我有2个表:User和Posts。用户可以有多个帖子,帖子不能有多个用户。如何在模型中建立关系以及如何在ActiveDataProvider中加入我在我的Posts表中有user_id,并希望在我的gridview中显示数据,如Posts(id,title,text)和User(name)我该如何做到这一点?我需要在我的模型中建立关系我该如何使用它?文章模型:
<?php
namespace app'models;
use Yii;
/**
* This is the model class for table "posts".
*
* @property integer $id
* @property integer $user_id
* @property string $post_title
* @property string $post_text
*/
class Posts extends 'yii'db'ActiveRecord
{
/**
* @inheritdoc
*/
public static function tableName()
{
return 'posts';
}
/**
* @inheritdoc
*/
public function rules()
{
return [
[['user_id'], 'integer'],
[['post_title'], 'string', 'max' => 50],
[['post_text'], 'string', 'max' => 255],
];
}
/**
* @inheritdoc
*/
public function attributeLabels()
{
return [
'id' => 'ID',
'user_id' => 'User ID',
'post_title' => 'Post Title',
'post_text' => 'Post Text',
];
}
public function insertPost()
{
$userId = 'Yii::$app->user->identity->id;
$posts = new Posts();
$posts->user_id = $userId;
$posts->post_title = $this->post_title;
$posts->post_text = $this->post_text;
return $posts->save();
}
public function getUser()
{
return $this->hasOne(User::classname(),['user_id'=>'id']);
}
}
用户模式:* @property integer $id
* @property string $email
* @property string $password
* @property string $name
*/
class User extends 'yii'db'ActiveRecord
{
/**
* @inheritdoc
*/
public static function tableName()
{
return 'user';
}
/**
* @inheritdoc
*/
public function rules()
{
return [
[['email'], 'string', 'max' => 100],
[['password'], 'string', 'max' => 255],
[['name'], 'string', 'max' => 25],
];
}
/**
* @inheritdoc
*/
public function attributeLabels()
{
return [
'id' => 'ID',
'email' => 'Email',
'password' => 'Password',
'name' => 'Name',
];
}
public function setPassword($password)
{
$this->password = sha1($password);
}
public function validatePassword($password)
{
return $this->password === sha1($password);
}
public static function findIdentity($id)
{
return self::findOne($id);
}
public static function findIdentityByAccessToken($token, $type = null)
{
}
public function getId()
{
return $this->id;
}
public function getAuthKey()
{
}
public function validateAuthKey($authKey)
{
}
public function getPost()
{
return $this->hasMany(Posts::classname(),['id'=>'user_id']);
}
}
在User模型中已经有了User和Post之间的关系(函数getPost)
可以访问Post的值,例如:
$userModel = User::find()->where([ 'id' => $id])->one();
$myUserPost = $userModel->post;
$myUserPostAttribute = $userModel->post->attribute;
对于ActiveDataProvider,您可以使用
$dataProvider = User::find()->where([ 'id' => $id]);
并最终在用户模型中为单个属性添加getter,例如:
getMyPostAttribute1()
{
return $this->post->attribute1
}
所以你可以很容易地在gridview
中使用这个getter <?= GridView::widget([
'dataProvider' => $dataProvider,
......
'columns' => [
['class' => 'yii'grid'SerialColumn'],
'myPostAttribute1',
....