Java regex or expression which

Simple Logic or Regex Expresssion for a string

Is it possible to do it using regex in java. Logic I have thought of is using string manipulation (with substring,recursion etc). Is there a simple way of achieving this? I would prefer a regular expression which works in java. Other suggestions are also welcome. To give you a context The string above is coming as query parameter, I have to find out what all columns I need to select based on that. so all these individual items in the output will have a respective column name in property file. Thanks Pal

2 Answers 2

public class Main < public static void main(String[] args) < ; String input ="name,number,address(line1,test(city)),status,contact(id,phone(number,type),email(id),type),closedate"; Listlist = new ArrayList(Arrays.asList(input.split(","))); // We need a list for the iterator (or ArrayIterator) List result = new Main().parse(list); System.out.println(String.join(",", result)); > private List parse(List inputString) < Iteratorit = inputString.iterator(); ArrayList result = new ArrayList<>(); while(it.hasNext()) < String word = it.next(); if(! word.contains("("))< result.add(word); >else < // if we come across a "(", start the recursion and parse it till we find the matching ")" result.addAll(buildDistributedString(it, word,"")); >> return result; > /* * recursivly parse the string * @param startword The first word of it (containing the new prefix, the ( and the first word of this prefic * @param prefix Concatenation of previous prefixes in the recursion */ private List buildDistributedString(Iterator it, String startword,String prefix) < ArrayListresult = new ArrayList<>(); String[] splitted = startword.split("\\("); prefix += splitted[0]+"."; if(splitted[1].contains(")")) < //if the '(' is immediately matches, return only this one item result.add(prefix+splitted[1].substring(0,splitted[1].length()-1)); return result; >else < result.add(prefix+splitted[1]); >while(it.hasNext()) < String word = it.next(); if( word.contains("("))< // go deeper in the recursion ListstringList = buildDistributedString(it, word, prefix); if(stringList.get(stringList.size()-1).contains(")")) < // if multiple ")"'s were found in the same word, go up multiple recursion levels String lastString = stringList.remove(stringList.size()-1); stringList.add(lastString.substring(0,lastString.length() -1)); result.addAll(stringList); break; >result.addAll(stringList); > else if(word.contains(")")) < // end this recursion level result.add(prefix + word.substring(0,word.length()-1)); // ")" is always the last char break; >else < result.add(prefix+word); >> return result; > > 

I wrote a quick parser for this. There probably are some improvements possible, but this should give you an idea. It was just meant to get a working version asap.

Not sure, but I think it would also work with java 7. Even if not, I dont think it would be hard to rewrite the parts that aren’t.

String.join() was only introduced in java 8, and OP (or anybody else) might not know about it and assume your code doesn’t work

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Regular expressions in Java — Tutorial

A regular expression (regex) defines a search pattern for strings. The search pattern can be anything from a simple character, a fixed string or a complex expression containing special characters describing the pattern.

A regex can be used to search, edit and manipulate text, this process is called: The regular expression is applied to the text/string.

The regex is applied on the text from left to right. Once a source character has been used in a match, it cannot be reused. For example, the regex aba will match ababababa only two times (aba_aba__).

1.2. Regex examples

A simple example for a regular expression is a (literal) string. For example, the Hello World regex matches the «Hello World» string. . (dot) is another example for a regular expression. A dot matches any single character; it would match, for example, «a» or «1».

The following tables lists several regular expressions and describes which pattern they would match.

Matches exactly «this is text»

Matches the word «this» followed by one or more whitespace characters followed by the word «is» followed by one or more whitespace characters followed by the word «text».

^ defines that the patter must start at beginning of a new line. \d+ matches one or several digits. The ? makes the statement in brackets optional. \. matches «.», parentheses are used for grouping. Matches for example «5», «1.5» and «2.21».

2. Prerequisites

The following tutorial assumes that you have basic knowledge of the Java programming language.

Some of the following examples use JUnit Tutorial to validate the result. You should be able to adjust them in case if you do not want to use JUnit.

3. Rules of writing regular expressions

The following description is an overview of available meta characters which can be used in regular expressions. This chapter is supposed to be a references for the different regex elements.

