Question

我正在尝试小写一个反向引用的字符串，但它不会工作。 :(我只想降低root之后的第一个字符。

my_string = "<root>Here is July's best food</root>"

正确的应该是：

my_string = "<root>here is July's best food</root>"

这是我的问题：

UPDATE my_table
SET my_string = regexp_replace(my_string, '<root>(.)', lower(E'<root>\\1'))
WHERE my_id = 1;

Answer 1

使用split_part()和lower()

 select '<root>'||lower(split_part(split_part('<root>Here is July''s best food</root>',' ',1),'>',2)) || split_part('<root>Here is July''s best food</root>',split_part(split_part('<root>Here is July''s best food</root>',' ',1),'>',2),2)

结果：

 "<root>here is July's best food</root>"

UPDATE my_table
SET my_string =  '<root>'||lower(split_part(split_part(my_string,' ',1),'>',2)) || split_part(my_string,split_part(split_part(my_string,' ',1),'>',2),2) 
WHERE my_id = 1;

以下功能是可选的

您可以制作如下所示的功能

 create or replace function lower_first_word(txt text,del text) returns text as
$$
 select del||lower(split_part(split_part(txt ,' ',1),'>',2)) || split_part(txt ,split_part(split_part(txt ,' ',1),'>',2),2)
$$
language sql

并更新

UPDATE my_table
    SET my_string = lower_first_word('<root>Here is July''s best food</root>','<root>')
    WHERE my_id = 1;

<强> OR

使用regex和substring()

select lower(substring('<root>Here is July''s best food</root>',E'[A-Za-z<>]+'))||substring('<root>Here is July''s best food</root>',E'[^ ]* (.*)')

并且更新语句是

UPDATE my_table
SET my_string = lower(substring('<root>Here is July''s best food</root>', E '[A-Za-z<>]+')) || substring('<root>Here is July''s best food</root>', E '[^ ]* (.*)')
WHERE my_id = 1;

Postgres中不能小写反向引用的字符串

1 个答案: