我有一个简单的代码，该代码在Yii2中使用数据库事务，这将更新用户余额并将新记录添加到用户余额历史记录中。

//User model
public function changeBalance(UserBalanceHistory $balance)
{
    $balance->balance = $this->balance;
    $balance->user_id = $this->id;
    $this->balance    = $this->getBalance() + $balance->getDelta();

    $transaction = Yii::$app->db->beginTransaction();
    try {
        if ($balance->save() && $this->save()) {
            $transaction->commit();
            return true;
        }
    } catch (Exception $e) {
        Yii::error($e->getMessage());
    }

    $transaction->rollBack();
}

我应该经常使用数据库事务来保存数据完整性。但是像上面那样处理数据库事务需要很多代码行，因此我创建了以下函数来调动我的代码：

function dbTransaction(callable $callback)
{
    $transaction = Yii::$app->db->beginTransaction();

    try {
        //if callback returns true than commit transaction
        if (call_user_func($callback)) {
            $transaction->commit();
            Yii::trace('Transaction wrapper success');
        }
    } catch (\Exception $e) {
        $transaction->rollBack();
        throw $e;
    }
    $transaction->rollBack();
}

使用此功能，我可以处理如下交易：

//User …

我使用了Yii2多选下拉列表,它在创建时工作正常,但在更新时没有显示我选择的值!

形成:

       $form->field($model, 'categories[]')            
         ->dropDownList($model->CategoryDropdown,
         [
          'multiple'=>'multiple'
          'class'=>'chosen-select input-md required',              
         ]             
        )->label("Add Categories");    

模型:

public function getCategoryDropdown()
{
        $listCategory   = Category::find()->select('ID,name')
            ->where(['is_subcategory' => 'Yes'])
            ->andWhere(['status' => 'active','approved' => 'active'])
            ->all();
        $list   = ArrayHelper::map( $listCategory,'ID','name');

        return $list;
}

控制器:

 public function actionCreate(){
 ...
     $model->categories = implode(",",$_POST['Company']['categories']);
    ...
    return $this->render('create', [
            'model' => $model,           
        ]);
 }

public function actionUpdate($id)
{
    $model = $this->findModel($id);    

    echo $model->categories; //  1,2,4,5  values already assigned
    ...
    return $this->render('update', [
                'model' => $model,                    
            ]); 
  }

数据库:

1,2,4,5

当我更新我的recored时,如何在下拉列表中显示多个选定的值？

如何在Yii2中以一种形式使用多个模型？

我的情况:

在我的创建操作中,我可以保存到agenda_fiscalizacao表中,但在更新中,当我尝试加载表单时收到此错误:

Call to a member function formName() on array    

我的更新行动:

public function actionUpdate($id)
{
    $model = $this->findModel($id);
    $modelAgenda = AgendaFiscalizacao::findAll(['fiscalizacao_id' => $id]);

    if ($model->load(Yii::$app->request->post()) && Model::loadMultiple($modelAgenda, Yii::$app->request->post())) {
        $valid = $model->validate();
        $valid = $modelAgenda->validade() && $valid;

        if ($valid) {
            $model->save(false);
            $modelAgenda->save(false);
            return $this->redirect(['view', 'id' => $model->id]);
        }
    }

    return $this->render('update', [
        'model' => $model,
        'modelAgenda' => $modelAgenda
    ]);
}

我的表单视图

<?= $form->field($modelAgenda, 'agenda_id')->checkboxList(Agenda::combo(), ['class' => 'checkbox']) ?>
<?= $form->field($model, 'bioma_id')->dropDownList(Bioma::combo(), ['prompt' => $prompt]) ?>
<?= $form->field($model, …

基于https://github.com/wbraganca/yii2-dynamicform/wiki/Dynamic-Forms-With-Yii2-relation-trait-(VERY-EASY),我正在尝试实现动态表单.Create工作正常,但是在Update窗体中,如果我删除了任何动态表单元素,它就不会被删除,但如果我添加了Update操作,它就会被保存.这是我的更新代码

public function actionUpdate($id)
{
    $modelAlumni = $this->findModel($id);
    $modelsJob = $modelAlumni->jobs;


    if ($modelAlumni->loadAll(Yii::$app->request->post()) && $modelAlumni->saveAll()) {
        return $this->redirect(['view', 'id' => $modelAlumni->id]);
    } else {
        return $this->render('update', [
            'modelAlumni' => $modelAlumni,
            'modelsJob' => (empty($modelsJob)) ? [new Job] : $modelsJob
        ]);
   }
}

为什么不删除？

这是我的校友模特

<?php

namespace app\models;

use Yii;

/**
 * This is the model class for table "alumni".
 *
 * @property integer $id
 * @property string $name
 * @property integer $gender
 * @property integer $contact_number
 * @property string …

我可以在Yii2模型页面(http://www.yiiframework.com/doc-2.0/yii-base-model.html)中看到,在"字段"部分中,您可以根据某些字段设置"不同的字段列表"上下文信息.例如,根据$ scenario或当前应用程序用户的权限,您可以返回不同的可见字段集或过滤掉某些字段."

