MySQL查询分析结果并标记它们

时间:2014-04-08 15:32:50

标签: mysql

我有两个mysql表:发票和付款。我想在一段时间内列出所有的invoicerow以及是否付款的说明..

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

所得:

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

依旧...... 不行。这可以在(mysql)查询中完成吗?

2 个答案:

答案 0 :(得分:2)

您可以将JOINCASE

一起使用
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

另一种方法也可以使用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

答案 1 :(得分:0)

尝试使用CASE而不是IF。:

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

简单示例(fiddle)

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