Develop Reference

How to insert a syllable before a single vowel

I try to insert a syllable before each vowel with the following restrictions:
- Before each following vowel (a, e, i, o, u), insert the stray syllable "av".
- Unless the vowel is preceded by another vowel
Actually I've done this
public static String translate(String text) {
char [] charArray = text.toCharArray();
StringBuilder sb = new StringBuilder();
for(char c : charArray) {
if (charArray.length < 255){
if (isVowel(c)) {
sb.append(String.format("av%s" , c));
}
else {
sb.append(c);
}
}
}
return sb.toString();
}
static boolean isVowel(char c) {
return (c == 'a') || (c == 'e') ||
(c == 'i') || (c == 'o') ||
(c == 'u');
}
With words without double vowels it works perfectly:
Cat becomes
Cavat
But with double vowel it doesn't work
Meet becomes
Maveavet // Should return Meet
How to check if 2 successive letters are vowels in order not to add the syllable if it's the case ?
Thanks in advance

Here is a version that uses an ordinary for loop:
public static String translate(String text) {
char [] charArray = text.toCharArray();
StringBuilder sb = new StringBuilder();
Set<Character> vowels = new HashSet<>(Arrays.asList('a','e','i','o','u'));
char cFirst = charArray[0];
if (vowels.contains(cFirst) && !(vowels.contains(charArray[1]))) {
sb.append(String.format("av%s" , cFirst));
}
else sb.append(cFirst);
for (int i = 1; i < charArray.length-1; i++){
char c = charArray[i];
char cNext = charArray[i+1];
char cPrev = charArray[i-1];
if (vowels.contains(c) && !(vowels.contains(cNext) || vowels.contains(cPrev))) {
sb.append(String.format("av%s" , c));
}
else sb.append(c);
}
char cLast = charArray[charArray.length-1];
if (vowels.contains(cLast) && !(vowels.contains(charArray[charArray.length-2]))) {
sb.append(String.format("av%s" , cLast));
}
else sb.append(cLast);
return sb.toString();
}
Works for all the cases I could think of, but probably not the optimal way of doing this.

One approach is to manage the indexes yourself instead of using an enhanced for loop. This allows you to 'skip' over two vowel chunks by checking the next character if it exists. Something like this should work:
int i = 0;
while (i < charArray.length) {
char c = charArray[i];
if (isVowel(c)) {
if (i + i == charArray.length || isVowel(charArray[i+1])) {
sb.append("av");
} else {
i++;
}
}
sb.append(c);
i++;
}
Note that this does not solve for when there are three sequential vowels.

How do I unescape a JSON String using Java/Jackson?

I am using Jackson version 2.4.3 for converting my complex Java object into a String object, so below is what I'm getting in output. The output is like below (Fyi - I just printed some part of the output)
"{\"FirstName\":\"John \",\"LastName\":cena,\"salary\":7500,\"skills\":[\"java\",\"python\"]}";
Here is my code (PaymentTnx is a complex Java object)
ObjectMapper mapper = new ObjectMapper();
mapper.setVisibility(PropertyAccessor.FIELD, Visibility.ANY);
String lpTransactionJSON = mapper.writeValueAsString(paymentTxn);
I don't want to see \ slashes in my JSON string. What do I need to do to get a string like below:
"{"FirstName":"John ","LastName":cena,"salary":7500,"skills":["java","python"]}";

String test = "{\"FirstName\":\"John \",\"LastName\":cena,\"salary\":7500,\"skills\":[\"java\",\"python\"]}";
System.out.println(StringEscapeUtils.unescapeJava(test));
This might help you.

