Laravel left join攜帶多個條件

在laravel中使用leftJoin添加多個條件時，如select a.* from a left join b on a.id = b.pid and b.status = 1這種類似sql，發現框架自身封裝的leftJoin不支持多個參數傳遞（當然可用寫原生sql），laravel框架自身封裝的leftJoin方法如下：

 public function leftJoin($table, $first, $operator = null, $second = null)
 {
    return $this->join($table, $first, $operator, $second, 'left');
 }

 瀏覽下 \vendor\laravel\framework\src\Illuminate\Database\Query\Builder.php文件，發現join方法可用實現自己想要的left join攜帶多參數。laravel自身的join方法如下：

public function join($table, $one, $operator = null, $two = null, $type = 'inner', $where = false)
{
    // If the first "column" of the join is really a Closure instance the developer
    // is trying to build a join with a complex "on" clause containing more than
    // one condition, so we'll add the join and call a Closure with the query.
    if ($one instanceof Closure) {
        $join = new JoinClause($type, $table);

        call_user_func($one, $join);

        $this->joins[] = $join;

        $this->addBinding($join->bindings, 'join');
    }

    // If the column is simply a string, we can assume the join simply has a basic
    // "on" clause with a single condition. So we will just build the join with
    // this simple join clauses attached to it. There is not a join callback.
    else {
        $join = new JoinClause($type, $table);

        $this->joins[] = $join->on(
            $one, $operator, $two, 'and', $where
        );

        $this->addBinding($join->bindings, 'join');
    }

    return $this;
}

當左右連接攜帶多條件時，可以這樣寫（當join不傳left時，默認是inner）：

DB::table('app_a as a')
->join('app_b as b',function($join){
    $join->on('a.id','=','b.goodId')
        ->where('b.status','=','SUCCESS')
        ->where('b.type','=','UNLOCK');
}, null,null,'left')
->where('a.id','>',1)
->get();

//相當於
SELECT * FROM app_a as a 
LEFT JOIN app_b as b on a.id = b.goodId 
and b.status = 'SUCCESS' and b.type = 'UNLOCK' 
where a.id > 1;

參考文章：https://my.oschina.net/u/3403514/blog/1819202/