCHASE_ID and FK_dbo.T_PAYMENT_INVOICE_dbo.T_PURCHASE_PURCHASE_Id.
When I save PAYMENT_INVOICE I get an error:
Invalid column name 'PURCHASE_Id'
I do not need a second key, what is my mistake?
Kendo UI edit option showing some unexpected behaviour, as you can see in the image there is a text box below Server column and 2 below ServerIP column all containing the id of server "SQL" i selected. Problem is when ever i want to show Server IP column this behaviour occurs, both server and server IP are from the same table.
#(Html.Kendo().Grid<EnvironmentPOCO>()
.Name("Grid")
.Columns(columns =>
{
columns.Bound(d => d.EnvironmentName).Width(200).Title("EnvirontmentName");
columns.ForeignKey(d => d.EnvironmentTypeID, (List<EnvironmentTypePOCO>)ViewData["EnvironmentType"], "EnvironmentTypeID", "EnvironmentTypeCode").Width(150).Title("EnvironmentCode").EditorTemplateName("_EnvironmentCodeDropDown");
columns.ForeignKey(d => d.ServerID, (List<ServerPOCO>)ViewData["ServerDetails"], "ServerID", "ServerName").Width(200).Title("Server").EditorTemplateName("_ServerDropDown");
columns.ForeignKey(d => d.ServerID, (List<ServerPOCO>)ViewData["ServerDetails"], "ServerID", "ServerIP").Width(200).Title("ServerIP");
columns.ForeignKey(d => d.ProjectID, (List<ProjectPOCO>)ViewData["Projects"], "ProjectID", "ProjectName").Width(200).Title("ProjectName").EditorTemplateName("_ProjectNameDropDown");
// columns.ForeignKey(d => d.ProjectID, (List<ProjectPOCO>)ViewData["Projects"], "ProjectID", "ProjectDescription").Width(200).Title("ProjectDescription")/*.EditorTemplateName("_ProjectDescription")*/;
columns.Command(d =>
{
d.Edit();
d.Destroy();
}).Width(200).Title("Action");
})
.ToolBar(tools => tools.Create())
.Sortable()
.Pageable()
.Filterable()
.DataSource(dataSource => dataSource
.Ajax()
.Model(model =>
{
model.Id(m => m.EnvironmentID);
model.Field(m => m.EnvironmentName);
model.Field(m => m.EnvironmentTypeID);
model.Field(m => m.ProjectID);
model.Field(m => m.ServerID);
})
.Read(read => read.Url(ViewBag.ApiBaseUrl).Type(HttpVerbs.Get))
.Create(create => create.Url(ViewBag.ApiBaseUrl).Type(HttpVerbs.Post))
.Update(update => update.Url(ViewBag.ApiBaseUrl).Type(HttpVerbs.Put))
.Destroy(destroy => destroy.Url(ViewBag.ApiBaseUrl).Type(HttpVerbs.Delete))
)
)
I found a solution to this, actually if you want to show 2 fields of the foriegn key, You can make a property in your class.
public string ServerDetailsProperty
{
get
{
return string.Format(" Name: {0} || IP: {1}", ServerName, ServerIP);
}
}
then call it in your csHTML file like this.
columns.ForeignKey(d => d.ServerID, (List<ServerPOCO>)ViewData["ServerDetails"], "ServerID", "**ServerDetailsProperty**").Width(200).Title("ServerIP");
Now if you press Edit you want see the unexpected behiour as in the diagram.
Try Changing the id and name attribute of the Server IP columns as the Grid is not able to differentiate between the Server Column and Server IP column on Edit.
For your reference I have tried below:
columns.ForeignKey(d => d.ServerID, (List<ServerPOCO>)ViewData["ServerDetails"], "ServerID", "ServerIP").Width(200).Title("ServerIP").HtmlAttributes(new { #id="ServerIP_#=ServerIP#", #name="ServerIP_#=ServerIP#" })
Let me know if this doesn't solve your issue.
EDIT:
You can add the dropdown in client template as below:
columns.Bound(s => s.ServerID).ClientTemplate((#Html.Kendo().DropDownList()
.BindTo((List<ServerPOCO>)ViewData["ServerDetails"])
.Name("ServerIP#=ServerIP#")
.DataTextField("ServerIP")
.DataValueField("ServerID")
.ToClientTemplate()).ToHtmlString());
On Grid DataBound event set the grid scripts to load with document as below:
function onGridDataBound(e) {
$('#GridName script').appendTo(document.body);
}
Finally set the field to readonly in model meta:
model.Field(s => s.SensorID).Editable(false);
For further information have a look at the explaination: Dropdown in Column Client Template
Here’s my widget in the Form.Class:
$this->widgetSchema['schools'] = new sfWidgetFormChoice(array(
'choices' => Doctrine_Core::getTable('school')->getUsersSchools($userId),
'renderer_class' => 'sfWidgetFormSelectDoubleList',
'renderer_options' => array(
'label_unassociated' => 'Unassociated',
'label_associated' => 'Associated'
)));
The above works just fine, but the values that are stored are unassociated to the choices list referenced above. I need to store the ids of the array retrieved as the values. Instead, the list that is retrieved is chronological and the ids are ignored.
Here's the schoolTable query:
public function getUsersSchools($id){
$q =Doctrine_Query::create()
->select('id')
->from('school')
->where('user_id = ?', $id)
->execute();
return $q;
}
If I understand your question correctly you would like to store associated school ids.
Use the sfWidgetFormDoctrineChoice widget instead and it will work out of the box, as it using primary keys as ids.
$query = Doctrine_Core::getTable('school')->queryForSelect($userId);
$this->setWidget('schools', new sfWidgetFormDoctrineChoice(array(
'model' => 'school',
'query' => $query,
'multiple' => true,
'renderer_class' => 'sfWidgetFormSelectDoubleList',
'renderer_options' => array(
'label_unassociated' => 'Unassociated',
'label_associated' => 'Associated'
),
)));
$this->setValidator('schools', new sfValidatorDoctrineChoice(array(
'model' => 'schoool',
'query' => $query,
'multiple' => true,
)));
// in SchoolTable class
public function queryForSelect($userId)
{
return $this->createQuery('s')
->andWhere('s.user_id = ?', $userId)
;
}
If you has a proper schema (I presume the schools should be a many-to-many association), then the current from should has a schools_list field (properly defined in the generated base from) and then you can modify that field to be rendered by sfWidgetFormSelectDoubleList:
$this->widgetSchema['schools_list']->setOption('renderer_class', 'sfWidgetFormSelectDoubleList');
$this->widgetSchema['schools_list']->setOption('renderer_options', array(
'label_unassociated' => 'Unassociated',
'label_associated' => 'Associated'
));
Does anyone know how to sort the MVCContrib grid when using a complex object.
My grid is displaying a list of Person and I'm trying to sort on the Country property. The problem is that Country is a property an Address class which is a property of Person.
Person.Address.Country
<%Html.Grid(Model).Columns(column =>
{
column.For(x => x.Id);
column.For(x => x.FirstName);
column.For(x => x.LastName).Sortable(false);
column.For(x => x.Address.Country).Sortable(false);
column.For(x => x.Age).Sortable(true);
}).Render(); %>
Exception:
Property 'Country' is not defined for type '{Namespace}.Person'
var sourceProp = Expression.Property(sourceParam, this.SortBy);
\MVCContrib\UI\Grid\Sortable\ComparableSortList.cs Line: 41
Any suggestions would be helpful.
Thank you,
MG1
A workaround would be to expose Country as a property on Person and use that:
public string Country { get { return Address.Country; } }
#orip gave you an answer.
But if you want to use the sorting feature you need to use:
<%Html.Grid(Model).Columns(column =>
{
column.For(x => x.Id);
column.For(x => x.FirstName);
column.For(x => x.LastName).Sortable(false);
column.For(x => x.Address.Country).Sortable(false);
column.For(x => x.Age).Sortable(true);
}).RenderUsing(new SortableHtmlTableGridRenderer<Person>())
.Render(); %>
Source: http://www.jeremyskinner.co.uk/2009/02/23/rewriting-the-mvccontrib-grid-part-3-gridmodels-and-gridrenderers/
You need to use SortColumnName for this.
column.For(x => x.Address.Country).SortColumnName("Address.Country");
I have tested this and it works like a charm :)
If you are not able to access SortColumnName(), you can get the latest version of MVC contrib from
http://mvccontrib.codeplex.com/SourceControl/changeset/changes/7db1cecc938f