我有一个PostgreSQL的函数(或表),它给了我下面的输出:
Sl.no username Designation salary etc..
1 A XYZ 10000 ...
2 B RTS 50000 ...
3 C QWE 20000 ...
4 D HGD 34343 ...
现在,我想下面的输出:
Sl.no 1 2 3 4 ...
Username A B C D ...
Designation XYZ RTS QWE HGD ...
Salary 10000 50000 20000 34343 ...
这个怎么做?
立足于形式的表我的回答:
CREATE TABLE tbl (
sl_no int
, username text
, designation text
, salary int
);
在新的一列中的每一行的成绩来回报。使用动态返回类型是这样,它几乎没有可能使这一方面,对数据库的单一调用完全动态的。有两个步骤演示解决方案:
通常,这是由表可以容纳的最大列数的限制。所以不超过1600行(或更少)的表的选项。细节:
crosstab()
SELECT 'SELECT *
FROM crosstab(
''SELECT unnest(''' || quote_literal(array_agg(attname))
|| '''::text[]) AS col
, row_number() OVER ()
, unnest(ARRAY[' || string_agg(quote_ident(attname)
|| '::text', ',') || ']) AS val
FROM ' || attrelid::regclass || '
ORDER BY generate_series(1,' || count(*) || '), 2''
) t (col text, '
|| (SELECT string_agg('r'|| rn ||' text', ',')
FROM (SELECT row_number() OVER () AS rn FROM tbl) t)
|| ')' AS sql
FROM pg_attribute
WHERE attrelid = 'tbl'::regclass
AND attnum > 0
AND NOT attisdropped
GROUP BY attrelid;
可以打包成具有单个参数的功能... 生成形式的查询:
SELECT *
FROM crosstab(
'SELECT unnest(''{sl_no,username,designation,salary}''::text[]) AS col
, row_number() OVER ()
, unnest(ARRAY[sl_no::text,username::text,designation::text,salary::text]) AS val
FROM tbl
ORDER BY generate_series(1,4), 2'
) t (col text, r1 text,r2 text,r3 text,r4 text)
产生所期望的结果:
col r1 r2 r3 r4
-----------------------------------
sl_no 1 2 3 4
username A B C D
designation XYZ RTS QWE HGD
salary 10000 50000 20000 34343
unnest()
SELECT 'SELECT unnest(''{sl_no, username, designation, salary}''::text[] AS col)
, ' || string_agg('unnest('
|| quote_literal(ARRAY[sl_no::text, username::text, designation::text, salary::text])
|| '::text[]) AS row' || sl_no, E'\n , ') AS sql
FROM tbl;
生成形式的查询:
SELECT unnest('{sl_no, username, designation, salary}'::text[]) AS col
, unnest('{10,Joe,Music,1234}'::text[]) AS row1
, unnest('{11,Bob,Movie,2345}'::text[]) AS row2
, unnest('{12,Dave,Theatre,2356}'::text[]) AS row3
, unnest('{4,D,HGD,34343}'::text[]) AS row4
同样的结果。
crosstab()
如果你可以用这个。击败休息。
SELECT 'SELECT *
FROM crosstab(
$ct$SELECT u.attnum, t.rn, u.val
FROM (SELECT row_number() OVER () AS rn, * FROM '
|| attrelid::regclass || ') t
, unnest(ARRAY[' || string_agg(quote_ident(attname)
|| '::text', ',') || '])
WITH ORDINALITY u(val, attnum)
ORDER BY 1, 2$ct$
) t (attnum bigint, '
|| (SELECT string_agg('r'|| rn ||' text', ', ')
FROM (SELECT row_number() OVER () AS rn FROM tbl) t)
|| ')' AS sql
FROM pg_attribute
WHERE attrelid = 'tbl'::regclass
AND attnum > 0
AND NOT attisdropped
GROUP BY attrelid;
与attnum
,而不是实际的列名操作。简单和快捷。加入结果pg_attribute
一次或像皮克9.3例如整合列名。
生成形式的查询:
SELECT *
FROM crosstab(
$ct$SELECT u.attnum, t.rn, u.val
FROM (SELECT row_number() OVER () AS rn, * FROM tbl) t
, unnest(ARRAY[sl_no::text,username::text,designation::text,salary::text])
WITH ORDINALITY u(val, attnum)
ORDER BY 1, 2$ct$
) t (attnum bigint, r1 text, r2 text, r3 text, r4 text);
它使用的先进功能的整个范围。只是过多的解释。
unnest()
一个unnest()
现在可以利用多个阵列在平行于UNNEST。
SELECT 'SELECT * FROM unnest(
''{sl_no, username, designation, salary}''::text[]
, ' || string_agg(quote_literal(ARRAY[sl_no::text, username::text, designation::text, salary::text])
|| '::text[]', E'\n, ')
|| E') \n AS t(col,' || string_agg('row' || sl_no, ',') || ')' AS sql
FROM tbl;
结果:
SELECT * FROM unnest(
'{sl_no, username, designation, salary}'::text[]
,'{10,Joe,Music,1234}'::text[]
,'{11,Bob,Movie,2345}'::text[]
,'{12,Dave,Theatre,2356}'::text[])
AS t(col,row1,row2,row3,row4)
SQL Fiddle在第9.3运行。
SELECT
unnest(array['Sl.no', 'username', 'Designation','salary']) AS "Columns",
unnest(array[Sl.no, username, value3Count,salary]) AS "Values"
FROM view_name
ORDER BY "Columns"
如果(像我一样),你需要从一个bash脚本此信息,注意,是PSQL把这件事告诉输出表列作为行简单的命令行开关:
psql mydbname -x -A -F= -c "SELECT * FROM foo WHERE id=123"
该-x
选项是敲门砖的psql输出的列行。
我有一个比欧文更简单的方法上文所指出,该名工人对我的Postgres(我认为应该与所有主要的关系数据库,其支持SQL标准工作)
您可以使用简单UNION而不是交叉表:
SELECT text 'a' AS "text" UNION SELECT 'b';
text
------
a
b
(2 rows)
当然,这取决于你打算将此情况。考虑到你事先知道哪些领域你需要,你可以采取这种方法,即使查询不同的表。即:
SELECT 'My first metric' as name, count(*) as total from first_table UNION
SELECT 'My second metric' as name, count(*) as total from second_table
name | Total
------------------|--------
My first metric | 10
My second metric | 20
(2 rows)
这是一个更易于维护的方法,恕我直言。看看这个页面了解更多信息:https://www.postgresql.org/docs/current/typeconv-union-case.html
有一个在普通的SQL或PL / pgSQL中这样做不正确的方法。
这将是更好的方式来做到这一点的应用,从数据库获取数据。