正则表达式,选择在点号之间且不包含括号的分组。

huangapple go评论119阅读模式
英文:

Regex to select groups between dots without brackets

问题

以下是您要翻译的部分:

我有这样的字符串:

    test.as123[13].tdfv
    Parent.child1
    test.d12[0].asdf

我需要一个正则表达式,可以给我匹配到这样的组:

- test, as123, tdfv
- Parent, child1
- test, d12, asdf

所以我需要在句点处分割这些组,并删除名称中的括号。

我尝试使用否定字符类,就像这样:

    ([^\W\d]+)

但结果会是"child"而不是"child1"。

我的问题是,我不确定如何忽略特定的字符串`\[\d\]`并继续匹配`\d`。

我尝试使用负向先行断言,但我觉得我做错了什么。
这是我的尝试:

    \w+(?!\[[0-9]*?\])
这确实匹配了,例如:

- test, 13, tdfv

请注意,我已经删除了不需要的部分,只返回了翻译好的内容。如果您有任何其他问题,请随时提出。

英文:

I got Strings like this:

test.as123[13].tdfv
Parent.child1
test.d12[0].asdf

I need a regex which gets me groups like this:

  • test, as123, tdfv
  • Parent, child1
  • test, d12, asdf

So I need to divide the groups on the dot and also remove the brackets in its name.

I tried using negation like this:

([^\W\d]+)

But there will be child as result and not child1.

My problem is that I'm not sure on how to ignore the specific String \[\d\] and keep on matching \d

I tried using negative lookahead, but I think I'm doing something wrong.
Here's my attemp:

\w+(?!\[[0-9]*?\])

This does match for example:

  • test, 13, tdfv

答案1

得分: 3

// You can remove all occurrences of digits inside square brackets and then split on a dot:

.replaceAll("\\[\\d+]", "").split("\\.")

// Or, you may directly split on the `(?:\\[\\d+])?\\.` pattern:

.split("(?:\\[\\d+])?\\.")

// It matches

// - `(?:\\[\\d+])?` - an optional occurrence of a `[`, 1+ digits, `]`
// - `\\.` - a dot.

// See the [Java demo online][1]:

String s = "test.as123[13].tdfv";
String results[] = s.replaceAll("\\[\\d+]", "").split("\\.");
System.out.println(Arrays.toString(results));
// => [test, as123, tdfv]

String results2[] = s.split("(?:\\[\\d+])?\\.");
System.out.println(Arrays.toString(results2));
// => [test, as123, tdfv]
英文:

You can remove all occurrences of digits inside square brackets and then split on a dot:

.replaceAll("\\[\\d+]", "").split("\\.")

Or, you may directly split on the (?:\[\d+])?\. pattern:

.split("(?:\\[\\d+])?\\.")

It matches

  • (?:\[\d+])? - an optional occurrence of a [, 1+ digits, ]
  • \. - a dot.

See the Java demo online:

String s = "test.as123[13].tdfv";
String results[] = s.replaceAll("\\[\\d+]", "").split("\\.");
System.out.println(Arrays.toString(results));
// => [test, as123, tdfv]

String results2[] = s.split("(?:\\[\\d+])?\\.");
System.out.println(Arrays.toString(results2));
// => [test, as123, tdfv]

huangapple
  • 本文由 发表于 2020年8月27日 17:25:44
  • 转载请务必保留本文链接:https://go.coder-hub.com/63613019.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定