2019独角兽企业重金招聘Python工程师标准>>> 
在laravel中使用leftJoin添加多个条件时，如select a.* from a left join b on a.id = b.pid and b.status = 1这种类似sql，发现框架自身封装的leftJoin不支持多个参数传递（当然可用写原生sql），laravel框架自身封装的leftJoin方法如下：
public function leftJoin($table, $first, $operator = null, $second = null)
{
return $this->join($table, $first, $operator, $second, 'left');
}
浏览下 \vendor\laravel\framework\src\Illuminate\Database\Query\Builder.php文件，发现join方法可用实现自己想要的left join携带多参数。laravel自身的join方法如下：
public function join($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 $this;
}
当左右连接携带多条件时，可以这样写：
DB::table('app_a as a')
->join('app_b as b',function($join){
$join->on('a.id','=','b.goodId')
->where('b.status','=','SUCCESS')
->where('b.type','=','UNLOCK');
}, null,null,'left')
->where('a.id','>',1)
->get();
//相当于
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;当join不传left时，默认是inner。