Question

如何在DataExtension中检索原始类的类名（字符串）？

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  var html = [];
  $.each( data, function( index, hees ) {
      $.each( hees, function( att, val ) {
        html.push( "<li id='" + att + "'>" + val + "</li>" );
      });
  });

  $( "<ul/>", {
    "class": "my-new-list",
    html: html.join( "" )
  }).appendTo( "#locations" );
});

因此，例如，当将此扩展应用于class FooExtension extends DataExtension { // This returns "FooExtension" (but I need the original-class name, not the extension's) private $foo1 = self::class; // This returns nothing public function Foo2 { return $this->owner; } // This returns "FooExtension", surprisingly public function Foo3 { return get_class($this->owner); } // This returns nothing either public function Foo4 { return $this->owner->class; } }时，我想在扩展类的某处返回或分配 Page （字符串）。

Answer 1

您可以通过在DataExtension中调用$this->owner->ClassName来获取所有者的课程。

SilverStripe通过DataExtension检索原始类名

1 个答案: