Question

我有3个班级Persona，Alumno和factura。我需要使用gridview和搜索功能在factura视图上显示person fullname。全名在Persona中。 Factura只通过id_alumno与alumno相关，alumno通过id_persona与persona相关。我尝试通过创建一个新的变量fullname和一个函数来解决这个问题，以便在类factura上获得角色。

class Factura extends \yii\db\ActiveRecord{
     public $full_name;
}

public static function tableName()
{
    return 'factura';
}

public function rules()
{
return [
  [['id_factura_reemplazo', 'id_obra_social', 'id_alumno',], 'integer'],
  [['id_obra_social', 'id_alumno'], 'required'],
  [['fecha_factura','fullname','id_persona'], 'safe'],
];
}

public function attributeLabels()
{
return [
  'id_factura' => 'Id Factura',
  'id_factura_reemplazo' => 'Id Factura Reemplazo',
  'id_obra_social' => 'Id Obra Social',
  'id_alumno' => 'Id Alumno',
  'fullName'=>Yii::t('app', 'Nombre y Apellido'),
  ];
}

public function getIdAlumno()
{
    return $this->hasOne(Alumno::className(), ['id_alumno' => 'id_alumno']);
}

public function getIdPersona() {
   return $this->hasOne(Persona::className(), ['id_persona' => 'id_persona'])
   ->via('idAlumno');
}

public function getFullname(){
  if($this->idPersona)
     return $this->idPersona->nombre." ".$this->idPersona->apellido;
  return null;
}

我得到：无效的调用 - yii \ base \ InvalidCallException

设置只读属性：app \ models \ Facturasearch :: fullName

 if (method_exists($this, 'get' . $name)) {
        throw new InvalidCallException('Setting read-only property: ' .       get_class($this) . '::' . $name);
    } else {
        throw new UnknownPropertyException('Setting unknown property: ' . get_class($this) . '::' . $name);
    }

它就像它没有吸气剂......当它试图做$ this-＆gt; load（$ params）;

时问题从线上跳了出来

class Facturasearch extends Factura{    
public function rules()
{
    return [
        [['id_factura', 'id_factura_reemplazo',  'id_alumno', 'numero'], 'integer'],
        [['id_obra_social','fecha_factura', 'observacion','id_alumno','fullName','id_persona'], 'safe'],
    ];
}
public function scenarios()
{
    // bypass scenarios() implementation in the parent class
    return Model::scenarios();
}
public function search($params)
{
    $query = Factura::find();

    $dataProvider = new ActiveDataProvider([
        'query' => $query,
    ]);

    $this->load($params);
    ...

Answer 1

您应该在FacturaSearch中添加pubblic var以获取全名

class Facturasearch extends Factura{  

  public $fullname;


  public function rules()
  {
      return [
          [['id_factura', 'id_factura_reemplazo',  'id_alumno', 'numero'], 'integer'],
          [['id_obra_social','fecha_factura', 'observacion','id_alumno','fullName','id_persona'], 'safe'],
      ];
  }
  public function scenarios()
  {
      // bypass scenarios() implementation in the parent class
      return Model::scenarios();
  }
  public function search($params)
  {
      $query = Factura::find();

      $dataProvider = new ActiveDataProvider([
          'query' => $query,
      ]);

      $this->load($params);
      ...

如何从非直接相关的表中获取数据并在gridview上使用它与modelSearch（）Yii2

1 个答案: