Yii：在加入之前执行where条件

问题描述

我目前的查询有问题

public class Game {
    private Scanner input = new Scanner(system.in); // Scanner reading from system.in
    private int rounds;    // 5-30
    private int playerAmount; // Indicating how many players take part
    private ArrayList<Player> players = new ArrayList<>(); // Storing player instances

    public Game() {
        System.out.println("Welcome to The Farm! Please enter how many will be playing today (1-4 players) by entering your names. When finished please enter menu to begin");

        for(int i = 0; i < 4; i++) {
            String name = input.readLine();
            if (name.equals("menu") || players.size() >= 4) {
                break; // User entered menu or the there are already 4 players in the ArrayList -> break out of the for loop
            }
            newPlayer(name); // New player gets added if the if-condition above isn't executed
        }
        playerAmount = players.size();
    }
}

SQL查询的通常执行顺序是"menu"，然后是TableA::find() ->select('*') ->joinWith(['TableB']) ->joinWith(['TableC']) ->joinWith(['TableD']) ->where([TableA.attribute1=1 or TableA.attribute2=1]) ->andWhere([(further possible conditions on the other Tables)]) ->all() 。

是否有一种方法可以在联接之前执行第一个where条件，以减少联接的行数？像

From (with joins)

解决方法

您可以修改用于联接表的条件。

在SQL中，它看起来像这样：

SELECT 
  * 
FROM 
  `tableA`
  JOIN `tableB` ON (
    `tableA`.`FK` = `tableB`.`PK` 
    AND `tableA`.`attr1` = 'someValue'
  ) 
  JOIN tableC ON (`tableB`.`FK` = `tableC`.`PK`)

要在Yii中执行此操作，可以使用ActiveQuery::onCondition()方法。如果只想对此查询应用此条件，则可以使用joinWith()方法中的回调来修改用于联接的查询。

$query = TableA::find()
    ->joinWith([
        'tableB' => function(\yii\db\ActiveQuery $query) {
            $query->onCondition(['tableA.attr1' => 'someValue']);
        }
    ])
    //... the rest of query

其他选项将在SQL查询的FROM部分中使用子查询，如下所示：

SELECT 
  * 
FROM 
  (
    SELECT 
      * 
    FROM 
      `tableA` 
    WHERE 
      `attr1` = 'someValue'
  ) AS `tableA` 
  JOIN `tableB` ON (`tableA`.`FK` = `tableB`.`PK`)

在yii中：

$subQuery = TableA::find()
    ->select('*')
    ->where(['attr1' => 'someValue']);

$query = TableA::find()
    ->select('*')
    ->from(['tableA' => $subQuery])
    ->joinWith('tableB')
    // ... the rest of query

此方法的主要缺点是子查询中的临时表将没有任何索引，因此联接和其他条件将变慢。如果tableA有很多行，并且在联接之前要应用的条件将显着减少行数，那么仍然可以使用这种方法。

,

嗯，我认为这是最好的方法。但是，如果您不打印来自关系TableB或TabelC的数据（仅从条件中获取数据），则可以这样设置关系：

TableA::find()
->joinWith('TableB',false) //dont load data from this relational table
->joinWith('TableC',false) //dont load data from this relational table
->joinWith(['TableD']) //load data from this relational table
->where([TableA.attribute1=1 or TableA.attribute2=1])
->andWhere([(further possible conditions on the other Tables)])
->all()