Php 当没有返回行时，从select查询中获取预定义值

我的问题是：

SELECT 
    col1, col2, sum(col3), ref
FROM tbl1
LEFT JOIN tbl2 on tbl2.ref = tbl1.ref
WHERE ref in (1,2,3,4,5,...);

PHP中的结果如下所示：

[
    0 => [
        'col1' => 'x',
        'col2' => 'y',
        'col3' => 1234,
        'ref'  => 1,
    ],
    1 => [
        'col1' => 'x',
        'col2' => 'y',
        'col3' => 1234,
        'ref'  => 3,
    ],
    2 => [
        'col1' => 'x',
        'col2' => 'y',
        'col3' => 1234,
        'ref'  => 3,
    ],
    3 => [
        'col1' => 'x',
        'col2' => 'y',
        'col3' => 1234,
        'ref'  => 5,
    ],
]

由于tbl中没有参考文献2和4的数据，因此仅返回了1,3,5

但是如果没有找到那些

ref

中的行，我需要一些默认数据（至少一行）

比如：

[
    0 => [
        'col1' => 'x',
        'col2' => 'y',
        'col3' => 'z',
        'ref'  => 1,
    ],
    1 => [
        'col1' => null,
        'col2' => null,
        'col3' => null,
        'ref'  => 2,
    ],
    2 => [
        'col1' => 'x',
        'col2' => 'y',
        'col3' => 1234,
        'ref'  => 3,
    ],
    3 => [
        'col1' => 'x',
        'col2' => 'y',
        'col3' => 1234,
        'ref'  => 3,
    ],
    4 => [
        'col1' => null,
        'col2' => null,
        'col3' => null,
        'ref'  => 4,
    ],
    5 => [
        'col1' => 'x',
        'col2' => 'y',
        'col3' => 1234,
        'ref'  => 5,
    ],
]

---

这可能吗？

您可以使用这些值创建一个派生表，并使用

连接：

SELECT tbl.col1, tbl.col2, tbl.col3, r.ref
FROM (SELECT 1 as ref UNION ALL
      SELECT 2 as ref UNION ALL
      SELECT 3 as ref UNION ALL
      SELECT 4 as ref UNION ALL
      SELECT 5 as ref
     ) r LEFT JOIN
     tbl
     ON r.ref = tbl.ref;

实际上，这个列表不仅是（1,2,3,4,5），它应该是（1,2,3，…）@Amin。只需扩展表中的值。