使用正则表达式获取两个不同字符之间的字符串

I'm struggling to find the right syntax for the following string:

我正在努力为以下字符串找到正确的语法：

^Lorem/ipsum dolor sit/amet consectetur adipiscing/elit

Here, I want to extract the word between ^ and / and the first word after the first /. That is: Lorem and ipsum.

在这里，我想提取^和/之间的单词和第一个/之后的第一个单词。那就是：Lorem和ipsum。

Moreover, how do I change the syntax if I want to extract the word after the second /? This would be amet.

此外，如果我想在第二个/之后提取单词，如何更改语法？这将是一个很好的结果。

2 个解决方案

#1

There are two parts:

有两个部分：

First you want to match everything after a ^ and before a /:

首先，你希望在^之后和/之前匹配所有内容：

/\^([^\/]*)/

/ \ ^（[^ \ /] *）/
- \^ matches the character ^ literally
- \ ^字面上匹配字符^
- 1st Capturing group ([^\/]*)
  - [^\/] match a single character not present in the list below
  - [^ \ /]匹配下面列表中不存在的单个字符
  - Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
  - 量词：*在零和无限次之间，尽可能多次，根据需要回馈[贪心]
- 第一个捕获组（[^ \ / * *] [^ \ /]匹配量化器下面列表中不存在的单个字符：*在零和无限次之间，尽可能多次，根据需要返回[贪婪]
- \/ matches the character / literally
- \ /匹配字符/字面
Second you want to match every word after a /:

其次你要匹配/后的每个单词：

/\/([^\s\/]*)/g

/ \ /（[^ \ S \ /] *）/克
- \/ matches the character / literally
- \ /匹配字符/字面
- 1st Capturing group ([^\s\/]*)
  - [^\s\/] match a single character not present in the list below
  - [^ \ s \ /]匹配下面列表中不存在的单个字符
  - Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
  - 量词：*在零和无限次之间，尽可能多次，根据需要回馈[贪心]
  - \s match any white space character [\r\n\t\f ]
  - \ s匹配任何空格字符[\ r \ n \ t \ f]
  - \/ matches the character / literally
  - \ /匹配字符/字面
- 第一个捕获组（[^ \ s \ /] *）[^ \ s \ /]匹配Quantifier下面列表中不存在的单个字符：*在零和无限次之间，尽可能多次，根据需要返回[ \ greedy] \ s匹配任何空格字符[\ r \ n \ t \ f] \ /匹配字符/字面意思
- g modifier: global. All matches (don't return on first match)
- g修饰符：全局。所有比赛（首场比赛不返回）

Code (PHP):

代码（PHP）：

$str = "^Lorem/ipsum dolor sit/amet consectetur adipiscing/elit"; 

// Part one:
preg_match("/\\^([^\\/]*)/", $str, $matches);

$matches[1] == 'Lorem'; // True    

// Part two:
preg_match_all("/\\/([^\\s\\/]*)/", $str, $matches); 

$matches[1] == 'ipsum';
$matches[2] == 'amet';
$matches[3] == 'elit';

Code and explaination from wonderful tool regex101.

精彩工具regex101的代码和解释。

#2

The 2 words are in capture group 1 and 2. Regex101.

这两个单词位于捕获组1和2中.Regex101。

\^(\w+)\/(\w+)

If you want to capture amet, use this:

如果要捕获amet，请使用以下命令：

\^(\w+)\/.+?\/(\w+)

#1