正则表达式中的运算符优先级 [英] Operator precedence in regular expressions

问题描述

当 Oracle 的正则表达式不包含括号时，它们的默认运算符优先级是多少?

What is the default operator precedence in Oracle's regular expressions when they don't contain parentheses?

例如，给定

 H|ha+

它会被评估为 H|h 然后像 ((H|h)a) 那样连接到 a，还是会H 与 ha 交替，如 (H|(ha))?

would it be evaluated as H|h and then concatenated to a as in ((H|h)a), or would the H be alternated with ha as in (H|(ha))?

另外，+ 什么时候开始，等等?

Also, when does the + kick in, etc.?

推荐答案

鉴于 Oracle 文档:

表 4-2 列出了支持在传递给 SQL 正则表达式函数和条件的正则表达式中使用的元字符列表.这些元字符符合 POSIX 标准；行为与标准的任何差异都在描述"列中注明.

Table 4-2 lists the list of metacharacters supported for use in regular expressions passed to SQL regular expression functions and conditions. These metacharacters conform to the POSIX standard; any differences in behavior from the standard are noted in the "Description" column.

然后查看该表中的 | 值:

And taking a look at the | value in that table:

表达式 a|b 匹配字符 a 或字符 b.

The expression a|b matches character a or character b.

另外看看POSIX 文档:

运算符优先级of 运算符的优先顺序如下:

Operator precedence The order of precedence for of operators is as follows:

1. 与排序规则相关的括号符号 [==] [::] [..]

1. Collation-related bracket symbols [==] [::] [..]

转义字符 \

字符集(括号表达式)[]

Character set (bracket expression) []

分组()

单字符 ERE 重复 * + ?{m,n}

Single-character-ERE duplication * + ? {m,n}

串联

锚定 ^$

交替|

我会说 H|ha+ 与 (?:H|ha+) 相同.

I would say that H|ha+ would be the same as (?:H|ha+).

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