object $DB_DataObject->tableName (
string $name
)
引数なしの場合、オブジェクトが扱うテーブル名を返します。 文字列を渡した場合、そのオブジェクトが扱うテーブル名をセットします。
This function can not be called statically.
テーブル名の取得と設定
<?php
$person = new DataObjects_Person;
echo $person->tableName();
// echo's person
// now use the same object to query the extra_people table
$person->tableName('extra_people');
$person->id = 12;
$person->find(true);
// you can also use this in conjunction with table(), to create dataobjects for random tables..
$d = new DB_DataObject;
$d->tableName('person');
$d->table(array(
'id' => DB_DATAOBJECT_INT,
'name' => DB_DATAOBJECT_STRING,
));
$d->keys(array('id'));
$d->id = 12;
$d->find(true);
// should do the same as above..!
?>