【发布时间】:2016-03-25 04:07:25
【问题描述】:
我需要具有给定属性 ID 的用户,例如Property.id = $id
我想构建这个查询。
$sql = "SELECT p.address1, p.address2, p.address3, p.postcode,
u.email, u.fullname
c.company_name,
tc.created,
tn.active
FROM property AS p
LEFT JOIN tenancy AS tc ON tc.property_id = p.id
LEFT JOIN tenant AS tn ON tn.tenancy_id = tc.id
LEFT JOIN users AS u ON u.id = tn.user_id
WHERE p.id = $id
AND p.active = 1
AND tc.active = 1
AND tn.active = 1
AND u.active = 1
";
这是我的联想:
// Tenants
class TenantTable extends Table
{
public function initialize(array $config)
{
$this->table('tenant');
$this->displayField('id');
$this->primaryKey('id');
$this->addBehavior('Timestamp');
$this->belongsTo('Users', [
'foreignKey' => 'user_id',
'joinType' => 'INNER'
]);
$this->belongsTo('Tenancies', [
'foreignKey' => 'tenancy_id',
'className' => 'tenancy',
'joinType' => 'INNER'
]);
// Users
class UsersTable extends Table
{
public function initialize(array $config)
{
$this->table('users');
$this->displayField('name');
$this->hasMany('Tenants', [
'foreignKey' => 'user_id',
'foreignKey' => 'tenants'
]);
}
// Tenancies
class TenancyTable extends Table
{
public function initialize(array $config)
{
parent::initialize($config);
$this->table('tenancy');
$this->displayField('id');
$this->primaryKey('id');
$this->addBehavior('Timestamp');
$this->belongsTo('Properies', [
'foreignKey' => 'property_id',
'className' => 'property'
]);
$this->belongsTo('Company', [
'foreignKey' => 'company_id'
]);
$this->hasMany('Tenant', [
'foreignKey' => 'tenancy_id',
'className' => 'tenant'
]);
// Properties
class PropertyTable extends Table
{
public function initialize(array $config)
{
parent::initialize($config);
$this->table('property');
$this->displayField('id');
$this->primaryKey('id');
$this->addBehavior('Timestamp');
$this->belongsTo('Company', [
'foreignKey' => 'company_id'
]);
$this->hasMany('Tenancies', [
'foreignKey' => 'property_id',
'className' => 'tenancy'
]);
}
我可以获取 hasMany 关联,但在获取用户的 belongsTo 时,它告诉我租户与用户没有关联。关联是如何设置的并且它们是正确的。我正在从 PropertyTable 运行以下 ORM。
$query = $this->find()
->select([
'Property.id', 'Property.company_id', 'Property.address1', 'Property.address2', 'Property.address3','Property.postcode',
'Tenancies.id', 'Tenancies.property_id', 'Tenancies.created', 'Tenancies.stage', 'Tenancies.landlord_offer_sent',
'Company.id', 'Company.company_name',
'Tenants.id', 'Tenants.user_id', 'Tenants.stage',
'Users.id', 'Users.email', 'Users.fullname'
])
->where(['Property.id' => $id])
->contain(['Company', 'Tenancies'])
->leftJoinWith('Tenancies', function(\Cake\ORM\Query $query) {
return $query->where([
'Tenancies.active' => 1,
]);
})
->contain(['Tenancies.Tenants'])
->leftJoinWith('Tenancies.Tenants', function(\Cake\ORM\Query $query) {
return $query->where([
'Tenants.active' => 1,
]);
})
/*======= The problem starts from here =========*/
->contain(['Tenancies.Tenants.Users'])
->leftJoinWith('Tenancies.Tenants.Users', function(\Cake\ORM\Query $query) {
return $query->where([
'Users.active' => 1,
]);
});
请帮忙
【问题讨论】:
-
仅仅因为您认为关联是正确的,并不意味着这一定是正确的。他们可能是正确的,也可能不是,这里的人不看他们就无法分辨。在编程的世界里,文字不值钱,代码才是最重要的,所以请始终包含所有相关代码!另外请分享您的调试尝试(例如您是否尝试过删除其他包含/加入?通过租户表包含/加入用户是否也会失败?等等...)。最后但并非最不重要的一点是,请始终提及您的确切 CakePHP 版本。
-
谢谢你,伙计。当然我会更新答案
-
@ndm 关联已更新
标签: cakephp cakephp-3.x