I have a list of alphanumeric strings as below
["nG5wnyPVNxS6PbbDNNbRsK5zanG94Et6Q4y74","GgQoDWqP7KtxXeePyyebu5EnNp8XxPC1odeNv","GgQoDWqP7KtxXeePyyebu5EnNp8XxPC1o12NN"]
I need to mask all elements with last 4 characters visible and [ " must not be masked as below.
["XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX4y74","XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXdeNv","XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX12NN"]
I have tried using
(\\W+)(\\W+)(\\w+)(\\w+)(\\w+)(\\w+)(\\w+)(\\W+)(\\W+)
as the key and $1$2XXXXXXXXXX$4$5$6$7$8$9 as the value in
maskedValue = maskedValue.replaceAll("(\\W+)(\\W+)(\\w+)(\\w+)(\\w+)(\\w+)(\\w+)(\\W+)(\\W+)", "$1$2XXXXXXXXXX$4$5$6$7$8$9")
but this only masked the first element.
["XXXXXXXXXXdeNv","nG5wnyPVNxS6PbbDNNbRsK5zanG94Et6Q4y74"]
Any leads are appreciated. Thanks in advance.
For a single value, you could use an assertion to match a word character asserting 4 characters at the end of the string.
\w(?=\w*\w{4}$)
Regex demo | Java demo
String values[] = {"nG5wnyPVNxS6PbbDNNbRsK5zanG94Et6Q4y74","GgQoDWqP7KtxXeePyyebu5EnNp8XxPC1odeNv","GgQoDWqP7KtxXeePyyebu5EnNp8XxPC1o12NN"};
for (String element : values)
System.out.println(element.replaceAll("\\w(?=\\w*\\w{4}$)", "X"));
Output
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX4y74
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXdeNv
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX12NN
For the whole string, you might use a finite quantifier in a positive lookbehind to match the opening " followed by a number of word characters. Then match all the characters that have 4 character before the closing "
"(?<=\"{0,100})\\w(?=\\w*\\w{4}\")"
Regex demo | Java demo
String regex = "(?<=\"{0,100})\\w(?=\\w*\\w{4}\")";
String string = "[\"nG5wnyPVNxS6PbbDNNbRsK5zanG94Et6Q4y74\",\"GgQoDWqP7KtxXeePyyebu5EnNp8XxPC1odeNv\",\"GgQoDWqP7KtxXeePyyebu5EnNp8XxPC1o12NN\"] ";
System.out.println(string.replaceAll(regex, "X"));
Output
["XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX4y74","XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXdeNv","XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX12NN"]
Using a stream:
List<String> terms = Arrays.asList(new String[] {
"nG5wnyPVNxS6PbbDNNbRsK5zanG94Et6Q4y74",
"GgQoDWqP7KtxXeePyyebu5EnNp8XxPC1odeNv",
"GgQoDWqP7KtxXeePyyebu5EnNp8XxPC1o12NN"
});
List<String> termsOut = terms.stream()
.map(t -> String.join("", Collections.nCopies(t.length() - 4, "x")) +
t.substring(t.length() - 4))
.collect(Collectors.toList());
System.out.println(termsOut);
This prints:
[xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx4y74,
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxdeNv,
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx12NN]
Note that this solution does not even use regex, which means it may outperform a regex based solution.
Assuming each of these strings will start and end with quotes
Algo:
Use a flag or stack data structure to know if it's a starting quote or ending quote.
For example:
Traverse the string. Initially flag will be false. When you encounter a new quote you have to flip flag and keep traversing till you find other quote. You can do the same with
Stack stack = new Stack<>();
Sample workflow:
String str="random";
boolean flag = false;
int idx = 0;
List<Pair<Integer, Integer>> indices = new ArrayList<>();
StringBuilder string = new StringBuilder(); // for final string
int start;
int end;
while(idx < str.length()){
if (str.charAt(idx) == '"' && !flag){
// start index of string
string.append(s.charAt(idx));
start = idx;
flag = true;
}
else if (str.charAt(idx) == '"' && !flag){
// end index of string
flag = false;
end = idx;
char[] mask = new char[end-3-start];
Arrays.fill(mask, 'x');
string.append(new String(mask)); // need to put 'x' in place
}
if (!flag){
string.append(s.charAt(idx));
}
idx++;
}
Complexity: O(n)
I want to split the string "004-034556" into two strings by the delimiter "-":
part1 = "004";
part2 = "034556";
That means the first string will contain the characters before '-', and the second string will contain the characters after '-'.
