2017-05-09 75 views
0

我在CakePHP的3.4组通过在CakePHP的相关模型

工作,我有两个型号skillsskill_categories和他们的协会都像

skill_categories->hasMany('skills'); 
skills->belongsTo('SkillCategories', 'joinType' => 'INNER') 

skills正在跟users其中

关联
skills->belongsTo('users') 
users->hasMany('Skills') 

我必须从中选择用户的所有关联skills通过表分组,这将导致产生像

Skill Category 1 
|-- Skill 11 
|-- Skill 12 
|-- Skill 13 
Skill Category 2 
|-- Skill 21 
|-- Skill 22 

或类似

'skill_categories' => [ 
     'title' => 'Skill Category 1', 
     'id' => 1, 
     'skills' => [ 
      0 => [ 
       'title' => 'Skill 11', 
       'id' => 4, 
      ], 
      1 => [ 
       'title' => 'Skill 12', 
       'id' => 6, 
      ] 
     ], 
] 

我在做什么是:方法1

$user_skills = $this->Skills->find() 
     ->select(['Skills.skill_category_id', 'Skills.title', 'Skills.measure', 'SkillCategories.title', 'Skills.id']) 
     ->where(['Skills.user_id' => $user->id, 'Skills.deleted' => false, 'Skills.status' => 0]) 
     ->contain(['SkillCategories']) 
     ->group(['Skills.skill_category_id']); 

     foreach($user_skills as $s)debug($s); 

但这投掷误差

Error: SQLSTATE[42000]: Syntax error or access violation: 1055 Expression #2 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'profPlus_db_new.Skills.title' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by

方法2

$user_skills = $this->SkillCategories->find() 
     ->where(['Skills.user_id' => $user->id, 'Skills.deleted' => false, 'Skills.status' => 0]) 
     ->contain(['Skills']) 
     ->group(['SkillCategories.id']); 

     foreach($user_skills as $s)debug($s); 

但是这给了错误的

Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'Skills.user_id' in 'where clause'


EDIT 2

技能架构

CREATE TABLE IF NOT EXISTS `skills` (
    `id` CHAR(36) NOT NULL, 
    `user_id` CHAR(36) NOT NULL, 
    `skill_category_id` CHAR(36) NOT NULL, 
    `title` VARCHAR(250) NOT NULL, 
    `measure` INT NOT NULL DEFAULT 0, 
    `status` INT NULL DEFAULT 0, 
    `deleted` TINYINT(1) NULL DEFAULT 0, 
    `created` TIMESTAMP NULL DEFAULT CURRENT_TIMESTAMP, 
    `modified` DATETIME NULL, 
    PRIMARY KEY (`id`) 
) 

skill_categories架构

CREATE TABLE IF NOT EXISTS `skill_categories` (
    `id` CHAR(36) NOT NULL, 
    `title` VARCHAR(200) NOT NULL, 
    `status` INT NULL DEFAULT 0, 
    `deleted` TINYINT(1) NULL DEFAULT 0, 
    `created` TIMESTAMP NULL DEFAULT CURRENT_TIMESTAMP, 
    `modified` DATETIME NULL, 
    PRIMARY KEY (`id`)) 

控制器代码

$user_skills = $this->SkillCategories->find() 
    ->contain([ 
     'Skills' => function($q) use($user) { 
      return $q 
      ->select(['id', 'skill_category_id', 'title', 'measure', 'user_id', 'deleted', 'status']) 
      ->where(['user_id' => $user->id, 'deleted' => false, 'status' => 0]); 
    }]) 
    ->group(['SkillCategories.id']); 

    foreach($user_skills as $s)debug($s); 

调试输出

object(App\Model\Entity\SkillCategory) { 

'id' => '581cd4ac-28a7-4016-b535-b34a27d47c0d', 
'title' => 'Programming', 
'status' => (int) 0, 
'deleted' => false, 
'skills' => [ 
    (int) 0 => object(App\Model\Entity\Skill) { 

     'id' => '36f16f7f-b484-4fd8-bfc5-4408ce97ff23', 
     'skill_category_id' => '581cd4ac-28a7-4016-b535-b34a27d47c0d', 
     'title' => 'PHP', 
     'measure' => (int) 92, 
     'user_id' => '824fbcef-cba8-419e-8215-547bd5d128ad', 
     'deleted' => false, 
     'status' => (int) 0, 
     '[repository]' => 'Skills' 

    }, 
    (int) 1 => object(App\Model\Entity\Skill) { 

     'id' => '4927e7c1-826a-405d-adbe-e8c084c2b9ef', 
     'skill_category_id' => '581cd4ac-28a7-4016-b535-b34a27d47c0d', 
     'title' => 'CakePHP', 
     'measure' => (int) 90, 
     'user_id' => '824fbcef-cba8-419e-8215-547bd5d128ad', 
     'deleted' => false, 
     'status' => (int) 0, 
     '[repository]' => 'Skills' 
    } 
], 
'[repository]' => 'SkillCategories' 

} 

object(App\Model\Entity\SkillCategory) { 

'id' => 'd55a2a95-05a0-410e-9a55-1a1509f76b8c', 
'title' => 'Office', 
'status' => (int) 0, 
'deleted' => false, 
'skills' => [], 
'[repository]' => 'SkillCategories' 

} 

注意:请参阅第2物体与标题办公室没有与用户

+0

您是否试图将条件传递给包含?(''id','user_id'); $'user_skills = $ this-> SkillCategories-> find() - > contains(['Skills'= function($ q)use($ user){return $ q-> select ,'deleted','status']) - > where(['user_id'=> $ user-> id,'deleted'=> false,'status'=> 0]);}]) - > group([ 'SkillCategories.id']);' – chrisShick

+0

谢谢,它完成了这项工作。作出回答,以便我可以将其标记为已接受 –

+0

我只是将它作为答案:D – chrisShick

回答

1

相关skills您可以通过条件进入包含! :D

$user_skills = $this->SkillCategories->find() 
    ->contain(['Skills' => function($q) use($user) { 
     return $q 
     ->select(['id', 'user_id', 'deleted', 'status', 'skill_category_id']) 
     ->where(['user_id' => $user->id, 'deleted' => false, 'status' => 0]); 
    }]) 
    ->group(['SkillCategories.id']); 
+0

嘿,这个脚本有一个问题。即使用户未填写或未与特定用户关联,它也会列出所有“技能类别”。防爆。如果有50个技能类别,并且用户只有10个技能类别相关的技能,那么即使这样,它也会列出其中40个无用的所有50个技能类别。 –

+0

@AnujTBE你确定吗?因为我运行了这个确切的查询,它适用于我。 – chrisShick

+0

是的,我很确定。我从管理面板添加了2个技能类别。然后在一个技能类别中添加2个技能,使用演示用户面板留下第二个类别。当我执行这个脚本时,它会显示两个类别。 –