Regex: how to substitute a string with n occurrences of a substring

Question

As a premise, I have an HTML text, with some <ol> elements. These have a start attribute, but the framework I'm using is not capable to interpret them during a PDF conversion. So, the trick I am trying to apply is to add a number of invisible <li> elements at the beginning.

As an example, suppose this input text:

<ol start="3">
   <li>Element 1</li>
   <li>Element 2</li>
   <li>Element 3</li>
</ol>

I want to produce this result:

<ol>
   <li style="visibility:hidden"></li>
   <li style="visibility:hidden"></li>
   <li>Element 1</li>
   <li>Element 2</li>
   <li>Element 3</li>
</ol>

So, adding n-1 invisible elements into the ordered list. But I'm not able to do that from Java in a generalized way.

Supposing the exact case in the example, I could do this (using replace, so - to be honest - without regex):

htmlString = htmlString.replace("<ol start=\"3\">",
            "<ol><li style=\"visibility:hidden\"></li><li style=\"visibility:hidden\"></li>");

But, obviously, it just applies to the case with "start=3". I know that I can use groups to extract the "3", but how can I use it as a "variable" to specify the string <li style=\"visibility:hidden\"></li> n-1 number of times? Thanks for any insight.

Using Python, you could use `re.sub` with a callback method: `re.sub(r'
', lambda m: '
' + ('\n
' * int(m.group(1))), s)` — tobias_k, Sep 06 '18 at 14:09
@SL5net: thank you, but unfortunately this does not solve my problem. In fact, I don't know in advance the number of occurrences. — Andrea, Sep 06 '18 at 14:10
@tobias_k: That's the power of Phyton :) in my case, I am developing a webapp with Java, so it's useless to me. But I hope your comment could help others. — Andrea, Sep 06 '18 at 14:11
@Andreaジーティーオー Looks like Java has a replace-with-callback function now, too. :-) — tobias_k, Sep 06 '18 at 14:24

The Coding Monk · Answer 1 · 2018-09-06T14:01:45.647

4

You cannot do this using regular expressions, or even if you find some hack to do this it's going to be a suboptimal solution..

The right way to do this is to use an HTML parsing library (e.g. Jsoup) and then add the <li> tags as children to the <ol>, specifically using the Element#prepend method. (With Jsoup you can also read the start attribute value in order to compute how many elements to add)

edited Sep 06 '18 at 14:01

answered Sep 06 '18 at 13:56

The Coding Monk

7,684
12
41
56

1

Thank you so much for your answer. To be honest, I suspected that it was too much, even for a powerful tool like regex. Thanks for the suggestions as well. I'm going to investigate further using Jsoup. – Andrea Sep 06 '18 at 14:03
1

Or if you could rely upon the HTML to be well-formed XHTML (the example is consistent with this), then you could do the job with an XSL transform. The bottom line is that you really need to parse and manipulate HTML / XHTML input with an appropriate tool. Regex is not such a tool in general, although it may be applicable to certain narrow, well-defined problems in this general area. – John Bollinger Sep 06 '18 at 14:04
I downvoted because you wrote the solution but did not provide an example snippet. – Paolo Sep 06 '18 at 14:20
@UnbearableLightness: I respect your comment, but I disagree with you. I don't see in this a valid reason for a downvote, since my question has been addressed (and it wasn't Jsoup-related at the beginning). – Andrea Sep 06 '18 at 14:25
In my opinion the answer is not complete. On SO one should always include a snippet which solves the problem rather than just providing documentation links (which could be unavailable in the future, therefore making the answer useless for future readers). – Paolo Sep 06 '18 at 14:33
2

I understand that in general link-only answers are not welcome but: 1) this is not a link-only answer since I detailed a general procedure (prepend and append are standard methods of DOM manipulation library), providing links to a specific implementation as a plus; 2) your reasoning does not apply in this case because if the link is not available it probably means that the library is not available, so good luck using that code snippet then. – The Coding Monk Sep 06 '18 at 14:40
You are not correct about your second point. As an example, I have often found MATLAB documentation pages which were no longer available despite the function still being valid. – Paolo Sep 06 '18 at 15:19

score 3 · Accepted Answer · edited Sep 06 '18 at 15:29

Since Java 9, there's a Matcher.replaceAll method taking a callback function as a parameter:

String text = "<ol start=\"3\">\n\t<li>Element 1</li>\n\t<li>Element 2</li>\n\t<li>Element 3</li>\n</ol>";

String result = Pattern
        .compile("<ol start=\"(\\d)\">")
        .matcher(text)
        .replaceAll(m -> "<ol>" + repeat("\n\t<li style=\"visibility:hidden\" />", 
                                         Integer.parseInt(m.group(1))-1));

To repeat the string you can take the trick from here, or use a loop.

public static String repeat(String s, int n) {
    return new String(new char[n]).replace("\0", s);
}

Afterwards, result is:

<ol>
    <li style="visibility:hidden" />
    <li style="visibility:hidden" />
    <li>Element 1</li>
    <li>Element 2</li>
    <li>Element 3</li>
</ol>

If you are stuck with an older version of Java, you can still match and replace in two steps.