I also want to check if the string has '-' in it.
Use the appropriately named method String#split().
String string = "004-034556";
String[] parts = string.split("-");
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556
Note that split's argument is assumed to be a regular expression, so remember to escape special characters if necessary.
there are 12 characters with special meanings: the backslash \, the caret ^, the dollar sign $, the period or dot ., the vertical bar or pipe symbol |, the question mark ?, the asterisk or star *, the plus sign +, the opening parenthesis (, the closing parenthesis ), and the opening square bracket [, the opening curly brace {, These special characters are often called "metacharacters".
For instance, to split on a period/dot . (which means "any character" in regex), use either backslash \ to escape the individual special character like so split("\\."), or use character class [] to represent literal character(s) like so split("[.]"), or use Pattern#quote() to escape the entire string like so split(Pattern.quote(".")).
String[] parts = string.split(Pattern.quote(".")); // Split on the exact string.
To test beforehand if the string contains certain character(s), just use String#contains().
if (string.contains("-")) {
// Split it.
} else {
throw new IllegalArgumentException("String " + string + " does not contain -");
}
Note, this does not take a regular expression. For that, use String#matches() instead.
If you'd like to retain the split character in the resulting parts, then make use of positive lookaround. In case you want to have the split character to end up in left hand side, use positive lookbehind by prefixing ?<= group on the pattern.
String string = "004-034556";
String[] parts = string.split("(?<=-)");
String part1 = parts[0]; // 004-
String part2 = parts[1]; // 034556
In case you want to have the split character to end up in right hand side, use positive lookahead by prefixing ?= group on the pattern.
String string = "004-034556";
String[] parts = string.split("(?=-)");
String part1 = parts[0]; // 004
String part2 = parts[1]; // -034556
If you'd like to limit the number of resulting parts, then you can supply the desired number as 2nd argument of split() method.
String string = "004-034556-42";
String[] parts = string.split("-", 2);
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556-42
An alternative to processing the string directly would be to use a regular expression with capturing groups. This has the advantage that it makes it straightforward to imply more sophisticated constraints on the input. For example, the following splits the string into two parts, and ensures that both consist only of digits:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
class SplitExample
{
private static Pattern twopart = Pattern.compile("(\\d+)-(\\d+)");
public static void checkString(String s)
{
Matcher m = twopart.matcher(s);
if (m.matches()) {
System.out.println(s + " matches; first part is " + m.group(1) +
", second part is " + m.group(2) + ".");
} else {
System.out.println(s + " does not match.");
}
}
public static void main(String[] args) {
checkString("123-4567");
checkString("foo-bar");
checkString("123-");
checkString("-4567");
checkString("123-4567-890");
}
}
As the pattern is fixed in this instance, it can be compiled in advance and stored as a static member (initialised at class load time in the example). The regular expression is:
(\d+)-(\d+)
The parentheses denote the capturing groups; the string that matched that part of the regexp can be accessed by the Match.group() method, as shown. The \d matches and single decimal digit, and the + means "match one or more of the previous expression). The - has no special meaning, so just matches that character in the input. Note that you need to double-escape the backslashes when writing this as a Java string. Some other examples:
([A-Z]+)-([A-Z]+) // Each part consists of only capital letters
([^-]+)-([^-]+) // Each part consists of characters other than -
([A-Z]{2})-(\d+) // The first part is exactly two capital letters,
// the second consists of digits
Use:
String[] result = yourString.split("-");
if (result.length != 2)
throw new IllegalArgumentException("String not in correct format");
This will split your string into two parts. The first element in the array will be the part containing the stuff before the -, and the second element in the array will contain the part of your string after the -.
