MySQL查詢分析結果並標記它們

Question

我有兩個mysql表：發票和付款。 我想在一段時間內列出所有的invoicerow，並注明是否付款。

SELECT invoices.i_id, invoices.sum, IF EXISTS (Select * from payments WHERE payments.i_id = invoices.i_id) THEN 'PAID' ELSE 'NOT PAID')
FROM invoices
WHERE invoices.date >= 2014-01-01
AND invoices.date <= 2014-01-31

由於：

1252515  122,50 PAID
1252514  150,40 PAID
1257425 1180,40 NOT PAID

等等......不行。 這可以在（mysql）查詢中完成嗎？

Answer 1

您可以將JOIN與CASE

SELECT i.i_id, i.`sum`,
(CASE WHEN p.i_id IS NOT NULL THEN 'PAID' ELSE 'NOT PAID' END) `status`
FROM invoices i
LEFT JOIN payments p ON(p.i_id = i.i_id)
WHERE i.date >= 2014-01-01
AND i.date <= 2014-01-31

另一種方式你也可以使用IF

SELECT i.i_id, i.`sum`,
   IF(p.i_id IS NULL, 'NOT PAID', 'PAID')  `status`
FROM invoices i
LEFT JOIN payments p ON(p.i_id = i.i_id)
WHERE i.date >= 2014-01-01
AND i.date <= 2014-01-31

Answer 2

嘗試使用CASE而不是IF：

CASE WHEN EXISTS (Select * from payments WHERE payments.i_id = invoices.i_id) THEN 'PAID' ELSE 'NOT PAID' END

簡單的例子（小提琴） ：

CREATE TABLE tbl1 (id INT);
CREATE TABLE tbl2 (id INT);

INSERT INTO tbl1 VALUES (1),(2);
INSERT INTO tbl2 VALUES (1);

SELECT id
      ,CASE WHEN EXISTS (SELECT * FROM tbl2 WHERE tbl2.id = tbl1.id) THEN 'PAID' ELSE 'NOT PAID' END
  FROM tbl1