php - Yii2 QueryBuilder 更新与加入

Question

我有以下原始 SQL 查询:

UPDATE `pay_audit`
JOIN `invoice_items`
ON `invoice_items`.`mdn` = `pay_audit`.`account_id` 
AND `invoice_items`.`unitprice` = `pay_audit`.`payment`
AND `invoice_items`.`producttype_name` LIKE 'PAYMENT'
AND DATE_FORMAT(`invoice_items`.`created`, '%Y-%m-%d') = '2015-02-21'
SET `pay_audit`.`invoice_item_id` = `invoice_items`.`id`
WHERE `pay_audit`.`report_date` = '2015-02-21'

日期是php中的变量$date。

我如何将这个原始 SQL 查询“转换”为 Yii2 QueryBuilder？

[更新]

正如 Felipe 提到的那样，查询构建器是不可能的，所以我最终按照以下方式进行操作:

    $today = date('Y-m-d');
    $sql = "";
    $sql .= "UPDATE `pay_audit` ";
    $sql .= "JOIN `invoice_items` ";
    $sql .= "ON `invoice_items`.`mdn` = `pay_audit`.`account_id` ";
    $sql .= "AND `invoice_items`.`unitprice` = `qpay_audit`.`payment` ";
    $sql .= "AND `invoice_items`.`producttype_name` LIKE 'PAYMENT' ";
    $sql .= "AND DATE_FORMAT(`invoice_items`.`created`, '%Y-%m-%d') = '$today' ";
    $sql .= "SET `pay_audit`.`invoice_item_id` = `invoice_items`.`id` ";
    $sql .= "WHERE `pay_audit`.`report_date` = '$today'";

    $command = \Yii::$app->db->createCommand($sql);
    $command->execute();

Answer 1

我害怕Yii 2 Query Builder仅适用于选择查询。

对于更新查询，您至少有三个选项:

• 原始 SQL:

  \Yii::$app->db->createCommand('update user set status = 1 where age > 30')->execute();
  

• 带占位符的原始 SQL(以防止 SQL 注入(inject))

  \Yii::$app->db->createCommand('update user set status = :status where age > 30')->bindValue(':status','1')->execute();
  

• 更新() 方法

  // update user set status = 1 where age > 30
  \Yii::$app->db->createCommand()->update('user', ['status' => 1], 'age > 30')->execute();
  

更多信息在这里:

• Yii DAO

• Another question on SO