3.1. Common matching symbols

Finds regex that must match at the beginning of the line.

Finds regex that must match at the end of the line.

Set definition, can match the letter a or b or c.

Set definition, can match a or b or c followed by either v or z.

When a caret appears as the first character inside square brackets, it negates the pattern. This pattern matches any character except a or b or c.

Ranges: matches a letter between a and d and figures from 1 to 7, but not d1.

Finds X directly followed by Z.

Checks if a line end follows.

3.2. Meta characters

The following meta characters have a pre-defined meaning and make certain common patterns easier to use. For example, you can use \d as simplified definition for [0..9] .

A whitespace character, short for [ \t\n\x0b\r\f]

A non-whitespace character, short for

A word character, short for [a-zA-Z_0-9]

Several non-whitespace characters

Matches a word boundary where a word character is [a-zA-Z0-9_]

These meta characters have the same first letter as their representation, e.g., digit, space, word, and boundary. Uppercase symbols define the opposite.

3.3. Quantifier

A quantifier defines how often an element can occur. The symbols ?, *, + and <> are qualifiers.

Occurs zero or more times, is short for

X* finds no or several letter X, .* finds any character sequence

Occurs one or more times, is short for

X+ — Finds one or several letter X

Occurs no or one times, ? is short for .

X? finds no or exactly one letter X

Occurs X number of times, <> describes the order of the preceding liberal

\d searches for three digits, . for any character sequence of length 10.

Occurs between X and Y times,

\d means \d must occur at least once and at a maximum of four.

? after a quantifier makes it a reluctant quantifier. It tries to find the smallest match. This makes the regular expression stop at the first match.

3.4. Grouping and back reference

You can group parts of your regular expression. In your pattern you group elements with round brackets, e.g., () . This allows you to assign a repetition operator to a complete group.

In addition these groups also create a back reference to the part of the regular expression. This captures the group. A back reference stores the part of the String which matched the group. This allows you to use this part in the replacement.

Via the $ you can refer to a group. $1 is the first group, $2 the second, etc.

Let’s, for example, assume you want to replace all whitespace between a letter followed by a point or a comma. This would involve that the point or the comma is part of the pattern. Still it should be included in the result.

// Removes whitespace between a word character and . or , String pattern = "(\\w)(\\s+)([\\.,])"; System.out.println(EXAMPLE_TEST.replaceAll(pattern, "$1$3"));

This example extracts the text between a title tag.

// Extract the text between the two title elements pattern = "(?i)()(.+?)()"; String updated = EXAMPLE_TEST.replaceAll(pattern, "$2");

3.5. Negative look ahead

Negative look ahead provides the possibility to exclude a pattern. With this you can say that a string should not be followed by another string.

Negative look ahead are defined via (?!pattern) . For example, the following will match «a» if «a» is not followed by «b».

3.6. Specifying modes inside the regular expression

You can add the mode modifiers to the start of the regex. To specify multiple modes, simply put them together as in (?ismx).

• (?i) makes the regex case insensitive.
• (?s) for «single line mode» makes the dot match all characters, including line breaks.
• (?m) for «multi-line mode» makes the caret and dollar match at the start and end of each line in the subject string.

3.7. Backslashes in Java

The backslash \ is an escape character in Java Strings. That means backslash has a predefined meaning in Java. You have to use double backslash \\ to define a single backslash. If you want to define \w , then you must be using \\w in your regex. If you want to use backslash as a literal, you have to type \\\\ as \ is also an escape character in regular expressions.

4. Using regular expressions with String methods

4.1. Redefined methods on String for processing regular expressions

Strings in Java have built-in support for regular expressions. Strings have four built-in methods for regular expressions: * matches() , * split()) , * replaceFirst() * replaceAll()

The replace() method does NOT support regular expressions.

These methods are not optimized for performance. We will later use classes which are optimized for performance.

Evaluates if «regex» matches s . Returns only true if the WHOLE string can be matched.

Creates an array with substrings of s divided at occurrence of «regex» . «regex» is not included in the result.

Replaces first occurance of «regex» with «replacement .

Replaces all occurances of «regex» with «replacement .

