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Check for regional language in java

I have a scenario where i have to check if message contains any regional language then the type should be "2" or if it contains English(can contains special char, digits) language then the type should be "0".
I have tried using
for (char c: messages.get(0).message().toCharArray()) {
if (Character.UnicodeBlock.of(c) ==
Character.UnicodeBlock.DEVANAGARI) {
isHindi = true;
break;
}
}
but in java it checks for each specific language.
I want to write generalise code for that.
Thanks

English is all ASCII so you can do
if (c > 127 && Character.isLetterOrDigit(c)) // non English letter or digit.

Let's convert it into a method, it serves better.
class IndianTest {
public static void main(String args[])
{
String x = "यर";
if (textContainsIndian(x)) {
System.out.println("indian");
} else if (isEnglish(x)) {
System.out.println("english");
} else
System.out.println("another language");
}
public static boolean textContainsIndian(String text) {
for (char charac : text.toCharArray()) {
if (Character.UnicodeBlock.of(charac) == Character.UnicodeBlock.DEVANAGARI) {
return true;
}
}
return false;
}
public static boolean isEnglish(String name) {
return name.matches("[a-zA-Z0-9]+");
}
}
Now, you can modify the code as your purpose.

Validating input string with java

First off, I will admit that this is an assignment of mine. however, I am at my wits end. I tried need to validate that the user input a proper expression (ie: "7 + 5;") and I managed to do it with split methods but I was told that I can't do that. I'm sure there is a simple solution to the problem but I am not seeing it.
The code is rather lengthy so I won't post it but if I will if needed.
Thanks!
Edit to answer questions: I am writing in jGrasp, so they can do whatever is on the keyboard. I was told to "Find a creative way to use substrings" which I don't know what that means. The expression needs to be Number Space operand Space number semicolon
here is what I have for the validation... I am using arrays for each character in the expression
public static boolean validFormat(String expr)
{
String tokens[] = expr.substring()
if (tokens.length == 3)
{
if (tokens[0].equals(""))
{
return false;
}
if (tokens[1].equals(""))
{
return false;
}
if (tokens[2].length < 2)
{
return false;
}
else
{
if (tokens[2].endwith(";"));
{
return false;
}
else
{
return true;
}
}
}
return false;
}
I get an error with calling the substring as well as an "else without if" error

First, you should limits the input to 6 characters using an if statement. Then use the CharAt() method to return each character to check the condition.

I was told to "Find a creative way to use substrings".
As told, try using String.substring() for the same.
public class Demo {
public static void main (String[] args) {
String exp = "7 + 5;";
System.out.printf("%s\t%b%n", exp, validate(exp));
exp = "4 + d;";
System.out.printf("%s\t%b%n", exp, validate(exp));
}
public static boolean validate(String exp) {
String n1 = exp.substring(0,1);//operand
String n2 = exp.substring(4,5);//operand
String co = exp.substring(5);//semicolon
String s1 = exp.substring(1,2);//space
String s2 = exp.substring(3,4);//space
String op = exp.substring(2,3);//operator
return num(n1) && num(n2) && semi(co) && space(s1) && space(s2) && opt(op);
}
public static boolean num(String n) {
return "0123456789".contains(n);
}
public static boolean semi(String s) {
return ";".equals(s);
}
public static boolean space(String s) {
return " ".equals(s);
}
public static boolean opt(String s) {
return "-+*%/^".contains(s);
}
}
This solution uses RegExp:
public class Demo {
public static void main (String[] args) {
String exp = "7 + 5;";
System.out.printf("%s\t%b%n", exp, validate(exp));
exp = "4 + d;";
System.out.printf("%s\t%b%n", exp, validate(exp));
}
public static boolean validate(String exp) {
return exp.matches("\\d\\s(\\+|\\-|\\*|\\/|\\^|\\%)\\s\\d\\;");
}
}

recursive expand of strings

I want to create a method which recursively expands the input string with another string.
public class StringTest {
public static String addZeichenkette(String out, int i) {
out += "bla";
if (i > 0) {
i--;
addZeichenkette(out, i);
}
return out;
}
public static void main(String[] args) {
String out = "Hello";
out = addZeichenkette(out, 7);
System.out.println(out);
}
}
The output should be :
Helloblablablablablablabla
instead it is
Hellobla
The main method has to be untouched (except the method calling).

You're ignoring the return from the recursive call.
public static String addZeichenkette(String out, int i)
{
if (i > 0)
{
out += "bla";
i--;
out = addZeichenkette(out, i);
}
return out;
}
Alternatively, you could just return the result addZeichenkette(out, i);, but both ways are fine.
As you pointed out, you would need to modify it to avoid adding i+1 bla's. Since when i = 0, we don't want any bla's to be added, we instead need to check for this.
To do this, I moved the concatenation of the bla into the condition.

