Question

使用带有外键关系的下拉框我需要一些帮助。我有下拉填充正确的值，但唯一的问题是当我添加用户有一个外键约束。但是如果我只使用普通输入框并输入另一个表中存在的id，我可以创建用户。例如，当我在add.ctp中输入id时，它可以正常工作

echo $this->Form->input('location');

但是当我使用它时它不会

echo $this->Form->input('location_id', array('type' => 'select', 'options' => $CompanyLocations));

这是我在UsersController中的添加功能

public function add()
{
$user = $this->Users->newEntity();
if ($this->request->is('post')) {
$user = $this->Users->patchEntity($user, $this->request->data);
if ($this->Users->save($user)) {
$this->Flash->success(__('The user has been saved.'));

            return $this->redirect(['action' => 'index']);
        }
        $this->Flash->error(__('The user could not be saved. Please, try again.'));
    }
    $CompanyLocations= $this->Users->CompanyLocations->find('list'); 
    $this->set(compact('CompanyLocations'));
    $this->set(compact('user'));
    $this->set('_serialize', ['user']);

这是在我的UsersTable

中

    $this->belongsTo('CompanyLocations');

和我的CompanyLocationsTable

public function initialize(array $config)
{
parent::initialize($config);

    $this->table('company_locations');
    $this->displayField('location_name');
    $this->primaryKey('location_id');

    $this->belongsTo('Locations', [
        'foreignKey' => 'location_id',
        'joinType' => 'INNER'
    ]);
}

和我的mysql代码

CREATE TABLE IF NOT EXISTS southpac_team.company_locations (
location_id INT NOT NULL AUTO_INCREMENT,
location_name VARCHAR(45) NULL,
PRIMARY KEY (location_id))
ENGINE = InnoDB;

DROP TABLE IF EXISTS southpac_team.users ;

CREATE TABLE IF NOT EXISTS southpac_team.users (
id INT NOT NULL AUTO_INCREMENT,
username VARCHAR(20) NOT NULL,
password VARCHAR(255) NOT NULL,
name VARCHAR(255) NOT NULL,
department INT NULL,
mobile VARCHAR(255) NULL,
email VARCHAR(255) NULL,
extension INT NULL,
lame_number INT NULL,
spa_auth_number VARCHAR(15) NULL,
creation_date DATE NULL,
picture VARCHAR(255) NULL,
employed TINYINT(1) NOT NULL,
location INT NOT NULL,
PRIMARY KEY (id),
INDEX get location_idx (location ASC),
CONSTRAINT get location
FOREIGN KEY (location)
REFERENCES southpac_team.company_locations(location_id)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;

Answer 1

命名约定

您没有遵循命名约定，默认情况下，belongsTo关联的外键名称是关联别名的单数下划线变体，后缀为_id，因此在{{1}的情况下这将是CompanyLocations，而不仅仅是company_location_id。

location

持有列表的变量也应该使用驼峰外壳，那么你甚至不需要通过options参数指定它：

echo $this->Form->input('company_location_id');

更改关联默认值

如果您正在使用无法修改的旧数据库，则需要相应地配置CakePHP，即通过$companyLocations= $this->Users->CompanyLocations->find('list'); $this->set(compact('companyLocations'));的options参数指定自定义外键。

Table::belongsTo()

Bake感到困惑

$this->belongsTo('CompanyLocations', [ 'foreignKey' => 'location' ]);中的belongsTo关联看起来也很可疑，除非你真的有一个CompanyLocationsTable与LocationsTable相关联：

CompanyLocationsTable

我猜你已经通过bake创建了模型，它将company_locations.location_id > locations.primary_key视为外键，因为它匹配location_id关联的默认外键命名方案。

另见

的 Cookbook > CakePHP at a Glance > Conventions > Model and Database Conventions

的 Cookbook > Database Access & ORM > Associations - Linking Tables Together > BelongsTo Associations

的 Cookbook > Views > Helpers > Form > Creating Form Controls

的 Cookbook > Views > Helpers > Form > Creating Inputs for Associated Data

CakePHP 3使用Drop down时的外键约束

1 个答案:

命名约定

更改关联默认值

Bake感到困惑

另见