Regex从以预定义字符开头的字符串匹配序列和数字

Question

是否可能匹配最后一个系列和数字从一个字符串看起来像:

CUI: RO39890982                                   BZ 347

我想要的:BZ和347来自两个独立的正则表达式

我在尝试什么:CUI:[ ]*sS*s(.*)

结果:BZ 347

任何帮助都非常感谢!谢谢。

Answer 1

如果你想要两个独立的表达式:

正则表达式1使用(S+)捕获第1组中的1个或多个非空白字符:

CUI:s*S+s+(S+)

Regex演示

---

正则表达式2使用限定词{2}重复非捕获组，并为值

嵌套捕获组

CUI:s*S+(?:s+(S+)){2}

Regex演示

如果您想在更多字段中重用量词，使用量词可能会很方便。

Answer 2

对于显示的示例，请尝试以下正则表达式。

正则表达式1: 获取BZ值:

bCUI:s+S+s+(S+)

解释: 添加以上regex的详细说明:

bCUI:      ##Matching a word boundary with string CUI: here.
s+S+s+   ##Matching 1 or more spaces followed by 1 or more non-spaces, followed by 1 or more spaces here.
(S+)       ##Creating 1st capturing group which has 1 or more non-spaces here, which will catch BZ value here.

---

---

正则表达式2:获取347值(或类似模式值)与所示示例。

bCUI:s+S+s+S+s+(S+)

解释:为以上内容添加详细说明。

bCUI:     ##Matching a word boundary with string CUI: here.
s+S+s+  ##Matching 1 or more spaces followed by 1 or more non-spaces, followed by 1 or more spaces here.
S+s+     ##matching 1 or non-spaces followed by 1 or more spaces here.
(S+)      ##Creating 1st capturing group which has 1 or more non-spaces here, which will catch 347 value here.

正则表达式的在线演示

Answer 3

Get347with

CUI: *S+ +[A-Z]+ +(d+)

GetBZwith

CUI: *S+ +([A-Z]+)

正则表达式证明#1，正则表达式证明#2。

--------------------------------------------------------------------------------
CUI:                     'CUI:'
--------------------------------------------------------------------------------
*                       ' ' (0 or more times (matching the most
amount possible))
--------------------------------------------------------------------------------
S+                      non-whitespace (all but n, r, t, f,
and " ") (1 or more times (matching the
most amount possible))
--------------------------------------------------------------------------------
+                       ' ' (1 or more times (matching the most
amount possible))
--------------------------------------------------------------------------------
[A-Z]+                   any character of: 'A' to 'Z' (1 or more
times (matching the most amount possible))
--------------------------------------------------------------------------------
+                       ' ' (1 or more times (matching the most
amount possible))
--------------------------------------------------------------------------------
(                        group and capture to 1:
--------------------------------------------------------------------------------
d+                      digits (0-9) (1 or more times (matching
the most amount possible))
--------------------------------------------------------------------------------
)                        end of 1