If the array length is not 2, then the string was not in the format: string-string.
Check out the split() method in the String class.
This:
String[] out = string.split("-");
should do the thing you want. The string class has many method to operate with a string.
// This leaves the regexes issue out of question
// But we must remember that each character in the Delimiter String is treated
// like a single delimiter
public static String[] SplitUsingTokenizer(String subject, String delimiters) {
StringTokenizer strTkn = new StringTokenizer(subject, delimiters);
ArrayList<String> arrLis = new ArrayList<String>(subject.length());
while(strTkn.hasMoreTokens())
arrLis.add(strTkn.nextToken());
return arrLis.toArray(new String[0]);
}
With Java 8:
List<String> stringList = Pattern.compile("-")
.splitAsStream("004-034556")
.collect(Collectors.toList());
stringList.forEach(s -> System.out.println(s));
Use org.apache.commons.lang.StringUtils' split method which can split strings based on the character or string you want to split.
Method signature:
public static String[] split(String str, char separatorChar);
In your case, you want to split a string when there is a "-".
You can simply do as follows:
String str = "004-034556";
String split[] = StringUtils.split(str,"-");
Output:
004
034556
Assume that if - does not exists in your string, it returns the given string, and you will not get any exception.
The requirements left room for interpretation. I recommend writing a method,
public final static String[] mySplit(final String s)
which encapsulate this function. Of course you can use String.split(..) as mentioned in the other answers for the implementation.
You should write some unit-tests for input strings and the desired results and behaviour.
Good test candidates should include:
- "0022-3333"
- "-"
- "5555-"
- "-333"
- "3344-"
- "--"
- ""
- "553535"
- "333-333-33"
- "222--222"
- "222--"
- "--4555"
With defining the according test results, you can specify the behaviour.
For example, if "-333" should return in [,333] or if it is an error.
Can "333-333-33" be separated in [333,333-33] or [333-333,33] or is it an error? And so on.
To summarize: there are at least five ways to split a string in Java:
String.split():
String[] parts ="10,20".split(",");
Pattern.compile(regexp).splitAsStream(input):
List<String> strings = Pattern.compile("\\|")
.splitAsStream("010|020202")
.collect(Collectors.toList());
StringTokenizer (legacy class):
StringTokenizer strings = new StringTokenizer("Welcome to EXPLAINJAVA.COM!", ".");
while(strings.hasMoreTokens()){
String substring = strings.nextToken();
System.out.println(substring);
}
Google Guava Splitter:
Iterable<String> result = Splitter.on(",").split("1,2,3,4");
Apache Commons StringUtils:
String[] strings = StringUtils.split("1,2,3,4", ",");
So you can choose the best option for you depending on what you need, e.g. return type (array, list, or iterable).
Here is a big overview of these methods and the most common examples (how to split by dot, slash, question mark, etc.)
You can try like this also
String concatenated_String="hi^Hello";
String split_string_array[]=concatenated_String.split("\\^");
Assuming, that
you don't really need regular expressions for your split
you happen to already use apache commons lang in your app
The easiest way is to use StringUtils#split(java.lang.String, char). That's more convenient than the one provided by Java out of the box if you don't need regular expressions. Like its manual says, it works like this:
A null input String returns null.
StringUtils.split(null, *) = null
StringUtils.split("", *) = []
StringUtils.split("a.b.c", '.') = ["a", "b", "c"]
StringUtils.split("a..b.c", '.') = ["a", "b", "c"]
StringUtils.split("a:b:c", '.') = ["a:b:c"]
StringUtils.split("a b c", ' ') = ["a", "b", "c"]
I would recommend using commong-lang, since usually it contains a lot of stuff that's usable. However, if you don't need it for anything else than doing a split, then implementing yourself or escaping the regex is a better option.
