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Java - Find the the words before and after a given word in a string

Say I have a string
String str = "This problem sucks and is hard"
and I wanted to get the words before and after "problem", so "This" and "sucks". Is regex the best way to accomplish this (keeping in mind that I'm a beginner with regex), or does Java have some kind of library (i.e. StringUtils) that can accomplish this for me?

To find the words before and after a given word, you can use this regex:
(\w+)\W+problem\W+(\w+)
The capture groups are the words you're looking for.
In Java, that would be:
Pattern p = Pattern.compile("(\\w+)\\W+problem\\W+(\\w+)");
Matcher m = p.matcher("This problem sucks and is hard");
if (m.find())
System.out.printf("'%s', '%s'", m.group(1), m.group(2));
Output
'This', 'sucks'
If you want full Unicode support, add flag UNICODE_CHARACTER_CLASS, or inline as (?U):
Pattern p = Pattern.compile("(?U)(\\w+)\\W+problema\\W+(\\w+)");
Matcher m = p.matcher("Questo problema è schifoso e dura");
if (m.find())
System.out.printf("'%s', '%s'", m.group(1), m.group(2));
Output
'Questo', 'è'
For finding multiple matches, use a while loop:
Pattern p = Pattern.compile("(?U)(\\w+)\\W+problems\\W+(\\w+)");
Matcher m = p.matcher("Big problems or small problems, they are all just problems, man!");
while (m.find())
System.out.printf("'%s', '%s'%n", m.group(1), m.group(2));
Output
'Big', 'or'
'small', 'they'
'just', 'man'
Note: The use of \W+ allows symbols to occur between words, e.g. "No(!) problem here" will still find "No" and "here".
Also note that a number is considered a word: "I found 1 problem here" returns "1" and "here".

There is a StringUtils library by apache which does have the methods to substring before and after the string. Additionally there is java's own substring which you can play with to get what you need.
Apache StringUtils library API:
https://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html
The methods that you might need - substringBefore() and substringBefore().
https://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#substringBefore(java.lang.String,%20java.lang.String)
Check this out if you want to explore java's own api's
Java: Getting a substring from a string starting after a particular character

A bit verbose but this gets the job done accurately and quickly:
import java.io.*;
import java.util.*;
public class HelloWorld{
public static void main(String []args){
String EntireString="Hello World this is a test";
String SearchWord="World";
System.out.println(getPreviousWordFromString(EntireString,SearchWord));
}
public static String getPreviousWordFromString(String EntireString, String SearchWord) {
List<Integer> IndicesOfWords = new ArrayList();
boolean isWord = false;
int indexOfSearchWord=-1;
if(EntireString.indexOf(SearchWord)!=-1) {
indexOfSearchWord = EntireString.indexOf(SearchWord)-1;
} else {
System.out.println("ERROR: SearchWord passed (2nd arg) does not exist in string EntireString. EntireString: "+EntireString+" SearchWord: "+SearchWord);
return "";
}
if(EntireString.indexOf(SearchWord)==0) {
System.out.println("ERROR: The search word passed is the first word in the search string, so there are no words before it.");
return "";
}
for (int i = 0; i < EntireString.length(); i++) {
if (Character.isLetter(EntireString.charAt(i)) && i != indexOfSearchWord) {
isWord = true;
} else if (!Character.isLetter(EntireString.charAt(i)) && isWord) {
IndicesOfWords.add(i);
isWord = false;
} else if (Character.isLetter(EntireString.charAt(i)) && i == indexOfSearchWord) {
IndicesOfWords.add(i);
}
}
if(IndicesOfWords.size()>0) {
boolean isFirstWordAWord=true;
for (int i = 0; i < IndicesOfWords.get(0); i++) {
if(!Character.isLetter(EntireString.charAt(i))) {
isFirstWordAWord=false;
}
}
if(isFirstWordAWord==true) {
String firstWord = EntireString.substring(0,IndicesOfWords.get(0));
IndicesOfWords.add(0,0);
}
} else {
return "";
}
String ResultingWord = "";
for (int i = IndicesOfWords.size()-1; i >= 0; i--) {
if (EntireString.substring(IndicesOfWords.get(i)).contains(SearchWord)) {
if (i > 0) {
ResultingWord=EntireString.substring(IndicesOfWords.get(i-1),IndicesOfWords.get(i));
break;
}
if (i==0) {
ResultingWord=EntireString.substring(IndicesOfWords.get(0),IndicesOfWords.get(1));
}
}
}
return ResultingWord;
}

does text contains substring

How to check if some String contains a specific String like "ABC72961". So we search for String which starts with "ABC" following by 5 digits. I've implemented a algorithm but I want it with "matches" or somehow else and then check the speed of these two solutions.

