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Splitting string at parentheses in Java

Now,if I have a String like this:
String start = "(1374)(48.4%)(32)(100%)(290)(43.1%)";
How can I extract the six numbers 1374 48.4 32 100 290 43.1 or 1374 48.4% 32 100% 290 43.1%? Can it be done with a regex?

You can search for a regex identifying floating point numbers: ([+-]?(\d+\.)?\d+)
String start = "(1374)(48.4%)(32)(100%)(290)(43.1%)";
Pattern p = Pattern.compile("([+-]?(\\d+\\.)?\\d+)");
Matcher m = p.matcher(start);
while (m.find()) {
System.out.println(m.group(1));
}
Or use a regex that also makes sure that the brackets are there:
Pattern p = Pattern.compile("\\(([+-]?(\\d+\\.)?\\d+)\\%?\\)");

Let's do it without a regex!
int i = 0;
while (i < start.length()) {
while (i < start.length()) {
char ch = start.charAt(i);
// Maybe add other characters, e.g. %, if desired.
if (Character.isDigit(ch) || ch == '.') {
break;
}
++i;
}
int startOfBlock = i;
while (i < start.length()) {
char ch = start.charAt(i);
if (!Character.isDigit(ch) && ch != '.') {
break;
}
++i;
}
if (i > startOfBlock) {
System.out.println(start.substring(startOfBlock, i));
}
}

or u can try the following regex [\d\.%]+ it gives u the combinations of the strings containing the following items \d (digit) \. dot and % procentage sign one or more times click here for a live demo

String start = "(1374)(48.4%)(32)(100%)(290)(43.1%)";
for(String splitString : start.split("[()]")) {
System.out.print(splitString + " ");
}

Add separator in string using regex in Java

I have a string (for example: "foo12"), and I want to add a delimiting character in between the letters and numbers (e.g. "foo|12"). However, I can't seem to figure out what the appropriate code is for doing this in Java. Should I use a regex + replace or do I need to use a matcher?

A regex replace would be just fine:
String result = subject.replaceAll("(?<=\\p{L})(?=\\p{N})", "|");
This looks for a position right after a letter and right before a digit (by using lookaround assertions). If you only want to look for ASCII letters/digits, use
String result = subject.replaceAll("(?i)(?<=[a-z])(?=[0-9])", "|");

Split letters and numbers and concatenate with "|". Here is a one-liner:
String x = "foo12";
String result = x.replaceAll("[0-9]", "") + "|" + x.replaceAll("[a-zA-Z]", "");
Printing result will output: foo|12

Why even use regex? This isn't too hard to implement on your own:
public static String addDelimiter(String str, char delimiter) {
StringBuilder string = new StringBuilder(str);
boolean isLetter = false;
boolean isNumber = false;
for (int index = 0; index < string.length(); index++) {
isNumber = isNumber(string.charAt(index));
if (isLetter && isNumber) {
//the last char was a letter, and now we have a number
//so here we adjust the stringbuilder
string.insert(index, delimiter);
index++; //We just inserted the delimiter, get past the delimiter
}
isLetter = isLetter(string.charAt(index));
}
return string.toString();
}
public static boolean isLetter(char c) {
return 'A' <= c && c <= 'Z' || 'a' <= c && c <= 'z';
}
public static boolean isNumber(char c) {
return '0' <= c && c <= '9';
}
The advantage of this over regex is that regex can easily be slower. Additionally, it is easy to change the isLetter and isNumber methods to allow for inserting the delimiter in different places.

How to get with JAVA a specific value for one substring from string?

I have ONE string field which is in format:
"TransactionID=30000001197169 ExecutionStatus=6
additionalCurrency=KMK
pin= 0000"
So they are not separated with some ; оr , they are not seperated even with one blank space.
I want to get value for Execution Status and put it in some field?
How to achieve this?
Thanks for help

This works. But I am not sure this is the most optimal.It just solves your problem.
String s = "TransactionID=30000001197169ExecutionStatus=6additionalCurrency=KMKpin=0000";
if(s!=null && s.contains("ExecutionStatus="))
{
String s1[] = s.split("ExecutionStatus=");
if(s1!=null && s1.length>1)
{
String line = s1[1];
String pattern = "[0-9]+";
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher m = r.matcher(line);
if (m.find( )) {
System.out.println("Match");
System.out.println("Found value: " + m.group(0) );
} else {
System.out.println("NO MATCH");
}
}
}

In your example they are indeed seperated by blanks, but the following should be working without blanks, too. Assuming your String is stored in String arguments
String executionStatus;
String[] anArray = arguments.split("=");
for (int i; i < anArray.length; i++)
if (anArray[i].contains("ExecutionStatus")){
executionStatus = anArray[++i].replace("additionalCurrency","");
executionStatus = executionStatus.trim();
}
}

Check if it contains() ExecutionStatus=
If yes then split the string with ExecutionStatus=
Now take the Second string from array find the first occurance of non digit char and use substring()

Assuming all that white space is present in your string, this works.
String str = "\"TransactionID=30000001197169 ExecutionStatus=6\n" +
" additionalCurrency=\"KMK\"\n" +
" pin= \"0000\"\"";
int start = str.indexOf("ExecutionStatus=") + "ExecutionStatus=".length();
int status = 0;
if (start >= 0) {
String strStatus = str.substring(start, str.indexOf("additionalCurrency=") - 1);
try {
status = Integer.parseInt(strStatus.trim());
} catch (NumberFormatException e) {
}
}

At the risk of attracting "... and now you have two problems!" comments, this is probably easiest done with regexes (str is the String defined above):
Pattern p = Pattern.compile("ExecutionStatus\\s*=\\s*(\\d+)"); // Whitespace matching around equals for safety, capturing group around the digits of the status)
Matcher m = p.matcher(str);
String status = m.find() ? m.group(1) : null;

Find count of digits in string variable

I have a string which sometimes gives character value and sometimes gives integer value. I want to get the count of number of digits in that string.
For example, if string contains "2485083572085748" then total number of digits is 16.
Please help me with this.

A cleaner solution using Regular Expressions:
// matches all non-digits, replaces it with "" and returns the length.
s.replaceAll("\\D", "").length()

String s = "2485083572085748";
int count = 0;
for (int i = 0, len = s.length(); i < len; i++) {
if (Character.isDigit(s.charAt(i))) {
count++;
}
}

Just to refresh this thread with stream option of counting digits in a string:
"2485083572085748".chars()
.filter(Character::isDigit)
.count();

If your string gets to big and full of other stuff than digits you should try to do it with regular expressions. Code below would do that to you:
String str = "asdasd 01829898 dasds ds8898";
Pattern p = Pattern.compile("\d"); // "\d" is for digits in regex
Matcher m = p.matcher(str);
int count = 0;
while(m.find()){
count++;
}
check out java regex lessons for more.
cheers!

Loop each character and count it.
String s = "2485083572085748";
int counter = 0;
for(char c : s.toCharArray()) {
if( c >= '0' && c<= '9') {
++counter;
}
}
System.out.println(counter);

public static int getCount(String number) {
int flag = 0;
for (int i = 0; i < number.length(); i++) {
if (Character.isDigit(number.charAt(i))) {
flag++;
}
}
return flag;
}

in JavaScript:
str = "2485083572085748"; //using the string in the question
let nondigits = /\D/g; //regex for all non-digits
let digitCount = str.replaceAll(nondigits, "").length;
//counts the digits after removing all non-digits
console.log(digitCount); //see in console
Thanks --> https://stackoverflow.com/users/1396264/vedant for the Java version above. It helped me too.

int count = 0;
for(char c: str.toCharArray()) {
if(Character.isDigit(c)) {
count++;
}
}
Also see
Javadoc

Something like:
using System.Text.RegularExpressions;
Regex r = new Regex( "[0-9]" );
Console.WriteLine( "Matches " + r.Matches("if string contains 2485083572085748 then" ).Count );

Check if String contains only letters

The idea is to have a String read and to verify that it does not contain any numeric characters. So something like "smith23" would not be acceptable.

What do you want? Speed or simplicity? For speed, go for a loop based approach. For simplicity, go for a one liner RegEx based approach.
Speed
public boolean isAlpha(String name) {
char[] chars = name.toCharArray();
for (char c : chars) {
if(!Character.isLetter(c)) {
return false;
}
}
return true;
}
Simplicity
public boolean isAlpha(String name) {
return name.matches("[a-zA-Z]+");
}

Java 8 lambda expressions. Both fast and simple.
boolean allLetters = someString.chars().allMatch(Character::isLetter);

Or if you are using Apache Commons, [StringUtils.isAlpha()].

First import Pattern :
import java.util.regex.Pattern;
Then use this simple code:
String s = "smith23";
if (Pattern.matches("[a-zA-Z]+",s)) {
// Do something
System.out.println("Yes, string contains letters only");
}else{
System.out.println("Nope, Other characters detected");
}
This will output:
Nope, Other characters detected

I used this regex expression (".*[a-zA-Z]+.*"). With if not statement it will avoid all expressions that have a letter before, at the end or between any type of other character.
String strWithLetters = "123AZ456";
if(! Pattern.matches(".*[a-zA-Z]+.*", str1))
return true;
else return false

