使用.*，并确保您使用的实现相当于单行，以便在行尾匹配。

这里有一个很好的解释-> http://www.regular-expressions.info/dot.html

对于JavaScript，最好和最简单的答案似乎是/.\*/。

正如其他人所建议的/(.*?)/也可以，但是/。\*/更简单。()里面的图案是不需要的，就我所见也没有结局?匹配任何内容(包括空字符串)

的办法:

/[\s\ s]/不匹配空字符串，因此不是解决方案。 /[\s\ s]\*/ DOES也匹配空字符串。但它有一个问题:如果你在代码中使用它，那么你就不能注释掉这样的代码，因为*/被解释为注释结束。

/([\s\ s]\*)/可以工作并且不存在注释问题。但是它比/.*/要长，理解起来也更复杂。

Regex: /I bought.*sheep./ Matches - the whole string till the end of line I bought sheep. I bought a sheep. I bought five sheep. Regex: /I bought(.*)sheep./ Matches - the whole string and also capture the sub string within () for further use I bought sheep. I bought a sheep. I bought five sheep. I boughtsheep. I bought a sheep. I bought fivesheep. Example using Javascript/Regex 'I bought sheep. I bought a sheep. I bought five sheep.'.match(/I bought(.*)sheep./)[0]; Output: "I bought sheep. I bought a sheep. I bought five sheep." 'I bought sheep. I bought a sheep. I bought five sheep.'.match(/I bought(.*)sheep./)[1]; Output: " sheep. I bought a sheep. I bought five "

(.*?)匹配任何东西-我已经使用它很多年了。

我建议使用/(?=.*…)/g

例子

const text1 = 'I am using regex';
/(?=.*regex)/g.test(text1) // true

const text2 = 'regex is awesome';
/(?=.*regex)/g.test(text2) // true

const text3 = 'regex is util';
/(?=.*util)(?=.*regex)/g.test(text3) // true

const text4 = 'util is necessary';
/(?=.*util)(?=.*regex)/g.test(text4) // false because need regex in text

使用regex101进行测试

我用这个:(.|\n)+对我来说就像一个魅力!