I have not tried Jackson. I just have similar situation.
I used org.apache.commons.text.StringEscapeUtils.unescapeJson but it's not working for malformed JSON format like {\"name\": \"john\"}
So, I used this class. Perfectly working fine.
https://gist.githubusercontent.com/jjfiv/2ac5c081e088779f49aa/raw/8bda15d27c73047621a94359492a5a9433f497b2/JSONUtil.java
// BSD License (http://lemurproject.org/galago-license)
package org.lemurproject.galago.utility.json;
public class JSONUtil {
public static String escape(String input) {
StringBuilder output = new StringBuilder();
for(int i=0; i<input.length(); i++) {
char ch = input.charAt(i);
int chx = (int) ch;
// let's not put any nulls in our strings
assert(chx != 0);
if(ch == '\n') {
output.append("\\n");
} else if(ch == '\t') {
output.append("\\t");
} else if(ch == '\r') {
output.append("\\r");
} else if(ch == '\\') {
output.append("\\\\");
} else if(ch == '"') {
output.append("\\\"");
} else if(ch == '\b') {
output.append("\\b");
} else if(ch == '\f') {
output.append("\\f");
} else if(chx >= 0x10000) {
assert false : "Java stores as u16, so it should never give us a character that's bigger than 2 bytes. It literally can't.";
} else if(chx > 127) {
output.append(String.format("\\u%04x", chx));
} else {
output.append(ch);
}
}
return output.toString();
}
public static String unescape(String input) {
StringBuilder builder = new StringBuilder();
int i = 0;
while (i < input.length()) {
char delimiter = input.charAt(i); i++; // consume letter or backslash
if(delimiter == '\\' && i < input.length()) {
// consume first after backslash
char ch = input.charAt(i); i++;
if(ch == '\\' || ch == '/' || ch == '"' || ch == '\'') {
builder.append(ch);
}
else if(ch == 'n') builder.append('\n');
else if(ch == 'r') builder.append('\r');
else if(ch == 't') builder.append('\t');
else if(ch == 'b') builder.append('\b');
else if(ch == 'f') builder.append('\f');
else if(ch == 'u') {
StringBuilder hex = new StringBuilder();
// expect 4 digits
if (i+4 > input.length()) {
throw new RuntimeException("Not enough unicode digits! ");
}
for (char x : input.substring(i, i + 4).toCharArray()) {
if(!Character.isLetterOrDigit(x)) {
throw new RuntimeException("Bad character in unicode escape.");
}
hex.append(Character.toLowerCase(x));
}
i+=4; // consume those four digits.
int code = Integer.parseInt(hex.toString(), 16);
builder.append((char) code);
} else {
throw new RuntimeException("Illegal escape sequence: \\"+ch);
}
} else { // it's not a backslash, or it's the last character.
builder.append(delimiter);
}
}
return builder.toString();
}
}

With Jackson do:
toString(paymentTxn);
with
public String toString(Object obj) {
try (StringWriter w = new StringWriter();) {
new ObjectMapper().configure(SerializationFeature.INDENT_OUTPUT, true).writeValue(w, obj);
return w.toString();
} catch (Exception e) {
throw new RuntimeException(e);
}
}

This here is not valid JSON:
"{"FirstName":"John ","LastName":cena,"salary":7500,"skills":["java","python"]}";
This here is valid JSON, specifically a single string value:
"{\"FirstName\":\"John \",\"LastName\":cena,\"salary\":7500,\"skills\":[\"java\",\"python\"]}";
Given that you're calling writeValueAsString, this is the correct behaviour. I would suggest writeValue, perhaps?

How to use PropertyResourceBundle with keys containing whitespaces

I want to use properties files through PropertyResourceBundle for i18n. My current issue is that keys on the files I have can include white spaces, e.g. :
key number 1 = value number 1
key2 = value2
So, when I load the corresponding property file the first white space is used as the key-value delimiter instead of the '=' sign.
Then, my questions are: how can I use a key with white spaces in it without modifying the properties file (I'd like to avoid adding any slash or unicode character code)? Is there any way to override the default properties file delimiter so I can set '=' as the only one to be considered?

you will have to write your own Properties class, the one in the jdk considers white space as a separator, here is it's code. you'll find out that as soon as it encounter a white space it stop the key & start the value.
private void load0 (LineReader lr) throws IOException {
char[] convtBuf = new char[1024];
int limit;
int keyLen;
int valueStart;
char c;
boolean hasSep;
boolean precedingBackslash;
while ((limit = lr.readLine()) >= 0) {
c = 0;
keyLen = 0;
valueStart = limit;
hasSep = false;
//System.out.println("line=<" + new String(lineBuf, 0, limit) + ">");
precedingBackslash = false;
while (keyLen < limit) {
c = lr.lineBuf[keyLen];
//need check if escaped.
if ((c == '=' || c == ':') && !precedingBackslash) {
valueStart = keyLen + 1;
hasSep = true;
break;
} else if ((c == ' ' || c == '\t' || c == '\f') && !precedingBackslash) {
valueStart = keyLen + 1;
break;
}
if (c == '\\') {
precedingBackslash = !precedingBackslash;
} else {
precedingBackslash = false;
}
keyLen++;
}
while (valueStart < limit) {
c = lr.lineBuf[valueStart];
if (c != ' ' && c != '\t' && c != '\f') {
if (!hasSep && (c == '=' || c == ':')) {
hasSep = true;
} else {
break;
}
}
valueStart++;
}
String key = loadConvert(lr.lineBuf, 0, keyLen, convtBuf);
String value = loadConvert(lr.lineBuf, valueStart, limit - valueStart, convtBuf);
put(key, value);
}
}