You may want to use regex for this
^ABC[0-9]{5}$
^ : Beginning of the string
ABC : Matches ABC literally (case-sensitive)
[0-9]{5} : Matches 5x numbers from 0 to 9
$ : End of the string
And use String#matches to test it
Regex101
Example
String regex = "^ABC[0-9]{5}$";
String one = "ABC72961";
String two = "ABC2345";
String three = "AB12345";
String four = "ABABABAB";
System.out.println(one.matches(regex)); // true
System.out.println(two.matches(regex)); // false
System.out.println(three.matches(regex)); // false
System.out.println(four.matches(regex)); // false
EDIT
Seeing your comment, you want it to work for String one = "textABC72961text" also. For that to be possible, you should just erase ^ and $ that limit the String.
.*ABC[0-9]{5}.*
EDIT 2
Here is if you want to extract it
if (s.matches(".*ABC[0-9]{5}.*")) {
Matcher m = Pattern.compile("ABC[0-9]{5}").matcher(s);
m.find();
result = m.group();
}

str.contains("ABC72961");
Returns true if str contains the string. False if not.

public String getString() {
String str = extractString();
return str;
}
public boolean exists() {
return !getString().trim().equals("") ? false : true;
}
private List<Integer> getPositionsOfABC() {
List<Integer> positions = new ArrayList<>();
int index = text.indexOf("ABC");
while (index > 0) {
positions.add(index);
index = text.indexOf("ABC", index + 1);
}
return positions;
}
private static boolean isInteger(String str) {
boolean isValidInteger = false;
try {
Integer.parseInteger(str);
isValidInteger = true;
} catch (NumberFormatException ex) {
return isValidInteger;
}
return isValidInteger;
}
private String extractString() {
List<Integer> positions = getPositionsOfABC();
for (Integer position : positions) {
int index = position.intValue();
String substring = text.substring(index, index + LENGTH_OF_DIGITS);
String lastDigits = substring.substring(3, substring.length());
if (isInteger(lastDigits)) {
return substring;
}
}
return "";
}

Here's a simple code that checks whether a substring exists in a string without using library functions, regex or other complex data structures.
class SSC {
public static void main(String[] args) {
String main_str <-- MAIN STRING
String sub_str <-- SUBSTRING
String w; int flag=0;
for(int i=0;i<=main_str.length()-sub_str.length();i++){
w="";
for(int j=0;j<sub_str.length();j++){
w+=main_str.charAt(i+j);
}
if(w.equals(sub_str))
flag++;
}
if(flag>0)
System.out.print("exists "+flag+" times");
else
System.out.print("doesn't exist");
}
}
Hope this helps.

I think what you want to use is java.util.regex.Pattern.
Pattern p = Pattern.compile("ABC(\d*)");
Matcher m = p.matcher("ABC72961");
boolean b = m.matches();
or if it shall be exactly 5 digits after "ABC", you can use the regex ABC(\d{5})
https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html#compile(java.lang.String)
Another solution would be:
String stringToTest = "ABC72961"
boolean b = stringToTest.contains("ABC");
http://www.tutorialspoint.com/java/lang/string_contains.htm

You can use the String indexOf command like this:
int result = someString.indexOf("ABC72961")
result will be -1 if there are no matches.
If there is a match, the result will be the index where the match starts.

how to Split name file to count series in java?

i have a list of files in this form:
name_of_file_001.csv
name_of_file_002.csv
name_of_file_123.csv
or
name_of_file.csv
second_name_of_file.csv
i don't know if the file has 001 or not.
how to take name of file (only name_of_file) in java?

