C# 正则表达式获取模式的最后一次出现

Regex get last occurrence of the pattern

提问by Vazgun

I have a string and I need to select the last occurrence of the pattern. The string is:

我有一个字符串，我需要选择最后一次出现的模式。字符串是：

[[[1302638400000.0, 0], [1302724800000.0, 610.64999999999998], [1302811200000.0, 2266.6500000000001], [1303156800000.0, 4916.9300000000003], [1303329600000.0, 6107.3199999999997], [1303934400000.0, 9114.6700000000001]], [[1302638400000.0, 20000.0], [1302724800000.0, 20000.0], [1302811200000.0, 20000.0], [1303156800000.0, 20000.0], [1303329600000.0, 20000.0], [1303934400000.0, 20000.0]], [[1302638400000.0, 20000.0], [1302724800000.0, 20610.650000000001], [1302811200000.0, 22266.650000000001], [1303156800000.0, 24916.93], [1303329600000.0, 26107.32], [1303934400000.0, 29114.669999999998], [1304452800000.0, 30078.23]], [[1302718580000.0, 0.0], [1302772440000.0, 3.0532500000000073], [1303107093000.0, 11.333250000000007], [1303107102000.0, 21.753250000000008], [1303352295000.0, 24.584650000000003], [1303352311000.0, 26.8766], [1303815010000.0, 30.536599999999996], [1303815028000.0, 27.703349999999993]]];

[[[1302638400000.0，0]，[1302724800000.0，610.64999999999998]，[1302811200000.0，2266.6500000000001]，[1303156800000.0，4916.9300000000003]，[1303329600000.0，6107.3199999999997]，[1303934400000.0，9114.6700000000001]]，[[1302638400000.0，20000.0]，[1302724800000.0，20000.0 ]，[1302811200000.0，20000.0]，[1303156800000.0，20000.0]，[1303329600000.0，20000.0]，[1303934400000.0，20000.0]]，[[1302638400000.0，20000.0]，[1302724800000.0，20610.650000000001]，[1302811200000.0，22266.650000000001]，[1303156800000.0，24916.93 ], [1303329600000.0, 26107.32], [1303934400000.0, 29114.669999999998], [1304452800000.0, 30078.23]] [[1302718580000.0，0.0]，[1302772440000.0，3.0532500000000073]，[1303107093000.0，11.333250000000007]，[1303107102000.0，21.753250000000008]，[1303352295000.0，24.584650000000003]，[1303352311000.0，26.8766]，[1303815010000.0，30.536599999999996]，[1303815028000.0，27.703349999999993]]] ;

The pattern that I use is:

我使用的模式是：

\s\[\[(.*?)\]\]

Which unfortunately selects only 1st occurrence. The highlighted text is the desired result. It doesn't matter how many square brackets at the end, just need the last array set.

不幸的是，它只选择了第一次出现。突出显示的文本是所需的结果。最后有多少方括号无关紧要，只需要最后一个数组集。

UPDATE: If it can help you, then the coding is in c#

更新：如果它可以帮助你，那么编码是在 c#

回答by Lucian Enache

Try adding the $ to your pattern like this \s\[\[(.*?)\]\]\]\;$and let me know if it works.

尝试像这样将 $ 添加到您的模式中\s\[\[(.*?)\]\]\]\;$，让我知道它是否有效。

Currently I don't have bash under hand so I cannot check but it should do the trick.

目前我手头没有 bash，所以我无法检查，但它应该可以解决问题。

Edit : Right version \S+\s?+(?!((.*\[\[)))

编辑：正确的版本 \S+\s?+(?!((.*\[\[)))

which translates into :

翻译成：

\S   : all alfanumeric
\s?  : all 1 space occurences
?!   : not 
.*   : everything
\[\[ : until the last pattern of [[ (excluded)

Here is the Rubular example

这是 Rubular 示例

BTW great tool rubular, make me wanna look more into ruby and regular expressions :D

顺便说一句，很棒的工具 rubular，让我想更多地了解 ruby​​ 和正则表达式：D

回答by Chris Seymour

Do a greedy match up to the last set of [[and capture..

对最后一组进行贪婪匹配[[并捕获..

.*(\[\[.*\]\])\];

See it here.

在这里看到它。

回答by stema

There should be a global flag or a method that returns all matches in your language. Use this and take the last match.

应该有一个全局标志或一个方法，以您的语言返回所有匹配项。使用它并进行最后一场比赛。

In C# Matches()returns a MatchCollectioncontaining all found matches. So you can do for example this:

在 C# 中Matches()返回一个MatchCollection包含所有找到的匹配项。所以你可以这样做：

string source = "[[[1302638400000.0, 0], [1302724800000.0, 610.64999999999998], [1302811200000.0, 2266.6500000000001], [1303156800000.0, 4916.9300000000003], [1303329600000.0, 6107.3199999999997], [1303934400000.0, 9114.6700000000001]], [[1302638400000.0, 20000.0], [1302724800000.0, 20000.0], [1302811200000.0, 20000.0], [1303156800000.0, 20000.0], [1303329600000.0, 20000.0], [1303934400000.0, 20000.0]], [[1302638400000.0, 20000.0], [1302724800000.0, 20610.650000000001], [1302811200000.0, 22266.650000000001], [1303156800000.0, 24916.93], [1303329600000.0, 26107.32], [1303934400000.0, 29114.669999999998], [1304452800000.0, 30078.23]], [[1302718580000.0, 0.0], [1302772440000.0, 3.0532500000000073], [1303107093000.0, 11.333250000000007], [1303107102000.0, 21.753250000000008], [1303352295000.0, 24.584650000000003], [1303352311000.0, 26.8766], [1303815010000.0, 30.536599999999996], [1303815028000.0, 27.703349999999993]]];";
Regex r = new Regex(@"\s\[\[(.*?)\]\]");

MatchCollection result = r.Matches(source);

if (result.Count > 0) {
    Console.WriteLine(result[result.Count - 1]);
} else {
    Console.WriteLine("No match found!");
}
Console.ReadLine();

回答by Alan Moore

Use the RightToLeftoption:

使用RightToLeft选项：

Regex.Match(s, @"\[\[(.*?)\]\]", RegexOptions.RightToLeft)

This option is exclusive to the .NET regex flavor, and does exactly what you asked for: searches from the end of the input instead of the beginning. Of particular note, the non-greedy ?modifier works just as you expect; if you leave it off you'll get the whole input, but with it you get:

此选项是 .NET regex 风格独有的，并且完全符合您的要求：从输入的末尾而不是开头进行搜索。需要特别注意的是，非贪婪?修饰符的作用与您预期的一样；如果您不使用它，您将获得整个输入，但有了它，您将获得：

[[1302718580000.0, 0.0], [1302772440000.0, 3.0532500000000073], [1303107093000.0, 11.333250000000007], [1303107102000.0, 21.753250000000008], [1303352295000.0, 24.584650000000003], [1303352311000.0, 26.8766], [1303815010000.0, 30.536599999999996], [1303815028000.0, 27.703349999999993]]]

[[1302718580000.0, 0.0], [1302772440000.0, 3.0532500000000073], [1303107093000.0, 11.333250000000007], [1303107102000.0, 21.753250000000008], [1303352295000.0, 24.584650000000003], [1303352311000.0, 26.8766], [1303815010000.0, 30.536599999999996], [1303815028000.0, 27.703349999999993]]]