For simple use cases String.split() should do the job. If you use guava, there is also a Splitter class which allows chaining of different string operations and supports CharMatcher:
Splitter.on('-')
.trimResults()
.omitEmptyStrings()
.split(string);
The fastest way, which also consumes the least resource could be:
String s = "abc-def";
int p = s.indexOf('-');
if (p >= 0) {
String left = s.substring(0, p);
String right = s.substring(p + 1);
} else {
// s does not contain '-'
}
String Split with multiple characters using Regex
public class StringSplitTest {
public static void main(String args[]) {
String s = " ;String; String; String; String, String; String;;String;String; String; String; ;String;String;String;String";
//String[] strs = s.split("[,\\s\\;]");
String[] strs = s.split("[,\\;]");
System.out.println("Substrings length:"+strs.length);
for (int i=0; i < strs.length; i++) {
System.out.println("Str["+i+"]:"+strs[i]);
}
}
}
Output:
Substrings length:17
Str[0]:
Str[1]:String
Str[2]: String
Str[3]: String
Str[4]: String
Str[5]: String
Str[6]: String
Str[7]:
Str[8]:String
Str[9]:String
Str[10]: String
Str[11]: String
Str[12]:
Str[13]:String
Str[14]:String
Str[15]:String
Str[16]:String
But do not expect the same output across all JDK versions. I have seen one bug which exists in some JDK versions where the first null string has been ignored. This bug is not present in the latest JDK version, but it exists in some versions between JDK 1.7 late versions and 1.8 early versions.
There are only two methods you really need to consider.
Use String.split for a one-character delimiter or you don't care about performance
If performance is not an issue, or if the delimiter is a single character that is not a regular expression special character (i.e., not one of .$|()[{^?*+\) then you can use String.split.
String[] results = input.split(",");
The split method has an optimization to avoid using a regular expression if the delimeter is a single character and not in the above list. Otherwise, it has to compile a regular expression, and this is not ideal.
Use Pattern.split and precompile the pattern if using a complex delimiter and you care about performance.
If performance is an issue, and your delimiter is not one of the above, you should pre-compile a regular expression pattern which you can then reuse.
// Save this somewhere
Pattern pattern = Pattern.compile("[,;:]");
/// ... later
String[] results = pattern.split(input);
This last option still creates a new Matcher object. You can also cache this object and reset it for each input for maximum performance, but that is somewhat more complicated and not thread-safe.
You can split a string by a line break by using the following statement:
String textStr[] = yourString.split("\\r?\\n");
You can split a string by a hyphen/character by using the following statement:
String textStr[] = yourString.split("-");
public class SplitTest {
public static String[] split(String text, String delimiter) {
java.util.List<String> parts = new java.util.ArrayList<String>();
text += delimiter;
for (int i = text.indexOf(delimiter), j=0; i != -1;) {
String temp = text.substring(j,i);
if(temp.trim().length() != 0) {
parts.add(temp);
}
j = i + delimiter.length();
i = text.indexOf(delimiter,j);
}
return parts.toArray(new String[0]);
}
public static void main(String[] args) {
String str = "004-034556";
String delimiter = "-";
String result[] = split(str, delimiter);
for(String s:result)
System.out.println(s);
}
}
Please don't use StringTokenizer class as it is a legacy class that is retained for compatibility reasons, and its use is discouraged in new code. And we can make use of the split method as suggested by others as well.
String[] sampleTokens = "004-034556".split("-");
System.out.println(Arrays.toString(sampleTokens));
And as expected it will print:
[004, 034556]
In this answer I also want to point out one change that has taken place for split method in Java 8. The String#split() method makes use of Pattern.split, and now it will remove empty strings at the start of the result array. Notice this change in documentation for Java 8:
When there is a positive-width match at the beginning of the input
sequence then an empty leading substring is included at the beginning
of the resulting array. A zero-width match at the beginning however
never produces such empty leading substring.
It means for the following example:
String[] sampleTokensAgain = "004".split("");
System.out.println(Arrays.toString(sampleTokensAgain));
we will get three strings: [0, 0, 4] and not four as was the case in Java 7 and before. Also check this similar question.