You need to return the accumulated (recursed) string:
return addZeichenkette(out, i);
not just the one where i == 0.
public static String addZeichenkette(String out, int i) {
return i > 0 ? addZeichenkette(out + "bla", --i ) : out;
}

Java contains a string or not?

http://codingbat.com/prob/p126880
Given two strings, return true if either of the strings appears at the very end of the other string, ignoring upper/lower case differences (in other words, the computation should not be "case sensitive"). Note: str.toLowerCase() returns the lowercase version of a string.
I cannot get when it is true, it always gives false.
public boolean endOther(String a, String b)
{
//variables
a.toLowerCase();
b.toLowerCase();
String f1="";
String f2="";
int d=0;
int sum=0;
//Program code;
if(a.length()-b.length()>0)
{
(f1).equals(a);
(f2).equals(b);
d=a.length();
}
else if(a.length()-b.length()<0)
{
(f1).equals(b);
(f2).equals(a); //gett**ing bigger and lower String**
d=b.length();
}
else if((a).equals(b))
sum++;
// I think problem is because it is not enter the for.
for(int i=0; i>d; i++)
{
if((f1.substring(i,i+f2.length())).equals(f2))
sum++;
}
if(sum>0)
return true;
else
return false;
}

This is a working example of what you are trying to achieve to test in your Java IDE like Netbeans or Eclipse whatever. This is really simple, the String object has an endsWith method so why try to invent something yourself.
If you have any troubles reading this code hit me up, should be quite straight forward. You will just have to convert your string to lowercase, that's for you to add.
public class StringEnds {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.printf("String a: ");
String a = scanner.next();
System.out.printf("String b: ");
String b = scanner.next();
// Compare a and b
if (endsWith(a, b)) {
System.out.printf("Succes\n");
} else {
System.out.printf("Fail\n");
}
}
public static boolean endsWith(String firstString, String secondString) {
return firstString.endsWith(secondString) || secondString.endsWith(firstString);
}
}
Here's your codebat solution (it is quite short):
public boolean endOther(String a, String b) {
return a.toLowerCase().endsWith(b.toLowerCase()) || b.toLowerCase().endsWith(a.toLowerCase());
}

This is my answer. I tried both ways, hope it helps.
public boolean endOther(String a, String b) {
int small = Math.min(a.length(), b.length());
if (a.length()==b.length() && a.equalsIgnoreCase(b)) {
return true;
}
if (small==a.length()) {
if (b.substring(b.length()-small).equalsIgnoreCase(a)) {
return true;
}
// from here is the toLowerCase() method.
a = a.toLowerCase();
b = b.toLowerCase();
} else if (small==b.length()) {
if (a.endsWith(b)) {
return true;
}
}
return false;
}

How to find specific character in Java

I have a string like a>5 and b<6 or c>=7 in java
When I check whether string contains > then output is true for both > and >=
How can I restrict my check only to specific character?
How can I use matches function?

Your mistake is, you think of a lexical entity, >=, as of a "character." That will bite you more than once, as there actually are two characters, > and =, and > is indeed here. So depending on what you need, the answer may be different.
Why don't you want to see >= found?
What usage of > is of interest for you? Will e.g. <tag>some text</tag> be a proper string which you'd prefer to allow?

You want to discriminate between greater than and greater than or equal to. Why not write a method that returns the operator?
enum Operator {
...
}
public Operator getOperator(String s) {
if(s.contains(">=")) {
return Operator.GREATER_THAN_OR_EQUAL_TO;
} else if (s.contains(">") {
return Operator.GREATER_THAN;
}
}
If the input can be a complex expression that contains multiple operators, instead of contains, try using indexOf(...) and look ahead one character for the '='.

I just threw this together based on the updated specifications. Basically a real simple parser rather mechanically created: (And I'm not good with naming at 7 in the morning oh well)
public class Main {
public static void main(String[] args) {
String test = "a > b >= c > x";
Main m = new Main(test);
System.out.println(m.getTokenNumber());
test = "aasdfasdf asdfdasf";
m = new Main(test);
System.out.println(m.getTokenNumber());
}
private String input;
private int pos;
public Main(String input) {
this.input = input;
pos = 0;
}
public TokenNumber getTokenNumber() {
TokenNumber tokenNumber = new TokenNumber();
Token t = nextToken();
while (t != Token.NONE) {
tokenNumber.addToken(t);
t = nextToken();
}
return tokenNumber;
}
private Token nextToken() {
while (pos < input.length() && input.charAt(pos) != '>') pos++;
if (pos == input.length()) return Token.NONE;
pos++;
if (pos == input.length() || input.charAt(pos) != '=') return Token.GREATER;
return Token.GREATER_EQUAL;
}
enum Token {
GREATER, GREATER_EQUAL, NONE;
}
static class TokenNumber {
public int greater;
public int greater_than;
public void addToken(Token t) {
if (t == Token.GREATER) greater++;
else greater_than++;
assert t != Token.NONE;
}
public String toString() {
return String.format("Greater: %d%nGreater Than: %d", greater, greater_than);
}
}
}

"your string >".contains(">");// returns true

boolean value = Pattern.compile(">").matcher("a>5").find(); // returns true