Question

我有一个已知开头和结尾的字符串，但我想只匹配未知中心。

例如，假设你知道你会有字符串说 “我今天午饭吃了_______”，你只想匹配空白。

以下是我的尝试：

^I had (.*) for lunch today$

哪个匹配整个字符串，以及组，即空白。

因此，当“我今天午餐吃披萨”时，会产生两场比赛： “我今天午餐吃披萨”和“披萨”

有没有办法只匹配空白？有没有办法获得“披萨”？或者至少将“披萨”作为第一场比赛？

Answer 1

你构建了一个捕获组;用它。

当您使用括号中包含的内容创建正则表达式时，在您的案例(.*?)中，您已创建了capturing group。与捕获组匹配的任何内容都将与完整匹配一起返回。

String.prototype.match()返回一个数组。它的第一个元素是完全匹配。所有后续元素都是捕获结果。由于您只有一个捕获组，因此结果为matches[1]。

var sentences = [
    "I had pizza for lunch today",
    "I had hamburger for lunch today",
    "I had coconut for lunch today",
    "I had nothing for lunch today"
];

sentences.map(function(sentence) {
   console.log(sentence + " -> " + sentence.match(/^I had (.*?) for lunch today$/)[1])
})

Answer 2

(?<=^I had )(.*?)(?= for lunch today$)

（？＆lt; =）和（？=）在此描述：what is the difference between ?:, ?! and ?= in regex?

=＆GT;

(?<=^I had ): It starts with "I had " but this is not captured.
(?= for lunch today$): It ends with " for lunch today" but this is not captured

=＆GT;

/(?<=^I had )(.*?)(?= for lunch today$)/_/

应该有效

或者像这样，如果不支持正面观察：

/(^I had )(.*?)(?= for lunch today$)/$1_/

=＆GT;我今天吃午饭了

Answer 3

它不使用RegEx并且不是最漂亮的，但是如果您知道包围您想要找到的字符串的单词，这可以完美地工作。

const testString = 'I had pizza for lunch today';

const getUnknown = (str, startWord, endWord) => str.split(startWord)[1]
  .split(endWord)[0]
  .trim();

console.log(getUnknown(testString, 'had', 'for'));

正则表达式填空字符串

3 个答案: