我正在使用group_concat
函数来汇总产品的一些移动,这些移动已经通过(stockout)
用于右表left join
。 MySQL代码如下-
SELECT
p.serialno AS SISN,
p.in_quantity AS INQTY,
SUM(IFNULL(s.out_quantity, '0')) AS OUTQTY,
GROUP_CONCAT(
CONCAT(
COALESCE(IFNULL(s.out_quantity,'0'), ''),'|'
)
) details
FROM stockin p LEFT JOIN stockout s
ON p.serialno = s.serialno
WHERE p.productid = 'TF00123'
GROUP BY p.stockin_id, p.serialno, s.serialno
输出如下
+------+-------+--------+--------------------------------+
| SISN | INQTY | OUTQTY | details |
+------+-------+--------+--------------------------------+
| AAA1 | 500 | 740 | 300|,100|,50|,50|,20|,20|,200| |
| AAA2 | 500 | 0 | 0| |
| AAA3 | 1 | 3 | 1|,1|,1| |
| AAA3 | 1 | 3 | 1|,1|,1| |
| AAA1 | 200 | 740 | 300|,100|,50|,50|,20|,20|,200| |
| AAA3 | 1 | 3 | 1|,1|,1| |
| AAA1 | 100 | 740 | 300|,100|,50|,50|,20|,20|,200| |
+------+-------+--------+--------------------------------+
7 rows in set (0.00 sec)
现在,如果p.serialno是相同的,我想在另一列中
group_concat
保留详细信息,例如详细信息。如果我手动执行,请检查输出-
+------+-------+------------------+--------+--------------------------------+
| SISN | INQTY | details | OUTQTY | details |
+------+-------+------------------+--------+--------------------------------+
| AAA1 | 800 | 500|,200|,100| | 740 | 300|,100|,50|,50|,20|,20|,200| |
| AAA2 | 500 | 0| | 0 | 0| |
| AAA3 | 3 | 1|,1|,1| | 3 | 1|,1|,1| |
+------+-------+------------------+--------+--------------------------------+
3 rows in set (0.00 sec)
最佳答案
只需使用
GROUP_CONCAT(
CONCAT(
COALESCE(IFNULL(p.out_quantity,'0'), ''),'|'
)
) details1
像这样
SELECT
p.serialno AS SISN,
p.in_quantity AS INQTY,
GROUP_CONCAT(
CONCAT(
COALESCE(IFNULL(p.out_quantity,'0'), ''),'|'
)
) details1
SUM(IFNULL(s.out_quantity, '0')) AS OUTQTY,
GROUP_CONCAT(
CONCAT(
COALESCE(IFNULL(s.out_quantity,'0'), ''),'|'
)
) details
FROM stockin p LEFT JOIN stockout s
ON p.serialno = s.serialno
WHERE p.productid = 'TF00123'
GROUP BY p.stockin_id, p.serialno, s.serialno