postgres子字符串转义开括号字符

Question

我正在尝试 return postgres 中字符串字段中括号内的所有字符：

这个函数似乎运行良好：

 select substring('foo bar [foobar] foo' from '%#"f%b#"%' for '#');
 substring
-----------
 foob

它不适用于左括号字符：

select substring('foo bar [foobar] foo' from '%#"[%]#"%' for '#');
substring
-----------

右括号工作正常，我试过反斜杠、双反斜杠、句号、单引号

Answer 1

您应该转义方括号，因为它们是元字符：

select substring('foo bar [foobar] foo' from '%#"#[%#]#"%' for '#');

 substring 
-----------
 [foobar]
(1 row)

或

select substring('foo bar [foobar] foo' from '%#[#"%#"#]%' for '#');

 substring 
-----------
 foobar
(1 row)

来自the documentation：

In addition to these facilities borrowed from LIKE, SIMILAR TO supports these pattern-matching metacharacters borrowed from POSIX regular expressions:

...

A bracket expression [...] specifies a character class, just as in POSIX regular expressions.

postgres子字符串转义开括号字符

postgres substring escape open bracket character

postgresql

brackets

escaping

postgresql-9.3