Mysql 在CActiveDataProvider上使用4表进行yii查询

嗨，主人，我有四张桌子

-佩加维 -佩尼莱安 -大学 -朱鲁桑

我通常使用

    SELECT  u.nama_univ, j.singkatan, peg.* FROM pegawai AS peg LEFT JOIN penilaian AS pen ON pen.no_test=peg.no_test LEFT JOIN universitas AS u ON u.id=peg.univ_s1 LEFT JOIN jurusan AS j ON j.id=peg.bidang_s1

如何将此查询转换为yii

我试过这个代码，但效果不好

$dataProvider = new CActiveDataProvider('Pegawai', array(
    'criteria' => array(
        'select' => array(
            '`pen`.*',
            '`u`.`nama_univ` AS `nama_univ`',
            '`j`.`singkatan` AS `singkatan`'
        ),
        'join' => 'JOIN `Penilaian` AS `pen` ON `pen`.`no_test` = `t`.`no_test`',
        'join' => 'JOIN `Universitas` AS `u` ON `u`.`id` = `t`.`no_test`',
        'join' => 'JOIN `Jurusan` AS `j` ON `j`.`id` = `t`.`no_test`',
    )
));

---

但是只有一个连接有执行

做一件事你只需要像这样打印criteria对象

$dataProvider = new CActiveDataProvider('Pegawai', array(
    'criteria' => array(
        'select' => array(
            '`pen`.*',
            '`u`.`nama_univ` AS `nama_univ`',
            '`j`.`singkatan` AS `singkatan`'
        ),
        'join' => 'JOIN `Penilaian` AS `pen` ON `pen`.`no_test` = `t`.`no_test`',
        'join' => 'JOIN `Universitas` AS `u` ON `u`.`id` = `t`.`no_test`',
        'join' => 'JOIN `Jurusan` AS `j` ON `j`.`id` = `t`.`no_test`',
    )
));
echo "<pre>";
print_r($dataProvider);   //Check the object or try to post it below
exit();

---

您已为数组“join”的同一索引分配了3个值，因此只存储最后一个值

尝试：

---

你们为你们的桌子做了模型吗？是的，我做了。但是我把数组关系搞混了你试过什么？人们帮助你编写代码，他们通常不会为你编写代码。嗨，Mihai，我编辑了我的帖子。请检查一下

$dataProvider = new CActiveDataProvider('Pegawai', array(
'criteria' => array(
    'select' => array(
        '`pen`.*',
        '`u`.`nama_univ` AS `nama_univ`',
        '`j`.`singkatan` AS `singkatan`'
    ),
    'join' => 'JOIN `Penilaian` AS `pen` ON `pen`.`no_test` = `t`.`no_test` JOIN `Universitas` AS `u` ON `u`.`id` = `t`.`no_test` JOIN `Jurusan` AS `j` ON `j`.`id` = `t`.`no_test`',
)
));