0

我有以下 SQL

SELECT
  i.si_num AS `id`,
  DATE_FORMAT(i.si_date, '%d-%b-%Y') AS `date`,
  i.si_tr AS `tr`,
  d.dl_name AS `customer`,
  i.si_net_value AS `net`,
  DATEDIFF(CURDATE(), si_date) AS Days,
  t.value AS tval,
  t.label AS label
FROM
  invoices AS i
LEFT JOIN
  dealer AS d ON i.si_tr = d.dl_id
LEFT JOIN
  transactions AS t ON i.si_num = t.invoice
WHERE
  i.si_tr = 'TR580494'
ORDER BY `si_num` DESC;

电流输出

+-----+-----------+----------+---------------------+---------+------+-------+-------+
| id  |   date    |    tr    |      customer       |   net   | Days | tval  | label |
+-----+-----------+----------+---------------------+---------+------+-------+-------+
| 404 | 18-Feb-17 | TR580494 | STARSHIP ENTERPRISE | 109790  |   55 | 96070 | acr   |
| 404 | 18-Feb-17 | TR580494 | STARSHIP ENTERPRISE | 109790  |   55 | 10080 | crn   |
| 404 | 18-Feb-17 | TR580494 | STARSHIP ENTERPRISE | 109790  |   55 | 3640  | crn   |
| 240 | 13-Feb-17 | TR580494 | STARSHIP ENTERPRISE | 0       |   60 | NULL  | NULL  |
| 239 | 13-Feb-17 | TR580494 | STARSHIP ENTERPRISE | 81975   |   60 | 30405 | acr   |
| 239 | 13-Feb-17 | TR580494 | STARSHIP ENTERPRISE | 81975   |   60 | 51570 | crn   |
| 132 | 3-Feb-17  | TR580494 | STARSHIP ENTERPRISE | 38132.5 |   70 | 33282 | acr   |
+-----+-----------+----------+---------------------+---------+------+-------+-------+

如您所见，生成了一些重复的行，例如 id: 404 & 239 这些行的唯一区别是“tval”和“label”列中的值

'tval' 和 'label' 列可以为每个 id 填充多次，我想要实现的是对于每个重复的记录，我想查看标签是 'crn' 还是 'arc' 并对各自的值求和到 id 并生成一行并将标签转置为列。请参阅下面的预期输出。

预期产出

+-----+-----------+----------+---------------------+---------+------+-------+-------+
| id  |   date    |    tr    |      customer       |   net   | days |  crn  |  acr  |
+-----+-----------+----------+---------------------+---------+------+-------+-------+
| 404 | 18-Feb-17 | TR580494 | STARSHIP ENTERPRISE | 109790  |   55 | 13720 | 96070 |
| 240 | 13-Feb-17 | TR580494 | STARSHIP ENTERPRISE | 0       |   60 | NULL  | NULL  |
| 239 | 13-Feb-17 | TR580494 | STARSHIP ENTERPRISE | 81975   |   60 | 51570 | 30405 |
| 132 | 3-Feb-17  | TR580494 | STARSHIP ENTERPRISE | 38132.5 |   70 | NULL  | 33282 |
+-----+-----------+----------+---------------------+---------+------+-------+-------+
4

1 回答 1

0

您可以使用case内部 a做到这一点sum：

SELECT
  i.si_num AS `id`,
  DATE_FORMAT(i.si_date, '%d-%b-%Y') AS `date`,
  i.si_tr AS `tr`,
  d.dl_name AS `customer`,
  i.si_net_value AS `net`,
  DATEDIFF(CURDATE(), si_date) AS Days,
  sum(case when t.label = 'acr' then t.value else null end) as acr
  sum(case when t.label = 'crn' then t.value else null end) as crn
FROM
  invoices AS i
LEFT JOIN
  dealer AS d ON i.si_tr = d.dl_id
LEFT JOIN
  transactions AS t ON i.si_num = t.invoice
WHERE
  i.si_tr = 'TR580494'
GROUP BY  i.si_num,
          DATE_FORMAT(i.si_date, '%d-%b-%Y'),
          i.si_tr,
          d.dl_name,
          i.si_net_value,
          DATEDIFF(CURDATE(), si_date)
ORDER BY `si_num` DESC;
于 2017-04-14T13:21:38.263 回答