Writing a method to remove vowels in a Java String [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
I am a beginner of programming, and am writing a Java method to remove vowel in Strings, but I do not know how to fix this error: ";" expected :
public String disemvowel(String s) {
boolean isVowel(char c);
if (c == 'a') {
return true;
} else if if (c == 'e') {
return true;
} else if if (c == 'i') {
return true;
} else if if (c == 'o') {
return true;
} else if if (c == 'u') {
return true;
}
String notVowel = "";
int l = s.length();
for (int z = 0; z <= l; z++) {
if (isVowel == "false") {
char x = s.charAt(z);
notVowel = notVowel + x;
}
}
return notVowel;
}

String str= "Your String";
str= str.replaceAll("[AEIOUaeiou]", "");
System.out.println(str);

A much simpler approach would be to do the following:
String string = "A really COOL string";
string = string.replaceAll("[AaEeIiOoUu]", "");
System.out.println(string);
This will apply the regular expression, [AaEeIiOoUu] to string. This expression will match all vowels in the character group [AaEeIiOoUu] and replace them with "" empty string.

You've got a lot of syntax errors.
boolean isVowel(char c); - not sure what you're doing with this. if you want it as a separate method, separate it out (and don't place a semicolon after it, which would be invalid syntax.
else if if is invalid syntax. If you're doing an else if, then you only need the one if.
Even if the code would compile, for (int z = 0; z <= l; z++) will cause you to step off of the String. Remove the <= in favor of <.
isVowel == "false" is never going to work. You're comparing a String to a boolean. You want !isVowel instead.
Putting the syntax errors aside, think of it like this.
You have a string that contains vowels. You wish to have a string that doesn't contain vowels.
The most straightforward approach is to iterate over the String, placing all non-vowel characters into a separate String, which you then return.
Interestingly enough, the half-method you have there can accomplish the logic of determining whether something is or isn't a vowel. Extract that to its own method. Then, call it in your other method. Do take into account capital letters though.
I leave the rest as an exercise to the reader.

Here is your code, without changing any logic, but unscrambling the isVowel method:
public String disemvowel(String s) {
// Removed the "isVowel" method from here and moved it below
String notVowel = "";
int l = s.length();
for (int z = 0; z <= l; z++) {
// Note that the "isVowel" method has not been called.
// And note that, when called, isVowel returns a boolean, not a String.
// (And note that, as a general rule, you should not compare strings with "==".)
// So this area needs a lot of work, but we'll start with this
boolean itIsAVowel = isVowel(s.charAt(z));
// (I made the variable name "itIsAVowel" to emphasize that it's name has nothing to do with the method name.
// You can make it "isVowel" -- the same as the method -- but that does not in any way change the function.)
// Now take it from there...
if (isVowel == "false") {
char x = s.charAt(z);
notVowel = notVowel + x;
}
}
return notVowel;
}
// You had this line ending with ";"
boolean isVowel(char c) {
if (c == 'a') {
return true;
// Note that you coded "if if" on the lines below -- there should be only one "if" per line, not two
} else if if (c == 'e') {
return true;
} else if if (c == 'i') {
return true;
} else if if (c == 'o') {
return true;
} else if if (c == 'u') {
return true;
}
// You were missing this final return
return false;
}
(Yes, I know this should be a comment, but you can't put formatted code in a comment.)

You could try something like this:
public static String removeVowels(final String string){
final String vowels = "AaEeIiOoUu";
final StringBuilder builder = new StringBuilder();
for(final char c : string.toCharArray())
if(vowels.indexOf(c) < 0)
builder.append(c);
return builder.toString();
}

How to determine a string is english or arabic?