One way to do this is to run through the String in a for-each loop and use the required split character.
public class StringSplitTest {
public static void main(String[] arg){
String str = "004-034556";
String split[] = str.split("-");
System.out.println("The split parts of the String are");
for(String s:split)
System.out.println(s);
}
}
Output:
The split parts of the String are:
004
034556
import java.io.*;
public class BreakString {
public static void main(String args[]) {
String string = "004-034556-1234-2341";
String[] parts = string.split("-");
for(int i=0;i<parts.length;i++) ｛
System.out.println(parts[i]);
}
}
}
You can use Split():
import java.io.*;
public class Splitting
{
public static void main(String args[])
{
String Str = new String("004-034556");
String[] SplittoArray = Str.split("-");
String string1 = SplittoArray[0];
String string2 = SplittoArray[1];
}
}
Else, you can use StringTokenizer:
import java.util.*;
public class Splitting
{
public static void main(String[] args)
{
StringTokenizer Str = new StringTokenizer("004-034556");
String string1 = Str.nextToken("-");
String string2 = Str.nextToken("-");
}
}
Here are two ways two achieve it.
WAY 1: As you have to split two numbers by a special character you can use regex
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class TrialClass
{
public static void main(String[] args)
{
Pattern p = Pattern.compile("[0-9]+");
Matcher m = p.matcher("004-034556");
while(m.find())
{
System.out.println(m.group());
}
}
}
WAY 2: Using the string split method
public class TrialClass
{
public static void main(String[] args)
{
String temp = "004-034556";
String [] arrString = temp.split("-");
for(String splitString:arrString)
{
System.out.println(splitString);
}
}
}
You can simply use StringTokenizer to split a string in two or more parts whether there are any type of delimiters:
StringTokenizer st = new StringTokenizer("004-034556", "-");
while(st.hasMoreTokens())
{
System.out.println(st.nextToken());
}
Check out the split() method in the String class on javadoc.
https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String)
String data = "004-034556-1212-232-232";
int cnt = 1;
for (String item : data.split("-")) {
System.out.println("string "+cnt+" = "+item);
cnt++;
}
Here many examples for split string but I little code optimized.
String str="004-034556"
String[] sTemp=str.split("-");// '-' is a delimiter
string1=004 // sTemp[0];
string2=034556//sTemp[1];
I just wanted to write an algorithm instead of using Java built-in functions:
public static List<String> split(String str, char c){
List<String> list = new ArrayList<>();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < str.length(); i++){
if(str.charAt(i) != c){
sb.append(str.charAt(i));
}
else{
if(sb.length() > 0){
list.add(sb.toString());
sb = new StringBuilder();
}
}
}
if(sb.length() >0){
list.add(sb.toString());
}
return list;
}
You can use the method split:
public class Demo {
public static void main(String args[]) {
String str = "004-034556";
if ((str.contains("-"))) {
String[] temp = str.split("-");
for (String part:temp) {
System.out.println(part);
}
}
else {
System.out.println(str + " does not contain \"-\".");
}
}
}
To split a string, uses String.split(regex). Review the following examples:
String data = "004-034556";
String[] output = data.split("-");
System.out.println(output[0]);
System.out.println(output[1]);
Output
004
034556
Note:
This split (regex) takes a regex as an argument. Remember to escape the regex special characters, like period/dot.
String s = "TnGeneral|DOMESTIC";
String a[]=s.split("\\|");
System.out.println(a.toString());
System.out.println(a[0]);
System.out.println(a[1]);
Output:
TnGeneral
DOMESTIC
String s="004-034556";
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)=='-')
{
System.out.println(s.substring(0,i));
System.out.println(s.substring(i+1));
}
}
As mentioned by everyone, split() is the best option which may be used in your case. An alternative method can be using substring().