A quick way to do it is by:
public boolean isStringAlpha(String aString) {
int charCount = 0;
String alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
if (aString.length() == 0) {
return false; //zero length string ain't alpha
}
for (int i = 0; i < aString.length(); i++) {
for (int j = 0; j < alphabet.length(); j++) {
if (aString.substring(i, i + 1).equals(alphabet.substring(j, j + 1))
|| aString.substring(i, i + 1).equals(alphabet.substring(j, j + 1).toLowerCase())) {
charCount++;
}
}
if (charCount != (i + 1)) {
System.out.println("\n**Invalid input! Enter alpha values**\n");
return false;
}
}
return true;
}
Because you don't have to run the whole aString to check if it isn't an alpha String.

private boolean isOnlyLetters(String s){
char c=' ';
boolean isGood=false, safe=isGood;
int failCount=0;
for(int i=0;i<s.length();i++){
c = s.charAt(i);
if(Character.isLetter(c))
isGood=true;
else{
isGood=false;
failCount+=1;
}
}
if(failCount==0 && s.length()>0)
safe=true;
else
safe=false;
return safe;
}
I know it's a bit crowded. I was using it with my program and felt the desire to share it with people. It can tell if any character in a string is not a letter or not. Use it if you want something easy to clarify and look back on.

Faster way is below. Considering letters are only a-z,A-Z.
public static void main( String[] args ){
System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
System.out.println(bestWay("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
System.out.println(isAlpha("azAZpratiyushkumarsinghjdnfkjsaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa1aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"));
}
public static boolean bettertWay(String name) {
char[] chars = name.toCharArray();
long startTimeOne = System.nanoTime();
for(char c : chars){
if(!(c>=65 && c<=90)&&!(c>=97 && c<=122) ){
System.out.println(System.nanoTime() - startTimeOne);
return false;
}
}
System.out.println(System.nanoTime() - startTimeOne);
return true;
}
public static boolean isAlpha(String name) {
char[] chars = name.toCharArray();
long startTimeOne = System.nanoTime();
for (char c : chars) {
if(!Character.isLetter(c)) {
System.out.println(System.nanoTime() - startTimeOne);
return false;
}
}
System.out.println(System.nanoTime() - startTimeOne);
return true;
}
Runtime is calculated in nano seconds. It may vary system to system.
5748//bettertWay without numbers
true
89493 //isAlpha without numbers
true
3284 //bettertWay with numbers
false
22989 //isAlpha with numbers
false

Check this,i guess this is help you because it's work in my project so once you check this code
if(! Pattern.matches(".*[a-zA-Z]+.*[a-zA-Z]", str1))
{
String not contain only character;
}
else
{
String contain only character;
}

String expression = "^[a-zA-Z]*$";
CharSequence inputStr = str;
Pattern pattern = Pattern.compile(expression);
Matcher matcher = pattern.matcher(inputStr);
if(matcher.matches())
{
//if pattern matches
}
else
{
//if pattern does not matches
}

Try using regular expressions: String.matches

public boolean isAlpha(String name)
{
String s=name.toLowerCase();
for(int i=0; i<s.length();i++)
{
if((s.charAt(i)>='a' && s.charAt(i)<='z'))
{
continue;
}
else
{
return false;
}
}
return true;
}

Feels as if our need is to find whether the character are only alphabets.
Here's how you can solve it-
Character.isAlphabetic(c)
helps to check if the characters of the string are alphabets or not.
where c is
char c = s.charAt(elementIndex);

While there are many ways to skin this cat, I prefer to wrap such code into reusable extension methods that make it trivial to do going forward. When using extension methods, you can also avoid RegEx as it is slower than a direct character check. I like using the extensions in the Extensions.cs NuGet package. It makes this check as simple as:
Add the https://www.nuget.org/packages/Extensions.cs package to your project.
Add "using Extensions;" to the top of your code.
"smith23".IsAlphabetic() will return False whereas "john smith".IsAlphabetic() will return True. By default the .IsAlphabetic() method ignores spaces, but it can also be overridden such that "john smith".IsAlphabetic(false) will return False since the space is not considered part of the alphabet.
Every other check in the rest of the code is simply MyString.IsAlphabetic().

To allow only ASCII letters, the character class \p{Alpha} can be used. (This is equivalent to [\p{Lower}\p{Upper}] or [a-zA-Z].)
boolean allLettersASCII = str.matches("\\p{Alpha}*");
For allowing all Unicode letters, use the character class \p{L} (or equivalently, \p{IsL}).
boolean allLettersUnicode = str.matches("\\p{L}*");
See the Pattern documentation.

I found an easy of way of checking a string whether all its digit is letter or not.
public static boolean isStringLetter(String input) {
boolean b = false;
for (int id = 0; id < input.length(); id++) {
if ('a' <= input.charAt(id) && input.charAt(id) <= 'z') {
b = true;
} else if ('A' <= input.charAt(id) && input.charAt(id) <= 'Z') {
b = true;
} else {
b = false;
}
}
return b;
}
I hope it could help anyone who is looking for such method.

Use StringUtils.isAlpha() method and it will make your life simple.