Try the following:
int i=0;
while(!fullName.charAt(i).equals('.')&&!fullName.charAt(i).equals('0')){
i++;
}
String name=fullName.substring(0, i);
Take the string from the beginning of the fullName to the first appearance of . or 0.
EDIT:
Referring to the comments and the case of high numbers greater than 1.. and inspired from this answer:
int i=0;
String patternStr = "[0-9\.]";
Pattern pattern = Pattern.compile(patternStr);
Matcher matcher = pattern.matcher(fullName);
if(matcher.find()){
i=matcher.start(); //this will give you the first index of the regex
}
String name=fullName.substring(0, i);
EDIT2:
In the case where there's no Extension and the fullname doesn't match the regex(there's no numbers):
if(matcher.find()){
i=matcher.start(); //this will give you the first index of the regex
}else {
i=fullname.length();
}
String name=fullName.substring(0, i);
Or simply we will take all the name.

I modified chsdk's solution with respect to mmxx's comment:
int i=0;
while(i < fullName.length() && ".0123456789".indexOf(fullName.charAt(i)) == -1) {
i++;
}
String name=fullName.substring(0, i);
EDIT:
Added
i < fullName.length()

This little class solves the problem for all the examples shown in main:
public class Example {
private static boolean isNaturalNumber(String str)
{
return str.matches("\\d+(\\.\\d+)?");
}
public static String getFileName(String s) {
String fn = s.split("\\.")[0];
int idx = fn.lastIndexOf('_');
if (idx < 0) {
return fn;
}
String lastPart = fn.substring(idx+1);
System.out.println("last part = " + lastPart);
if (isNaturalNumber(lastPart)) {
return fn.substring(0,idx);
} else {
return fn;
}
}
public static void main(String []args){
System.out.println(getFileName("file_name_001.csv"));
System.out.println(getFileName("file_name_1234.csv"));
System.out.println(getFileName("file_name.csv"));
System.out.println(getFileName("file_name"));
System.out.println(getFileName("file"));
}
}
EDIT 1: Replaced the exception-based check with a regex check.
EDIT 2: Handling file names without any underscores.

i resolved the problem in this mode:
nameOfFile.split("\\.")[0].replaceall("_[0-9]*","");
split("\.")[0] remove ".csv" name_of_file_001.csv => name_of_file_001
.replaceall("_[0-9]*","") "remove, if there is, "_001" name_of_file_001 => name_of_file
the result is the name of file only

RegEx for dividing complex number String in Java

Looking for a Regular Expression in Java to separate a String that represents complex numbers. A code sample would be great.
The input string will be in the form:
"a+bi"
Example: "300+400i", "4+2i", "5000+324i".
I need to retrieve 300 & 400 from the String.'
I know we can do it crudely in this way.
str.substring(0, str.indexOf('+'));
str.substring(str.indexOf('+')+1,str.indexOf("i"));

I need to retrieve 300 & 400 from the String.
What about using String.split(regex) function:
String s[] = "300-400i".split("[\\Q+-\\Ei]");
System.out.println(s[0]+" "+s[1]); //prints 300 400

Regex that matches this is: /[0-9]{1,}[+-][0-9]{1,}i/
You can use this method:
Pattern complexNumberPattern = Pattern.compile("[0-9]{1,}");
Matcher complexNumberMatcher = complexNumberPattern.matcher(myString);
and use find and group methods on complexNumberMatcher to retrieve numbers from myString

Use this one:
[0-9]{1,}
It'll return the numbers.
Hope it helps.

