我有一个名称,姓氏,电子邮件,以前名字的用户模型。还有AdditionalEmail模型,我为用户存储其他电子邮件。每个用户可以有多个额外的电子邮件。每个模型中都有关系hasMany和belongsToMany。 我必须搜索姓氏,姓名,电子邮件和附加电子邮件。我尝试使用可包含的行为并为查询创建了下表:
$options['joins'] = array(
array(
'table' => 'users',
'alias' => 'User',
'type' => 'LEFT',
'conditions' => array('User.id = AdditionalEmail.user_id')
));
$options['conditions'] = array('OR' => array(
'User.surname LIKE' => $search,
'User.former_surnames LIKE' => $search,
'User.email LIKE' => $search,
'AdditionalEmail.email LIKE' => $search));
$options['recursive'] = -1;
$options['fields'] = 'DISTINCT User.username, User.name, User.surname, User.former_surnames';
$options['order'] = array('User.surname' => 'asc');
以上,但对我不起作用。我终于得到了查询:
SELECT DISTINCT users.* FROM users LEFT JOIN additional_emails ON ( users.id = additional_emails.user_id ) WHERE ((additional_emails.email LIKE \''. $search .'\') OR (users.email LIKE \''. $search .'\') OR (users.surname LIKE \''. $search .'\') OR (users.former_surnames LIKE \''. $search .'\')) ORDER BY users.surname
哪个有效。如何使用cakephp选项为find(或paginate)构建此查询。 问候 彼得
答案 0 :(得分:0)
$this->User->find('all',
array(
'fields' => array('DISTINCT *'),
'joins' => array(
array(
'table' => 'additional_emails',
'alias' => 'Email',
'type' => 'left',
'conditions' => array('User.id = Email.user_id')
),
),
'conditions' => array(
'OR' => array(
'Email.email LIKE' => $search,
'User.email LIKE' => $search,
'User.surname LIKE' => $search,
'User.former_surnames LIKE' => $search,
)
),
'order' => array('User.surname'),
)
);