但是,场景文档(http://www.yiiframework.com/doc-2.0/guide-structure-models.html#scenarios)表示场景用于为模型属性验证创建不同的上下文.

我正在使用Yii2 Restful API,我必须使用默认操作(actionIndex,actionView,...)从模型中获取数据并显示为API结果.我知道我可以覆盖那些方法(http://www.yiiframework.com/doc-2.0/guide-rest-controllers.html#extending-active-controller),但我怎么能说在那些方法中使用不同的方法字段(取决于不同的场景)？

我需要的是为actionIndex(项目列表)输出field1,field2,field3,但我想为actionView(项目列表)输出field1,field2,field3,field4.

我需要在GridView中对某些字段(asc,desc)进行排序,但计算相同的字段.看下面的代码: SearchModel:

class ObjectSearch extends Object {
use SearchModelTrait;

public function rules()
{
    return [
        ['id', 'integer', 'min' => 1],
    ];
}

public function search($params)
{
    $this->company_id = \Yii::$app->user->identity->companyId;
    $query = Object::find()->where(['company_id' => $this->company_id]);
    $dataProvider = new ActiveDataProvider([
        'query' => $query,
        'pagination' => false,
    ]);
    $dataProvider->setSort([
        'attributes' => [
            'id',
            'name',
            'lastReportResult' => [
                'asc' => ['lastReportResult' =>SORT_ASC ],
                'desc' => ['lastReportResult' => SORT_DESC],
                'default' => SORT_ASC
            ],
            'reportPercentDiff'
        ]
    ]);

    if (!($this->load($params,'ObjectSearch') && $this->validate())) {
        return $dataProvider;
    }

    $this->addCondition($query, …

我在主查询中有一个子查询，如下所示：

$subquery = (new Query())->select('COUNT(project_sup_id)')
    ->from('project_supervisors AS sup')
    ->where(['AND','sup.project_ref_id = p.project_id']);

$uQuery =(new Query())->select(['project_id','supcount' => $subquery])
    ->from('projects AS p')
    ->join('LEFT JOIN','project_supervisors AS sup','sup.project_ref_id = project_id')
    ->andWhere('IF(supcount>0, sup_project_status=1, project_status=1)');

我正在尝试根据逻辑编写 where 条件，即，如果从子查询获得的计数大于零，则 where 条件必须为sup_project_status=1else project_status=1。在普通的Mysql中，很容易在where里面写if条件，但在yii2中我不明白如何写。

我试图从RestoFoods模型中删除数据:

RestoFoods::deleteAll(["restaurant_id"=>$postData['resto_id'], 'food_id NOT IN'=> [1,2] ]);

我想要这个sql:

DELETE FROM `resto_foods` WHERE `restaurant_id`=1 AND (`food_id` NOT IN (1, 2));

如何在 Yii 2.0 中使用多对多关系过滤连接/连接表中的值的结果？

我有以下表格

Member (id INT, name, ...)
Event (id INT, name, date, ...)
EventMemberConnection (id INT, event_id INT, member_id INT, accepted TINYINT(1) DEFAULT 0, foreign keys....)

成员通过 EventMemberConnection 连接到事件，用一个整数告诉他们是被接受还是被拒绝

从模型文件 Event.php：

/**
 * @return \yii\db\ActiveQuery
 */
public function getAcceptedMembers() {
    return $this->hasMany(Member::className(), ['id' => 'member_id'])->viaTable('EventMemberConnection acceptedConnection', ['event_id' => 'id'])->onCondition(['acceptedConnection.accepted' => 1]);
}

/**
 * @return \yii\db\ActiveQuery
 */
public function getDeclinedMembers() {
    return $this->hasMany(Member::className(), ['id' => 'member_id'])->viaTable('EventMemberConnection declinedConnection', ['event_id' => 'id'])->onCondition(['declinedConnection.accepted' => 0]);
}

和查询，在休息控制器内：

public function prepareDataProvider() …

您可以在 yii2 的模型规则中输入密码模式吗？关于至少一个大写字符和至少一个数字的规则的提示？非常感谢

规则

['password', 'pattern' => '(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%]).{6,20}'],