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    <p>Whether or not regular expressions are well suited for this task is debatable. Most people would probably argue that it's not.</p> <p>As I understand it however, you have no choice as the API you're using takes a regular expression as argument, so here goes...</p> <h2>Code</h2> <pre><code>public class NumericRangeRegexGenerator { private static String baseRange(String num, boolean up, boolean leading1) { char c = num.charAt(0); char low = up ? c : leading1 ? '1' : '0'; char high = up ? '9' : c; if (num.length() == 1) return charClass(low, high); String re = c + "(" + baseRange(num.substring(1), up, false) + ")"; if (up) low++; else high--; if (low &lt;= high) re += "|" + charClass(low, high) + nDigits(num.length() - 1); return re; } private static String charClass(char b, char e) { return String.format(b==e ? "%c" : e-b&gt;1 ? "[%c-%c]" : "[%c%c]", b, e); } private static String nDigits(int n) { return nDigits(n, n); } private static String nDigits(int n, int m) { return "[0-9]" + String.format(n==m ? n==1 ? "":"{%d}":"{%d,%d}", n, m); } private static String eqLengths(String from, String to) { char fc = from.charAt(0), tc = to.charAt(0); if (from.length() == 1 &amp;&amp; to.length() == 1) return charClass(fc, tc); if (fc == tc) return fc + "("+rangeRegex(from.substring(1), to.substring(1))+")"; String re = fc + "(" + baseRange(from.substring(1), true, false) + ")|" + tc + "(" + baseRange(to.substring(1), false, false) + ")"; if (++fc &lt;= --tc) re += "|" + charClass(fc, tc) + nDigits(from.length() - 1); return re; } private static String nonEqLengths(String from, String to) { String re = baseRange(from,true,false) + "|" + baseRange(to,false,true); if (to.length() - from.length() &gt; 1) re += "|[1-9]" + nDigits(from.length(), to.length() - 2); return re; } public static String rangeRegex(int n, int m) { return rangeRegex("" + n, "" + m); } public static String rangeRegex(String n, String m) { return n.length() == m.length() ? eqLengths(n, m) : nonEqLengths(n, m); } } </code></pre> <h3>Usage</h3> <pre><code>// Generate expression for range 123 - 321 String regexp = NumericRangeRegexGenerator.rangeRegex(123, 321); </code></pre> <p><br/></p> <h2>Explanation</h2> <p>A brief explanation of the code follows.</p> <h3>Ranges on the shape <code>0000</code>-<code>abcd</code> and <code>abcd</code>-<code>9999</code></h3> <p>First we note that matching ranges such as <code>0000</code>-<code>abcd</code> is fairly easy.</p> <p>An expression covering for instance <code>000</code>-<code>527</code> can be expressed as</p> <ul> <li><code>[0-4]</code> followed by two arbitrary digits, or</li> <li><code>5</code> followed by <code>00</code>-<code>27</code> (which is resolved recursively!)</li> </ul> <p>Ranges on the shape <code>1000</code>-<code>abcd</code> and <code>abcd</code>-<code>9999</code> are just as easy.</p> <h3>Lower limit, upper limit of different lengths.</h3> <p>If the "from"-number is shorter than the "to"-number it is fairly straight forward.</p> <p>Assume for instance that the <code>from</code>-number has 3 digits and the <code>to</code>-number has 7 digits. The expression can then be composed as follows:</p> <ul> <li><code>from</code>-<code>999</code> (as described above),</li> <li>Any <code>4</code>, <code>5</code> or <code>6</code> digit number: <code>[1-9][0-9]{3-5}</code>, or</li> <li><code>1000000</code>-<code>to</code> (as described above)</li> </ul> <h3>Lower limit / upper limit of equal lengths.</h3> <p>This is the trickiest situation (still not <em>that</em> tricky though!)</p> <p>The solution is, again, best described by an example. Consider the range <code>273</code> - <code>548</code>. The expression can be composed by the following parts:</p> <ul> <li><code>2</code> followed by <code>73</code>-<code>99</code> (latter part described above),</li> <li><code>[34]</code> followed by any two digits, or</li> <li><code>5</code> followed by <code>00</code>-<code>48</code> (latter part described above)</li> </ul>
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    1. POHow to generate a regular expression at runtime to match a numeric range
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