Examples:
// A B C. -> A B C
// !A B C! -> !A B C
// A? B?? C??? -> A? B?? C
Here's what I have so far:
while (endsWithRegex(word, "\\p{P}")) {
word = word.substring(0, word.length() - 1);
}
public static boolean endsWithRegex(String word, String regex) {
return word != null && !word.isEmpty() &&
word.substring(word.length() - 1).replaceAll(regex, "").isEmpty();
}
This current solution works, but since it's already calling String.replaceAll within endsWithRegex, we should be able to do something like this:
word = word.replaceAll(/* regex */, "");
Any advice?
I suggest using
\s*\p{Punct}+\s*$
It will match optional whitespace and punctuation at the end of the string.
If you do not care about the whitespace, just use \p{Punct}+$.
Do not forget that in Java strings, backslashes should be doubled to denote literal backslashes (that must be used as regex escape symbols).
Java demo
String word = "!Words word! ";
word = word.replaceAll("\\s*\\p{Punct}+\\s*$", "");
System.out.println(word); // => !Words word
You can use:
str = str.replaceFirst("\\p{P}+$", "");
To include space also:
str = str.replaceFirst("[\\p{Space}\\p{P}]+$", "")
how about this, if you can take a minor hit in efficiency.
reverse the input string
keep removing characters until you hit an alphabet
reverse the string and return
I have modified the logic of your method
public static boolean endsWithRegex(String word, String regex) {
return word != null && !word.isEmpty() && word.matches(regex);
}
and your regex is : regex = ".*[^a-zA-Z]$";
Basically, I'm being passed a string and I need to tokenise it in much the same manner as command line options are tokenised by a *nix shell
Say I have the following string
"Hello\" World" "Hello Universe" Hi
How could I turn it into a 3 element list
Hello" World
Hello Universe
Hi
The following is my first attempt, but it's got a number of problems
It leaves the quote characters
It doesn't catch the escaped quote
Code:
public void test() {
String str = "\"Hello\\\" World\" \"Hello Universe\" Hi";
List<String> list = split(str);
}
public static List<String> split(String str) {
Pattern pattern = Pattern.compile(
"\"[^\"]*\"" + /* double quoted token*/
"|'[^']*'" + /*single quoted token*/
"|[A-Za-z']+" /*everything else*/
);
List<String> opts = new ArrayList<String>();
Scanner scanner = new Scanner(str).useDelimiter(pattern);
String token;
while ((token = scanner.findInLine(pattern)) != null) {
opts.add(token);
}
return opts;
}
So the incorrect output of the following code is
"Hello\"
World
" "
Hello
Universe
Hi
EDIT I'm totally open to a non regex solution. It's just the first solution that came to mind
If you decide you want to forego regex, and do parsing instead, there are a couple of options. If you are willing to have just a double quote or a single quote (but not both) as your quote, then you can use StreamTokenizer to solve this easily:
public static List<String> tokenize(String s) throws IOException {
List<String> opts = new ArrayList<String>();
StreamTokenizer st = new StreamTokenizer(new StringReader(s));
st.quoteChar('\"');
while (st.nextToken() != StreamTokenizer.TT_EOF) {
opts.add(st.sval);
}
return opts;
}
If you must support both quotes, here is a naive implementation that should work (caveat that a string like '"blah \" blah"blah' will yield something like 'blah " blahblah'. If that isn't OK, you will need to make some changes):
public static List<String> splitSSV(String in) throws IOException {
ArrayList<String> out = new ArrayList<String>();
StringReader r = new StringReader(in);
StringBuilder b = new StringBuilder();
int inQuote = -1;
boolean escape = false;
int c;
// read each character
while ((c = r.read()) != -1) {
if (escape) { // if the previous char is escape, add the current char
b.append((char)c);
escape = false;
continue;
}
switch (c) {
case '\\': // deal with escape char
escape = true;
break;
case '\"':
case '\'': // deal with quote chars
if (c == '\"' || c == '\'') {
if (inQuote == -1) { // not in a quote
inQuote = c; // now we are
} else {
inQuote = -1; // we were in a quote and now we aren't
}
}
break;
case ' ':
if (inQuote == -1) { // if we aren't in a quote, then add token to list
out.add(b.toString());
b.setLength(0);
} else {
b.append((char)c); // else append space to current token
}
break;
default:
b.append((char)c); // append all other chars to current token
}
}
if (b.length() > 0) {
out.add(b.toString()); // add final token to list
}
return out;
}
I'm pretty sure you can't do this by just tokenising on a regex. If you need to deal with nested and escaped delimiters, you need to write a parser. See e.g. http://kore-nordmann.de/blog/do_NOT_parse_using_regexp.html
There will be open source parsers which can do what you want, although I don't know any. You should also check out the StreamTokenizer class.
