我有4张桌子相互连接
人才表
+--------------------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| created | datetime | YES | | NULL | |
| user_id | int(10) unsigned | NO | | NULL | |
| firstname | varchar(128) | NO | | NULL | |
| lastname | varchar(128) | NO | | NULL | |
| phone_num | varchar(32) | NO | | NULL | |
+--------------------+------------------+------+-----+---------+----------------+
此表格将包含
等行+----+-----------+------------+
| id | firstname | lastname |
+----+-----------+------------+
| 1 | barney | stinson |
| 2 | Ted | Mosby |
+----+-----------+------------+
TalentCategory表
+----------------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+----------------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| created | datetime | NO | | NULL | |
| talent_id | int(10) unsigned | NO | | NULL | |
| talent_name_id | int(11) | NO | | NULL | |
| is_active | tinyint(1) | NO | | 1 | |
+----------------+------------------+------+-----+---------+----------------+
TalentName表
+--------------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| created | date | NO | | NULL | |
| name | varchar(128) | NO | | NULL | |
| slug | varchar(255) | NO | | NULL | |
| talent_count | int(11) | NO | | NULL | |
+--------------+------------------+------+-----+---------+----------------+
此表格将包含
等行+----+-------------------+-----------------+
| id | name | slug |
+----+-------------------+-----------------+
| 1 | actor / actress | actor-actress |
| 2 | dancer | dancer |
| 3 | model | model |
| 4 | singer / musician | singer-musician |
+----+-------------------+-----------------+
和TalentMedia表
+--------------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| created | datetime | NO | | NULL | |
| talent_id | int(10) unsigned | NO | | NULL | |
| media_id | int(10) unsigned | NO | | NULL | |
| is_cover | tinyint(1) | NO | | 0 | |
| is_avatar | tinyint(1) | NO | | 0 | |
| like_count | int(11) | NO | | 0 | |
| view_count | int(11) | NO | | 0 | |
| is_published | tinyint(1) | NO | | 0 | |
| is_deleted | tinyint(1) | NO | | 0 | |
| is_approved | tinyint(1) | NO | | 0 | |
| is_suspended | tinyint(1) | NO | | 0 | |
+--------------+------------------+------+-----+---------+----------------+
人才hasMany TalentCategory belongsTo TalentName
人才hasMany TalentMedia
我正在努力实现
SELECT
Talent.id,
Talent.firstname,
Talent.lastname,
TalentCategory.id,
TalentCategory.talent_id,
TalentCategory.talent_name_id,
TalentName.name,
TalentName.id,
TalentMedia.talent_id,
TalentMedia.media_id,
TalentMedia.is_suspended,
TalentMedia.is_avatar,
TalentMedia.is_cover
FROM talents AS Talent
JOIN talent_talents AS TalentCategory ON TalentCategory.talent_id = Talent.id
JOIN talent_names AS TalentName ON TalentName.id = TalentCategory.talent_name_id
JOIN talent_medias AS TalentMedia ON TalentMedia.talent_id = Talent.id
WHERE TalentName.id = 4 AND TalentMedia.is_suspended != 1 AND TalentMedia.is_cover !=1 AND TalentMedia.is_avatar = 1
GROUP BY Talent.id
或
select all talents which is a singer/musician that avatar is not suspended
这是描述所需输出的SqlFiddle
来自我的控制器，以便我可以在我的paginator设置中实现它。我一直在尝试一切，没有运气。
我尝试custom find types或custom query pagination，但我不太了解文档。
请帮我解决如何实现这个目标
答案 0 :(得分：0)
我设法通过创建自定义分页来实现这一点，我不知道这是否是正确的方法，至少我的解决方案现在适用
首先我覆盖Cake的paginate和paginateCount方法
public function paginate($conditions, $fields, $order, $limit, $page = 1, $recursive = null, $extra = array()) {
$recursive = -1;
$sql_query = $this->paginateQuery; //same query as [SqlFiddle][1]
if(!empty($conditions)){
$sql_query .= 'WHERE ';
foreach ($conditions as $key => $cond) {
$cond_str[] = $key .' '. $cond;
}
$sql_query .=implode(" AND ", $cond_str)." ";
}
$sql_query .= "GROUP BY Talent.id ";
$sql_query .= "ORDER BY Talent.id DESC ";
$sql_query .= "LIMIT " . (($page - 1) * $limit) . ', ' . $limit;
$results = $this->query($sql_query);
$this->virtualFields['fullname'] = 'CONCAT(Talent.firstname, " ", Talent.lastname)';
return $results;
}
public function paginateCount($conditions = null, $recursive = 0, $extra = array()) {
$recursive = -1;
$sql_query = $this->paginateQuery; //same query as [SqlFiddle][1]
if(!empty($conditions)){
$sql_query .= 'WHERE ';
foreach ($conditions as $key => $cond) {
$cond_str[] = $key .' '. $cond;
}
$sql_query .=implode(" AND ", $cond_str)." ";
}
$sql_query .= "GROUP BY Talent.id ";
$sql_query .= "ORDER BY Talent.id DESC ";
$this->recursive = $recursive;
$this->virtualFields['fullname'] = 'CONCAT(Talent.firstname, " ", Talent.lastname)';
$results = $this->query($sql_query);
return count($results);
}
$this->paginateQuery;上的值与SqlFiddle
我还需要改变$conditions，因为它是一个数组，所以我需要以某种方式将数组连接到一些可接受的条件。
接下来在我的控制器中，我打电话给
$this->Paginator->settings = array(
'Talent' => array(
'limit' => 1,
'conditions' => array(
'TalentCategory = ' => 3,
'TalentMedia.is_suspended !=' => 1,
'TalentMedia.is_cover !=' => 1,
'TalentMedia.is_avatar = ' => 1,
//and any other conditions related to the joined table
)
)
);
$this->set('talents', $this->Paginator->paginate('Talent'));