Create for the following example the Java project de.vogella.regex.test .

package de.vogella.regex.test; public class RegexTestStrings  public static final String EXAMPLE_TEST = "This is my small example " + "string which I'm going to " + "use for pattern matching."; public static void main(String[] args)  System.out.println(EXAMPLE_TEST.matches("\\w.*")); String[] splitString = (EXAMPLE_TEST.split("\\s+")); System.out.println(splitString.length);// should be 14 for (String string : splitString)  System.out.println(string); > // replace all whitespace with tabs System.out.println(EXAMPLE_TEST.replaceAll("\\s+", "\t")); > >

4.2. Examples

The following class gives several examples for the usage of regular expressions with strings. See the comment for the purpose.

If you want to test these examples, create for the Java project de.vogella.regex.string .

package de.vogella.regex.string; public class StringMatcher  // returns true if the string matches exactly "true" public boolean isTrue(String s) return s.matches("true"); > // returns true if the string matches exactly "true" or "True" public boolean isTrueVersion2(String s) return s.matches("[tT]rue"); > // returns true if the string matches exactly "true" or "True" // or "yes" or "Yes" public boolean isTrueOrYes(String s) return s.matches("[tT]rue|[yY]es"); > // returns true if the string contains exactly "true" public boolean containsTrue(String s) return s.matches(".*true.*"); > // returns true if the string contains of three letters public boolean isThreeLetters(String s) return s.matches("[a-zA-Z]"); // simpler from for // return s.matches("[a-Z][a-Z][a-Z]"); > // returns true if the string does not have a number at the beginning public boolean isNoNumberAtBeginning(String s) return s.matches("^[^\\d].*"); > // returns true if the string contains a arbitrary number of characters except b public boolean isIntersection(String s) return s.matches("([\\w&&[^b]])*"); > // returns true if the string contains a number less than 300 public boolean isLessThenThreeHundred(String s) return s.matches("[^0-9]*[12]?7[^0-9]*"); > >

And a small JUnit Test to validates the examples.

package de.vogella.regex.string; import org.junit.Before; import org.junit.Test; import static org.junit.Assert.assertFalse; import static org.junit.Assert.assertTrue; public class StringMatcherTest  private StringMatcher m; @Before public void setup() m = new StringMatcher(); > @Test public void testIsTrue()  assertTrue(m.isTrue("true")); assertFalse(m.isTrue("true2")); assertFalse(m.isTrue("True")); > @Test public void testIsTrueVersion2()  assertTrue(m.isTrueVersion2("true")); assertFalse(m.isTrueVersion2("true2")); assertTrue(m.isTrueVersion2("True"));; > @Test public void testIsTrueOrYes()  assertTrue(m.isTrueOrYes("true")); assertTrue(m.isTrueOrYes("yes")); assertTrue(m.isTrueOrYes("Yes")); assertFalse(m.isTrueOrYes("no")); > @Test public void testContainsTrue()  assertTrue(m.containsTrue("thetruewithin")); > @Test public void testIsThreeLetters()  assertTrue(m.isThreeLetters("abc")); assertFalse(m.isThreeLetters("abcd")); > @Test public void testisNoNumberAtBeginning()  assertTrue(m.isNoNumberAtBeginning("abc")); assertFalse(m.isNoNumberAtBeginning("1abcd")); assertTrue(m.isNoNumberAtBeginning("a1bcd")); assertTrue(m.isNoNumberAtBeginning("asdfdsf")); > @Test public void testisIntersection()  assertTrue(m.isIntersection("1")); assertFalse(m.isIntersection("abcksdfkdskfsdfdsf")); assertTrue(m.isIntersection("skdskfjsmcnxmvjwque484242")); > @Test public void testLessThenThreeHundred()  assertTrue(m.isLessThenThreeHundred("288")); assertFalse(m.isLessThenThreeHundred("3288")); assertFalse(m.isLessThenThreeHundred("328 8")); assertTrue(m.isLessThenThreeHundred("1")); assertTrue(m.isLessThenThreeHundred("99")); assertFalse(m.isLessThenThreeHundred("300")); > >

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