I am trying to remove all the leading zero from the number part of a string. I have came up with this code (below). From the given example it worked. But when I add a '0' in the begining it will not give the proper output. Anybody know how to achive this? Thanks in advance
input: (2016)abc00701def00019z -> output: (2016)abc701def19z -> resut: correct
input: 0(2016)abc00701def00019z -> output: (2016)abc71def19z -> result: wrong -> expected output: (2016)abc701def19z
EDIT: The string can contain other than english alphabet.
String localReference = "(2016)abc00701def00019z";
String localReference1 = localReference.replaceAll("[^0-9]+", " ");
List<String> lists = Arrays.asList(localReference1.trim().split(" "));
System.out.println(lists.toString());
String[] replacedString = new String[5];
String[] searchedString = new String[5];
int counter = 0;
for (String list : lists) {
String s = CharMatcher.is('0').trimLeadingFrom(list);
replacedString[counter] = s;
searchedString[counter++] = list;
System.out.println(String.format("Search: %s, replace: %s", list,s));
}
System.out.println(StringUtils.replaceEach(localReference, searchedString, replacedString));
str.replaceAll("(^|[^0-9])0+", "$1");
This removes any row of zeroes after non-digit characters and at the beginning of the string.
I tried doing the task using Regex and was able to do the required according to the two test cases you gave. Also $1 and $2 in the code below are the parts in the () brackets in preceding Regex.
Please find the code below:
public class Demo {
public static void main(String[] args) {
String str = "0(2016)abc00701def00019z";
/*Below line replaces all 0's which come after any a-z or A-Z and which have any number after them from 1-9. */
str = str.replaceAll("([a-zA-Z]+)0+([1-9]+)", "$1$2");
//Below line only replace the 0's coming in the start of the string
str = str.replaceAll("^0+","");
System.out.println(str);
}
}
java has \P{Alpha}+, which matches any non-alphabetic character and then removing the the starting Zero's.
String stringToSearch = "0(2016)abc00701def00019z";
Pattern p1 = Pattern.compile("\\P{Alpha}+");
Matcher m = p1.matcher(stringToSearch);
StringBuffer sb = new StringBuffer();
while(m.find()){
m.appendReplacement(sb,m.group().replaceAll("\\b0+",""));
}
m.appendTail(sb);
System.out.println(sb.toString());
output:
(2016)abc701def19z
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().
I am trying to get all words that begin with a letter from a long string. How would you do this is java? I don't want to loop through every letter or something inefficient.
EDIT: I also can't use any in built data structures (except arrays of course)- its for a cs class. I can however make my own data structures (which i have created sevral).
You could try obtaining an array collection from your String and then iterating through it:
String s = "my very long string to test";
for(String st : s.split(" ")){
if(st.startsWith("t")){
System.out.println(st);
}
}
You need to be clear about some things. What is a "word"? You want to find only "words" starting with a letter, so I assume that words can have other characters too. But what chars are allowed? What defines the start of such a word? Whitespace, any non letter, any non letter/non digit, ...?
e.g.:
String TestInput = "test séntènce îwhere I'm want,to üfind 1words starting $with le11ers.";
String regex = "(?<=^|\\s)\\pL\\w*";
Pattern p = Pattern.compile(regex, Pattern.UNICODE_CHARACTER_CLASS);
Matcher matcher = p.matcher(TestInput);
while (matcher.find()) {
System.out.println(matcher.group());
}
The regex (?<=^|\s)\pL\w* will find sequences that starts with a letter (\pL is a Unicode property for letter), followed by 0 or more "word" characters (Unicode letters and numbers, because of the modifier Pattern.UNICODE_CHARACTER_CLASS).
The lookbehind assertion (?<=^|\s) ensures that there is the start of the string or a whitespace before the sequence.
So my code will print:
test
séntènce ==> contains non ASCII letters
îwhere ==> starts with a non ASCII letter
I ==> 'm is missing, because `'` is not in `\w`
want
üfind ==> starts with a non ASCII letter
starting
le11ers ==> contains digits
Missing words:
,to ==> starting with a ","
1words ==> starting with a digit
$with ==> starting with a "$"
You could build a HashMap -
HashMap<String,String> map = new HashMap<String,String>();
example -
ant, bat, art, cat
Hashmap
a -> ant,art
b -> bat
c -> cat
to find all words that begin with "a", just do
map.get("a")
You can get the first letter of the string and check with API method that if it is letter or not.
String input = "jkk ds 32";
String[] array = input.split(" ");
for (String word : array) {
char[] arr = word.toCharArray();
char c = arr[0];
if (Character.isLetter(c)) {
System.out.println( word + "\t isLetter");
} else {
System.out.println(word + "\t not Letter");
}
}
Following are some sample output:
jkk isLetter
ds isLetter
32 not Letter
Scanner scan = new Scanner(text); // text being the string you are looking in
char test = 'x'; //whatever letter you are looking for
while(scan.hasNext()){
String wordFound = scan.next();
if(wordFound.charAt(0)==test){
//do something with the wordFound
}
}
this will do what you are looking for, inside the if statement do what you want with the word
Regexp way:
public static void main(String[] args) {
String text = "my very long string to test";
Matcher m = Pattern.compile("(^|\\W)(\\w*)").matcher(text);
while (m.find()) {
System.out.println("Found: "+m.group(2));
}
}
You can use split() method. Here is an example :
String string = "your string";
String[] parts = string.split(" C");
for(int i=0; i<parts.length; i++) {
String[] word = parts[i].split(" ");
if( i > 0 ) {
// ignore the rest words because don't starting with C
System.out.println("C" + word[0]);
}
else { // Check 1st excplicitly
for(int j=0; j<word.length; j++) {
if ( word[j].startsWith("c") || word[j].startsWith("C"))
System.out.println(word[j]);
}
}
}
where "C" is you letter. Just then loop around the array. For parts[0] you have to check if it starts with "C". It was my mistake to start looping from i=1. The correct is from 0.
I'm trying to remove a specific word from a certain string using the function replace() or replaceAll() but these remove all the occurrences of this word even if it's part of another word!
Example:
String content = "is not like is, but mistakes are common";
content = content.replace("is", "");
output: "not like , but mtakes are common"
desired output: "not like , but mistakes are common"
How can I substitute only whole words from a string?
What the heck,
String regex = "\\s*\\bis\\b\\s*";
content = content.replaceAll(regex, "");
Remember you need to use replaceAll(...) to use regular expressions, not replace(...)
\\b gives you the word boundaries
\\s* sops up any white space on either side of the word being removed (if you want to remove this too).
content = content.replaceAll("\\Wis\\W|^is\\W|\\Wis$", "");
You can try replacing " is " by " ". The is with a space before and one after, replaced by a single space.
Update:
To make it work for the first "is" in the sentence, also do another replace of "is " for "". Replacing the first is and the first space, with an empty string.
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
String input = s.nextLine();
char c = s.next().charAt(0);
System.out.println(removeAllOccurrencesOfChar(input, c));
}
public static String removeAllOccurrencesOfChar(String input, char c) {
String r = "";
for (int i = 0; i < input.length(); i ++) {
if (input.charAt(i) != c) r += input.charAt(i);
}
return r;
}
}