Is there a way to determine a string is English or Arabic?

Here is a simple logic that I just tried:
public static boolean isProbablyArabic(String s) {
for (int i = 0; i < s.length();) {
int c = s.codePointAt(i);
if (c >= 0x0600 && c <= 0x06E0)
return true;
i += Character.charCount(c);
}
return false;
}
It declares the text as arabic if and only if an arabic unicode code point is found in the text. You can enhance this logic to be more suitable for your needs.
The range 0600 - 06E0 is the code point range of Arabic characters and symbols (See Unicode tables)

Java in itself supports various language checks by unicode, Arabic is also supported. Much simpler and smallest way to do the same is by UnicodeBlock
public static boolean textContainsArabic(String text) {
for (char charac : text.toCharArray()) {
if (Character.UnicodeBlock.of(charac) == Character.UnicodeBlock.ARABIC) {
return true;
}
}
return false;
}

A minor change to cover all arabic characters and symbols range
private boolean isArabic(String text){
String textWithoutSpace = text.trim().replaceAll(" ",""); //to ignore whitepace
for (int i = 0; i < textWithoutSpace.length();) {
int c = textWithoutSpace.codePointAt(i);
//range of arabic chars/symbols is from 0x0600 to 0x06ff
//the arabic letter 'لا' is special case having the range from 0xFE70 to 0xFEFF
if (c >= 0x0600 && c <=0x06FF || (c >= 0xFE70 && c<=0xFEFF))
i += Character.charCount(c);
else
return false;
}
return true;
}

You can usually tell by the code points within the string itself. Arabic occupies certain blocks in the Unicode code space.
It's a fairly safe bet that, if a substantial proportion of the characters exist in those blocks (such as بلدي الحوامات مليء الثعابينة), it's Arabic text.

This answer is somewhat correct. But when we combine Farsi and English letters it returns TRUE!, which is not true.
Here I modified the same method so that it works well
public static boolean isProbablyArabic(String s) {
for (int i = 0; i < s.length();) {
int c = s.codePointAt(i);
if (!(c >= 0x0600 && c <= 0x06E0))
return false;
i += Character.charCount(c);
}
return true;
}

You could use N-gram-based text categorization (google for that phrase) but it is not a fail-proof technique, and it may require a not too short string.
You might also decide that a string with only ASCII letters is not Arabic.

English characters tend to be in these 4 Unicode blocks:
BASIC_LATIN
LATIN_1_SUPPLEMENT
LATIN_EXTENDED_A
GENERAL_PUNCTUATION
public static boolean isEnglish(String text) {
boolean onlyEnglish = false;
for (char character : text.toCharArray()) {
if (Character.UnicodeBlock.of(character) == Character.UnicodeBlock.BASIC_LATIN
|| Character.UnicodeBlock.of(character) == Character.UnicodeBlock.LATIN_1_SUPPLEMENT
|| Character.UnicodeBlock.of(character) == Character.UnicodeBlock.LATIN_EXTENDED_A
|| Character.UnicodeBlock.of(character) == Character.UnicodeBlock.GENERAL_PUNCTUATION) {
onlyEnglish = true;
} else {
onlyEnglish = false;
}
}
return onlyEnglish;
}

Just an adaptation of existing answer to Kotlin:
fun String.textContainsArabic(): Boolean =
any { Character.UnicodeBlock.of(it) == Character.UnicodeBlock.ARABIC }

I tried this with my code and it works fine.
Using codePointAt which is a method that returns the Unicode value of the character at the specified index in a string.
public static boolean isItArabic(String someText)
{
for(int i = 0; i<someText.length(); i++)
{
int point = someText.codePointAt(i);
if(!(point >= 1536 && point <= 1791)) {
return false;
}
}
return true;
}

Try This :
internal static bool ContainsArabicLetters(string text)
{
foreach (char character in text.ToCharArray())
{
if (character >= 0x600 && character <= 0x6ff)
return true;
if (character >= 0x750 && character <= 0x77f)
return true;
if (character >= 0xfb50 && character <= 0xfc3f)
return true;
if (character >= 0xfe70 && character <= 0xfefc)
return true;
}
return false;
}