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    <pre><code>private static String combineRE(String p1, String p2){ int groups1 = 0, groups2=0; StringBuilder newP = new StringBuilder("(?="); newP.append(p1); newP.append("$)(?="); Pattern capturingGroup = Pattern.compile("(?&lt;!\\\\)(\\\\\\\\)*\\((?!\\?)"); Matcher m = capturingGroup.matcher(p1); while(m.find()) groups1 ++; m = capturingGroup.matcher(p2); while(m.find()) groups2 ++; String new2 = p2; for(int i=1; i&lt;=groups2; i++) new2 = new2.replaceAll("(?&lt;!\\\\)\\\\"+i, "\\\\" + (i+groups1)); newP.append(new2); newP.append("$).*"); return newP.toString(); } </code></pre> <p>This function uses the basic structure <code>(?=p1$)(?=p2$).*</code>, while recounting the numbered backreferences int the second pattern. It uses a regular expression to count the number of capturing group openers (unescaped <code>(</code>s not followed by a <code>?</code>) in each pattern, then updates the backrefences in the second pattern before placing it in the resultant pattern. I've set up a testing environment with ideone: Please, add all the test cases you can think of, but I think this answers your question.</p> <p><a href="http://ideone.com/Wm8cRc" rel="nofollow">http://ideone.com/Wm8cRc</a></p> <p>Round 2:</p> <p>There's no good way to generate a Pattern that will <code>find()</code> a substring to match two patterns. I toyed briefly with <code>(?=p1(?&lt;p2)).*(?&lt;(?=p1)p2)</code> and other such nonsense before giving up and writing, instead, an algorithm. First, I slightly modified my CombineRE from before:</p> <pre><code>private static String combineRE(String p1, String p2, boolean anchors){ int groups1 = 0, groups2=0; StringBuilder newP = new StringBuilder((anchors)?"^(?=":"(?="); newP.append(p1); if (anchors) newP.append('$'); newP.append(")(?="); Pattern capturingGroup = Pattern.compile("(?&lt;!\\\\)(\\\\\\\\)*\\((?!\\?)"); Matcher m = capturingGroup.matcher(p1); while(m.find()) groups1 ++; m = capturingGroup.matcher(p2); while(m.find()) groups2 ++; String new2 = p2; for(int i=1; i&lt;=groups2; i++) new2 = new2.replaceAll("(?&lt;!\\\\)\\\\"+i, "\\\\" + (i+groups1)); newP.append(new2); if (anchors) newP.append('$'); newP.append(')'); if (anchors) newP.append(".*"); return newP.toString(); } </code></pre> <p>You'll see that it now supports optional anchors. I used this functionality in my new function:</p> <pre><code>private static String[] findAllCombinedRE(String p1, String p2, String haystack, boolean overlap){ ArrayList&lt;String&gt; toReturn = new ArrayList&lt;String&gt;(); Pattern pCombo = Pattern.compile(combineRE(p1,p2, false)); String pComboMatch = combineRE(p1,p2, true); Matcher m = pCombo.matcher(haystack); int s = 0; while (m.find(s)){ String match = haystack.substring(m.start()); s = m.start()+1; for (int i=match.length(); i&gt;0; i--){ String sMatch = match.substring(0,i); if (Pattern.matches(pComboMatch, sMatch)){ toReturn.add(sMatch); /** * Note that at this point we can caluclute match * object like Information: * * group() = sMatch; * start() = m.start(); * end() = m.start() + i; * * If it so suited us, we could pass this information * back in a wrapped object. */ if (!overlap){ s = m.start()+i; break; } } } } return toReturn.toArray(new String[]{}); } </code></pre> <p>It uses an anchor free version of the two Regular Expressions to find all strings that might match, then chops off one letter at a time until the string matches the anchored version. It also includes a boolean to control for overlapping matches.</p> <p><a href="http://ideone.com/CBoBN5" rel="nofollow">http://ideone.com/CBoBN5</a> Works pretty well.</p>
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    1. POJava regex patterns conjunction pattern
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    1. USFrankieTheKneeMan
    1. USFrankieTheKneeMan
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      1. This table or related slice is empty.
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    1. COhttp://ideone.com/6oH3rA : find result should be `acb`
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      1. USTimofey Gorshkov
    2. COYou really can't switch back and forth like this. First you asked for `match()`, now you're asking to `find()` the largest substring that `match()`es. I'm sorry, you're not going to be able to do that in a single regular expression. The best you can do is take two regular expressions, and use the second one to try to `match()` each one the first one `find()`s.
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      1. USFrankieTheKneeMan
    3. COUnfortunately seems to be so. There is `|` logical operator but no `&` in regular expressions…
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      1. USTimofey Gorshkov
 

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