Replace substring of text matching regexp

Question

I have text that looks like something like this:

1. Must have experience in Java 2. Team leader...

I want to render this in HTML as an ordered list. Now adding the </li> tag to the end is simple enough:

s = replace(s, ". ", "</li>");

But how do I go about replacing the 1., 2. etc with <li>?

I have the regular expression \d*\.$ which matches a number with a period, but the problem is is that is a substring so matching 1. Must have experience in Java 2. Team leader with \d*\.$ returns false.

@Reimeus the user isn't parsing HTML, they're trying to generate it. — ctwheels
– ctwheels, Commented Nov 16, 2017 at 15:07

ctwheels · Accepted Answer · 2017-11-16 15:06:23Z

3

Code

See regex in use here

\d+\.\s+(.*?)\s*(?=\d+\.\s+|$)

Replace

<li>$1</li>\n

Results

Input

Must have experience in Java 2. Team leader...

Output

<li>Must have experience in Java</li>
<li>Team leader...</li>

Explanation

\d+ Match one or more digits
\. Match the dot character . literally
\s+ Match one or more whitespace characters
(.*?) Capture any character any number of times, but as few as possible, into capture group 1
\s* Match any number of whitespace characters
(?=\d+\.\s+|$) Positive lookahead ensuring either of the following doesn't match
1. \d+\.\s+
  - \d+ Match one or more digits
  - \. Match the dot character . literally
  - \s+ Match one or more whitespace characters
2. $ Assert position at the end of the line

answered Nov 16, 2017 at 15:06

ctwheels

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Comments

Youcef LAIDANI · Accepted Answer · 2017-11-16 15:08:05Z

2

But how do I go about replacing the 1., 2. etc with <li>?

You can use String#replaceAll which can allow regex instead of replace :

s = s.replaceAll("\\d+\\.\\s", "</li>");

Note

You don't need to use $ in the end of your regex.
You have to escape dot . because it's mean any character in regex
You can use \s for one space or \s* for zero or more spaces or \s+ for one or more space

edited Nov 16, 2017 at 15:08

answered Nov 16, 2017 at 15:01

Youcef LAIDANI

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3 Comments

mohammedkhan Over a year ago

This works, (although in your answer your string replaces the regex match with </li> when it needs to be <li>) but @ctwheels answer is much more robust.

mohammedkhan Over a year ago

oh sorry, I misunderstood that comment. If you look at @ctwheels's answer under "output" it's exactly what I'm looking for. Though to be fair I was more interested in getting the opening <li> tag, the closing one was less complicated.

Youcef LAIDANI Over a year ago

its ok @mohammedkhan it was just a comment and i'm happy that ctwheels answer help you ;)

Joop Eggen · Accepted Answer · 2017-11-16 15:11:47Z

0

We want

<ol>
  <li>one</li>
  <li>two<li>
</ol>

This can be done as:

s = s.replaceAll("(?s)(\\d+\\.)\\s+(.*\\.)\\s*", "<li>$2</li></ol>");
s = s.replaceFirst("<li>", "<ol><li>");
s = s.replaceAll("(?s)</li></ol><li>", "</li>\n<li>");

The trick is to first add </li></ol> with a spurious </ol> that should only remain after the last list item.

(?s) is the DOTALL notation, causing . to also match line breaks.

In case of more than one numbered list this will not do. Also it assumes one single sentence per list item.

answered Nov 16, 2017 at 15:11

Joop Eggen

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