Matcher m = Pattern.compile("<ol start=\"(\\d)\">").matcher(text);
while (m.find()) {
    int n = Integer.parseInt(m.group(1));
    text = text.replace("<ol start=\"" + n + "\">", 
            "<ol>" + repeat("\n\t<li style=\"visibility:hidden\" />", n-1));
}

Update by Andrea ジーティーオー:

I modified the (great) solution above for including also <ol> that have multiple attributes, so that their tag do not end with start (example, <ol> with letters, as <ol start="4" style="list-style-type: upper-alpha;">). This uses replaceAll to deal with regex as a whole.

//Take something that starts with "<ol start=", ends with ">", and has a number in between
Matcher m = Pattern.compile("<ol start=\"(\\d)\"(.*?)>").matcher(htmlString);
while (m.find()) {
    int n = Integer.parseInt(m.group(1));
    htmlString = htmlString.replaceAll("(<ol start=\"" + n + "\")(.*?)(>)",
            "<ol $2>" + StringUtils.repeat("\n\t<li style=\"visibility:hidden\" />", n - 1));
}

Thank you, this is just great! But, once again, I can't use this :( the project I'm working on is a bit old one, and we upgraded to Java 7 just recently. Very appreciated, by the way, and useful for others of course. — Andrea, Sep 06 '18 at 14:28
I have no choice than accept this answer :) tested it, and works perfectly. I took the liberty of updating your answer, to include the more generalized solution I've ended up with in Java 7. Thank you so much again. — Andrea, Sep 06 '18 at 15:17

Eritrean · Answer 3 · 2018-09-06T14:57:05.783

Using Jsoup you can write something like:

import org.jsoup.Jsoup;
import org.jsoup.nodes.Document;
import org.jsoup.nodes.Element;
import org.jsoup.select.Elements;

class JsoupTest {
    public static void main(String[] args){
        String html = "<ol start=\"3\">\n" +
                        "   <li>Element 1</li>\n" +
                        "   <li>Element 2</li>\n" +
                        "   <li>Element 3</li>\n" +
                        "</ol>"
                + "<p>some other html elements</p>"
                + "<ol start=\"5\">\n" +
                        "   <li>Element 1</li>\n" +
                        "   <li>Element 2</li>\n" +
                        "   <li>Element 3</li>\n" +
                        "   <li>Element 4</li>\n" +
                        "   <li>Element 5</li>\n" +
                        "</ol>";

        Document doc = Jsoup.parse(html);
        Elements ols = doc.select("ol");
        for(Element ol :ols){
            int start = Integer.parseInt(ol.attr("start"));
            for(int i=0; i<start-1; i++){
                ol.prependElement("li").attr("style", "visibility:hidden");
            }  
            ol.attributes().remove("start");
            System.out.println(ol);
        }
    }
}

Thank you, I appreciate a ton! I think that your snippet does not consider multiple
, but I can adapt this in a more generalized way. — Andrea, Sep 06 '18 at 14:33

score 0 · Answer 4 · answered Sep 06 '18 at 14:57

You can try this.

String input="<ol    start=\"6\">"+
   "<li>Element 1</li>"+
   "<li>Element 2</li>"+
   "<li>Element 3</li>"+
   "<li>Element 4</li>"+
   "<li>Element 5</li>"+
   "<li>Element6</li>"+
"</ol>";

 Matcher match= Pattern.compile("<ol .*start.*=.*\\\"(.*)\\\"\\s*>(.*)(</ol>)").matcher(input);
    String resultString ="";
    if(match.find()){
    resultString =match.replaceAll("<ol>"+new String(new char[Integer.valueOf(match.group(1))-1]).replace("\0", "\n\t<li style=\"visibility:hidden\" />")+"$2$3");  

}

score -2 · Answer 5 · answered Sep 06 '18 at 14:36

-2

Please use java Matcher and Pattern to count the occurrence of li tag and use StringBuilder insert method to insert invisible elements.

Matcher m = Pattern.compile("<li>").matcher(s);
        while(m.find()){
           ++count;
        }

answered Sep 06 '18 at 14:36

Sudha mohan Panda

7
2

2

I don't see how this helps OP. First, OP is not counting `
` elements, but searching for a number in the `

tobias_k

Sep 06 '18 at 14:38

Yes, I see that it counts the `

`, but that's not what OP wants. OP wants to get the number from the `

` in the right place.

– tobias_k Sep 06 '18 at 14:59

Regex: how to substitute a string with n occurrences of a substring

5 Answers5

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