Regex
([-+]?\d+\.?\d*|[-+]?\d*\.?\d+)\s*\+\s*([-+]?\d+\.?\d*|[-+]?\d*\.?\d+)i
Example Regex
http://rubular.com/r/FfOAt1zk0v
Example Java
string regexPattern =
// Match any float, negative or positive, group it
#"([-+]?\d+\.?\d*|[-+]?\d*\.?\d+)" +
// ... possibly following that with whitespace
#"\s*" +
// ... followed by a plus
#"\+" +
// and possibly more whitespace:
#"\s*" +
// Match any other float, and save it
#"([-+]?\d+\.?\d*|[-+]?\d*\.?\d+)" +
// ... followed by 'i'
#"i";
Regex regex = new Regex(regexPattern);
Console.WriteLine("Regex used: " + regex);
while (true)
{
Console.WriteLine("Write a number: ");
string imgNumber = Console.ReadLine();
Match match = regex.Match(imgNumber);
double real = double.Parse(match.Groups[1].Value, CultureInfo.InvariantCulture);
double img = double.Parse(match.Groups[2].Value, CultureInfo.InvariantCulture);
Console.WriteLine("RealPart={0};Imaginary part={1}", real, img);
}

Try this one. As for me, it works.
public static void main(String[] args) {
String[] attempts = new String[]{"300+400i", "4i+2", "5000-324i", "555", "2i", "+400", "-i"};
for (String s : attempts) {
System.out.println("Parsing\t" + s);
printComplex(s);
}
}
static void printComplex(String in) {
String[] parts = in.split("[+-]");
int re = 0, im = 0, pos = -1;
for (String s : parts) {
if (pos != -1) {
s = in.charAt(pos) + s;
} else {
pos = 0;
if ("".equals(s)) {
continue;
}
}
pos += s.length();
if (s.lastIndexOf('i') == -1) {
if (!"+".equals(s) && !"-".equals(s)) {
re += Integer.parseInt(s);
}
} else {
s = s.replace("i", "");
if ("+".equals(s)) {
im++;
} else if ("-".equals(s)) {
im--;
} else {
im += Integer.parseInt(s);
}
}
}
System.out.println("Re:\t" + re + "\nIm:\t" + im);
}
Output:
Parsing 300+400i
Re: 300
Im: 400
Parsing 4i+2
Re: 2
Im: 4
Parsing 5000-324i
Re: 5000
Im: -324
Parsing 555
Re: 555
Im: 0
Parsing 2i
Re: 0
Im: 2

In theory you could use something like this:
Pattern complexNumberPattern = Pattern.compile("(.*)+(.*)");
Matcher complexNumberMatcher = complexNumberPattern.matcher(myString);
if (complexNumberMatcher.matches()) {
String prePlus = complexNumberMatcher.group(1);
String postPlus = complexNumberMatcher.group(2);
}
The advantage this would give you over selecting the numbers, is that it would allow you to read things like:
5b+17c as 5b and 17c
edit: just noticed you didn't want the letters, so never mind, but this would give you more control over it in case other letters appear in it.

Palindrome Checker Java boolean insight

So I'm making a palindrome checker in java, and i've seemed to hit a roadblock this is my code so far:
public class StringUtil
{
public static void main(String[] args)
{
System.out.println("Welcome to String Util.");
Scanner word = new Scanner(System.in);
String X = word.nextLine();
String R = palindrome(X);
System.out.println();
System.out.println("Original Word: " + X);
System.out.println("Palindrome: " + R);
}
public static boolean palindrome(String word)
{
int t = word.length(); //length of the word as a number
int r = 0;
if(word.charAt(t) == word.charAt(r))
{
return true;
}
else
return false;
}
so far I only want it to check if the first letter is the same as the last, but when i compile it i get an incompatible types error on "String R = palindrome(X);" How would i get it to print true or false on the output statement below it?

Your palindrome method returns a boolean, but you're attempting to assign it to a String. Change the definition of the R variable to boolean.

Since the characters are 0-indexed in a string (or array) in java, the last character is at length - 1
try
int t = word.length() - 1;
Oops, that's not the problem you're having. However, you would notice it immediately once the type error is resolved.

word.charAt(t - 1), otherwise you over counted the string, also return "true"; and return "false" if you'd like to use String as your result type;

import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexMatches {
public static void main(String args[]) {
// String to be scanned to find the pattern.
String line = "This order was placed for QT3000! OK?";
String pattern = "(.*)(\\d+)(.*)";
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher m = r.matcher(line);
if (m.find()) {
System.out.println("Found value: " + m.group(0));
System.out.println("Found value: " + m.group(1));
System.out.println("Found value: " + m.group(2));
} else {
System.out.println("NO MATCH");
}
}
}