To recap, you want to split on whitespace, except when surrounded by double quotes, which are not preceded by a backslash.
Step 1: tokenize the input: /([ \t]+)|(\\")|(")|([^ \t"]+)/
This gives you a sequence of SPACE, ESCAPED_QUOTE, QUOTE and TEXT tokens.
Step 2: build a finite state machine matching and reacting to the tokens:
State: START
SPACE -> return empty string
ESCAPED_QUOTE -> Error (?)
QUOTE -> State := WITHIN_QUOTES
TEXT -> return text
State: WITHIN_QUOTES
SPACE -> add value to accumulator
ESCAPED_QUOTE -> add quote to accumulator
QUOTE -> return and clear accumulator; State := START
TEXT -> add text to accumulator
Step 3: Profit!!
I think if you use pattern like this:
Pattern pattern = Pattern.compile("\".*?(?<!\\\\)\"|'.*?(?<!\\\\)'|[A-Za-z']+");
Then it will give you desired output. When I ran with your input data I got this list:
["Hello\" World", "Hello Universe", Hi]
I used [A-Za-z']+ from your own question but shouldn't it be just : [A-Za-z]+
EDIT
Change your opts.add(token); line to:
opts.add(token.replaceAll("^\"|\"$|^'|'$", ""));
The first thing you need to do is stop thinking of the job in terms of split(). split() is meant for breaking down simple strings like this/that/the other, where / is always a delimiter. But you're trying to split on whitespace, unless the whitespace is within quotes, except if the quotes are escaped with backslashes (and if backslashes escape quotes, they probably escape other things, like other backslashes).
With all those exceptions-to-exceptions, it's just not possible to create a regex to match all possible delimiters, not even with fancy gimmicks like lookarounds, conditionals, reluctant and possessive quantifiers. What you want to do is match the tokens, not the delimiters.
In the following code, a token that's enclosed in double-quotes or single-quotes may contain whitespace as well as the quote character if it's preceded by a backslash. Everything except the enclosing quotes is captured in group #1 (for double-quoted tokens) or group #2 (single-quoted). Any character may be escaped with a backslash, even in non-quoted tokens; the "escaping" backslashes are removed in a separate step.
public static void test()
{
String str = "\"Hello\\\" World\" 'Hello Universe' Hi";
List<String> commands = parseCommands(str);
for (String s : commands)
{
System.out.println(s);
}
}
public static List<String> parseCommands(String s)
{
String rgx = "\"((?:[^\"\\\\]++|\\\\.)*+)\"" // double-quoted
+ "|'((?:[^'\\\\]++|\\\\.)*+)'" // single-quoted
+ "|\\S+"; // not quoted
Pattern p = Pattern.compile(rgx);
Matcher m = p.matcher(s);
List<String> commands = new ArrayList<String>();
while (m.find())
{
String cmd = m.start(1) != -1 ? m.group(1) // strip double-quotes
: m.start(2) != -1 ? m.group(2) // strip single-quotes
: m.group();
cmd = cmd.replaceAll("\\\\(.)", "$1"); // remove escape characters
commands.add(cmd);
}
return commands;
}
output:
Hello" World
Hello Universe
Hi
This is about as simple as it gets for a regex-based solution--and it doesn't really deal with malformed input, like unbalanced quotes. If you're not fluent in regexes, you might be better off with a purely hand-coded solution or, even better, a dedicated command-line interpreter (CLI) library.