CODEWITHSUNDEEP
javaandroidunicodeunicode-string
Noor Hossain
Noor Hossain

Reputation: 1841

Convert mixed Unicode Numbers/Values/Strings (The whole Line contains also string) to a String

I have a line "Example: \u09b8\u09be\u09b0\u09cd\u09ac\u09bf\u09df\u09be\u09df \u0993\u09df\u09be\u09b0\u09cd\u0995\u09aa\u09be\u09b0\u09ae\u09bf\u099f \u09ad\u09bf\u09b8\u09be," What I want to, get a string from this line as follows : "Example: সার্বিয়ায় ওয়ার্কপারমিট ভিসা,"

I assume that it could be found by Integer.parseInt() method, then what will be the procedure?

Upvotes: -1

Views: 81

Answers (3)

Basil Bourque
Basil Bourque

Reputation: 340158

Java compiler automatically unescapes

Your input is a String literal containing escaped Unicode code point numbers. This syntax is understood directly by the Java compiler.

So there is no need to unescape.

String input = "ইউরোপের\u09b8\u09be\u09b0\u09cd\u09ac\u09bf\u09df\u09be\u09df \u0993\u09df\u09be\u09b0\u09cd\u0995\u09aa\u09be\u09b0\u09ae\u09bf\u099f \u09ad\u09bf\u09b8\u09be," ;
System.out.println( input ) ;

See this code run at Ideone.com.

ইউরোপেরসার্বিয়ায় ওয়ার্কপারমিট ভিসা,

Perhaps you want to generate Java source code without the escapes. That is legitimate as the Java language fully supports Unicode characters in source code.

We can add a double-quote character to each end of your string. Since Java syntax uses the double-quote character, we must escape like this: "\"".

// Generate source code. 
String sourceCode = "\"" + input +  "\"" ;
System.out.println( sourceCode ) ;

"ইউরোপেরসার্বিয়ায় ওয়ার্কপারমিট ভিসা,"

Upvotes: 1

g00se
g00se

Reputation: 4292

If my comment HERE is correct then this approach is simpler:

        Matcher m = Pattern.compile("\\\\u[0-9a-f]{4}").matcher(s);
        StringBuilder sb = new StringBuilder();
        while (m.find()) {
            m.appendReplacement(sb, Character.toString(Integer.parseInt(m.group(), 16)));
        }
        m.appendTail(sb);
        System.out.println(sb.toString());

Upvotes: 0

Noor Hossain
Noor Hossain

Reputation: 1841

The main approach for this method is

int hexValue = Integer.parseInt(letter, 16);

So, I have found the result with following functions :

public static  String unicodeStringToString(String unicode){
        System.out.println("ddd first Response 3:unicode " + unicode);

        if(TextUtils.isEmpty(unicode)){
            return  "";
        }

        StringBuilder txt = new StringBuilder();

        String [] words = unicode.split(" ");
        for (String word: words){
            if(word.contains("\\u")) {
                System.out.println("ddd first Response 4:word " + word);
                word = word.replace("\\", "");
                String[] letters = word.split("u");
                for (String letter : letters) {
                    System.out.println("ddd first Response 5:letter " + letter);
                    if(letter.contains(",")) {
                        letter= letter.replace(",", "");
                        String s = unicodeNumberToString(letter);
                        txt.append(s);
                        txt.append(",");
                    }else {
                        String s = unicodeNumberToString(letter);
                        txt.append(s);
                    }
                    txt.append(" ");
                }
            }else {
                txt.append(word);
                txt.append(" ");
            }

        }

        return txt.toString();

        // uses : textView.setText(Html.fromHtml(unicodeStringToString(unicode)));

    }

public static  String unicodeNumberToString(String letter){
        

        if(isInteger(letter)) {
            int hexValue = 0;
            try {
                hexValue = Integer.parseInt(letter, 16);
            } catch (NumberFormatException e) {
                e.printStackTrace();
            }
            letter = String.valueOf(((char) hexValue));
        }
        return  letter ;
    }

public static boolean isInteger(String s) {
        try {
            Integer.parseInt(s);
        } catch (NumberFormatException e) {
            System.out.println("integer Problem: isInteger:  "+e.toString());
            return false;
        }
        return true;
    }

Upvotes: -1

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