Question

我有一个prefs表，以下是相关列：

mydb=> SELECT pref_id, pref_name, pref_value FROM prefs;
 pref_id |  pref_name   |   pref_value
---------+--------------+----------------
       1 | PagerNumber  | 2125551234
       2 | PagerCarrier | @att.com
       3 | PagerCarrier | @something.com

我想生产这样的东西：

 section |  pager_number  | pager_carrier
---------+----------------+---------------
       1 | 2125551234     |
       2 |                | @att.com
       3 |                | @something.com

所以我使用了交叉表，在stackoverflow上跟随这个例子：PostgreSQL Crosstab Query

SELECT row_name AS section,
       category_1::text AS pager_number,
       category_2::text AS pager_carrier
FROM crosstab('select pref_id::bigint, pref_name::text, pref_value::text
    FROM prefs')
AS ct (row_name bigint, category_1 text, category_2 text);

所有值都进入pager_number，pager_carrier为空：

 section |  pager_number  | pager_carrier
---------+----------------+---------------
       1 | 2125551234     |
       2 | @att.com       |
       3 | @something.com |

有谁能看到发生了什么？

Answer 1

测试用例（提供样本数据的首选方式）：

CREATE TEMP TABLE prefs (pref_id int, pref_name text, pref_value text);

INSERT INTO prefs VALUES 
 (1, 'PagerNumber' , '2125551234')
,(2, 'PagerCarrier', '@att.com')
,(3, 'PagerCarrier', '@something.com');

查询：

SELECT *
FROM   crosstab(
       'SELECT pref_id, pref_name, pref_value
        FROM   prefs
        ORDER  BY 1, 2',

       $$VALUES ('PagerNumber'::text), ('PagerCarrier')$$
       )
AS x (section text, pager_number bigint, pager_carrier text);

准确返回问题中描述的结果。如果PagerNumber可能不是有效的bigint号码，请改用text。

您在问题中提到的答案已经过时，从一开始就不正确。我添加了proper answer with explanation and links over there.

Answer 2

而不是：

SELECT row_name AS section, category_1::text AS pager_number, category_2::text
AS pager_carrier
FROM crosstab('select pref_id::bigint, pref_name::text, pref_value::text
    FROM prefs')
AS ct (row_name bigint, category_1 text, category_2 text);

尝试：

SELECT *
FROM crosstab('select pref_id::bigint, pref_name::text, pref_value::text
FROM prefs ORDER BY 1,2')
AS prefs (row_name bigint, carrier_1 text, carrier_2 text);

如果你有：

    pref_id |  pref_name   |   pref_value
   ---------+--------------+----------------
   1 | PagerNumber  | 2125551234
   2 | PagerCarrier | @att.com
   3 | PagerCarrier | @something.com
   2 | PageNumber   | 2332323232
   3 | PagerCarrier | @somethingelse.com

你会得到：

    row_name |  carrier_1   |   carrier_2
   -----+--------------+----------------
   1 | 2125551234      |
   2 | @att.com        | 2332323232
   3 | @something.com  | @somethingelse.com

Postgress Crosstab Reference

PostgreSQL交叉表/枢轴问题

2 个答案: