ThinkPHP5.1 where多条件查询

2019-05-09T09:34:50
if ($keyword) {
            if ($sotype == "id") {
                $where[$stype] = $keyword;
            } else {
                $where = [
                    ['title', 'like', "%" . $keyword . "%"],
                ];
            }
        }
$where['level'] = 1;
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