鉴于字符串:
‘I think that Postgresql is nifty’
我想对该字符串中的单个单词进行操作.从本质上讲,我有一个单独的,我可以得到单词的详细信息,并希望在这本字典中加入该字符串的unnested数组.
到目前为止我有:
select word,meaning,partofspeech from unnest(string_to_array('I think that Postgresql is nifty',' ')) as word from table t join dictionary d on t.word = d.wordname;
这实现了我希望做的基本原理,但它不保留原始的单词顺序.
Postgres 9.4或更高版本中的ORDINALITY
原文链接:https://www.f2er.com/postgresql/192346.htmlSELECT * FROM regexp_split_to_table('I think postgres is nifty',' ') WITH ORDINALITY x(word,rn);
或者,应用于表格:
SELECT * FROM tbl t,regexp_split_to_table(t.my_column,rn);
细节:
> How to preserve the original order of elements in an unnested array?
关于隐式LATERAL连接:
> What is the difference between LATERAL and a subquery in PostgreSQL?
Postgres 9.3或更早 – 更一般的解释
对于单个字符串
您可以应用窗口函数row_number()
来记住元素的顺序.但是,使用通常的row_number()OVER(ORDER BY col),您可以根据排序顺序获取数字,而不是字符串中的原始位置.
您可以尝试简单地省略ORDER BY以“按原样”获取位置:
SELECT *,row_number() OVER () AS rn FROM ( SELECT regexp_split_to_table('I think postgres is nifty',' ') AS word ) x;
regexp_split_to_table()的性能随长字符串而降低. unnest(string_to_array(…))更好地扩展:
SELECT *,row_number() OVER () AS rn FROM ( SELECT unnest(string_to_array('I think postgres is nifty',' ')) AS word ) x;
然而,虽然这通常有效,并且我从未在简单查询中看到过它,但Postgresql没有关于没有显式ORDER BY的行的顺序.
要保证原始字符串中元素的序数,请使用generate_subscript()
(通过@deszo评论改进):
SELECT arr[rn] AS word,rn FROM ( SELECT *,generate_subscripts(arr,1) AS rn FROM ( SELECT string_to_array('I think postgres is nifty',' ') AS arr ) x ) y;
对于一个字符串表
将PARTITION BY id添加到OVER子句中……
演示表:
CREATE TEMP TABLE strings(string text); INSERT INTO strings VALUES ('I think postgres is nifty'),('And it keeps getting better');
我使用ctid作为主键的ad-hoc替代品.如果您有一个(或任何唯一列),请改用它.
SELECT *,row_number() OVER (PARTITION BY ctid) AS rn FROM ( SELECT ctid,unnest(string_to_array(string,' ')) AS word FROM strings ) x;
这没有任何明确的ID:
SELECT arr[rn] AS word,rn FROM ( SELECT *,generate_subscripts(arr,1) AS rn FROM ( SELECT string_to_array(string,' ') AS arr FROM strings ) x ) y;
回答问题
SELECT z.arr,z.rn,z.word,d.meaning --,partofspeech -- ? FROM ( SELECT *,arr[rn] AS word FROM ( SELECT *,1) AS rn FROM ( SELECT string_to_array(string,' ') AS arr FROM strings ) x ) y ) z JOIN dictionary d ON d.wordname = z.word ORDER BY z.arr,z.rn;