This question already has answers here:
How do I split a string in Java?
(39 answers)
Closed 3 years ago.
I am trying to parse data from the [] in a String in java
String text = "some text [Karan] some text";
I want to the computer to read Karan present inside the brackets
I have tried String.split() method but it packs it into an array which I don't want.
Is there any way to do this. Thank you
You can use regex like this:
String text = "some text [Karan] some [test2] text [test3] [test4] 22[test5]";
Pattern pattern = Pattern.compile("(?<=\\[).*?(?=\\])");
//or use this regex,it works well too
//Pattern pattern = Pattern.compile("(?<=\\[)[^\\[\\]]*(?=\\])");
Matcher matcher = pattern.matcher(text);
while(matcher.find()){
System.out.println(matcher.group());
}
and the result is Karan,test2,test3,test4,test5.
Regex is useful for processing text.This improved version of regular is using "Positive and Negative Lookbehind".Thanks for Matthieu's suggestions.
You might want to use the String.substring method in combination with String.indexOf.
Here is a short description of how you would extract text between brackets [] using these two methods:
Get the index of the first bracket [ character using String.indexOf method. (let's call this value start)
Get the index of the second bracket ] character using String.indexOf method (let's call this value end)
Call .substring on the string you are extracting information from with
text.substring(start+1, end)
Note how the first index start+1 is inclusive and end is exclusive.
Looks like this then:
String text = "some text [Karen] some text";
text = text.substring(text.indexOf("[") + 1, text.indexOf("]"));
System.out.println(text);
This question already has answers here:
ASCII non readable characters 28, 29 31
(3 answers)
Closed 3 years ago.
I'm using JAVA and I'm trying to add the ASCII character 29(Group Separator) to a String(alphanumeric) as part of my algorithm. But I'm unable to verify the output since it doesnt get printed.
If its a non-printable character, is there any other way I can verify that it does get added.
Tried 1)Printing it like any other ASCII character 2)Printing its HEX value(0x1D)
System.out.println("Test1====="+Character.toString((char)0x1D));
System.out.println("Test3====="+String.valueOf(Character.toChars(29)));
Expected Result:Verify its printed.
Actual Result:Unable to verify.
Maybe write a function that traverses a string and compare every char to
Character.toChars(29)? Something along the lines of:
String str = "Foo Bar" + yourCharacter29ToString;
for(int i=0;i<str.length();i++){
if(Character.toChars(29) == str.charAt(i)){
return true;
}
}
return false;
This could be enough as a proof of concept. (i did not check above code - read it as pseudo-code please)
To see which codepoints are in a String, you can use Character.getName(codepoint)
int[] codepoints = ("Test3==🚲==="+String.valueOf(Character.toChars(29)))
.codePoints()
.toArray(); // optionally, set up for traditional for loop
for (int codepoint : codepoints) {
char[] utf16 = Character.toChars(codepoint); // always one or two code units
if (utf16.length == 2) {
System.out.println(
String.format("U+%04X \\u%04X\\u%04X %s",
codepoint, (int)utf16[0], (int)utf16[1], Character.getName(codepoint)));
} else {
System.out.println(
String.format("U+%04X \\u%04X %s",
codepoint, (int)utf16[0], Character.getName(codepoint)));
}
}
The UTF-16 character encoding encodes a codepoint from the Unicode character set with one or two code units (char).
(Not sure how the existence of the ASCII character set is relevant to this project—or most any project. If you have bytes for ASCII-encoded text or need bytes for ASCII-encoded text, that's a different question. But, Java uses UTF-16 for text datatypes.)
I'm trying to create a regex that will allow only digits followed by only one character after every digit within a Textfield
Regex that needs to match - \d*\+{1}
Regex in case it does not match - [^\d*\+){1}] will replace with "" (removes everything else)
final String regexFinalInteger = "\\d*\\+{1}";
numberElements.textProperty().addListener((observable, oldValueE, newValueE) -> {
if (!newValueE.matches(regexFinalInteger)) {
numberElements.setText(newValueE.replaceAll("[^\\d*\\+){1}]", ""));
}
});
I will expect an output of 122+1+3 but the actual output can be 1++2+++4+123 (multiple +)
If I understood correctly, you want to replace multiple +s with only one.
I believe this would enable what you're looking for:
String regex = "[+](?=[+])";
String text = "122+1+3";
assertEquals("122+1+3", text.replaceAll(regex, ""));
text = "1++2+++4+123";
assertEquals("1+2+4+123", text.replaceAll(regex, ""));
That is my first Java program, I'm sorry if it offends someone.
Thanks in advance for your patience. This is my problem.
I'm writing a program in Java that works best with a big set of different characters.
I have to store all the characters in a String. I started with
private static final String values = "0123456789";
Then I added A-Z, a-z and all the commons symbols.
But they are still too few, so I tought that maybe Unicode could be the solution.
The problem is now: what is the best way to get all the unicode characters that can be displayed in Eclipse (my algorithm will probably fail if there are unrecognized characters - those displayed like little rectangles). Is it possible to build a string (or some strings) with all the characters present here (en.wikipedia.org/wiki/List_of_Unicode_characters) correctly displayed?
I can do a rough copy-paste from http://www.terena.org/activities/multiling/euroml/tests/test-ucspages1ucs.html or http://zenoplex.jp/tools/unicoderange_generator.html, but I would appreciate some cleaner solution.
I don't know if there is a way to extract characters fron a font (the Unifont one). Or maybe I should parse this (www. utf8-chartable.de/unicode-utf8-table.pl) webpage.
Moreover, by adding all the characters into a String I will probably get the error:
"The type generates a string that requires more than 65535 bytes to encode in Utf8 format in the constant pool" (discussed in this question on SO: /questions/10798769/how-to-process-a-string-with-823237-characters).
Hybrid solutions can be accepted. I can remove duplicates following this question on SO questions/4989091/removing-duplicates-from-a-string-in-java)
Finally: every solution to get the longest only-different-characters string is accepted.
Thanks!
You are mixing some things up. The question whether a character can be displayed in Eclipse depends on the font you have chosen; and whether the source file can be processed correctly depends on which character encoding you have set up for the source file. When choosing UTF-8 and a good unicode font you can use and display almost any character, at least more than fit into a single String literal.
But is it really required to show the character in Eclipse? You can use the unicode escapes, e.g. \u20ac to refer to characters, regardless of whether they can be displayed or if the file encoding can handle them.
And if it is not a requirement to blow up your source code, it’s easy to create a String containing all existing characters:
// all chars (i.e. UTF-16 values)
StringBuilder sb=new StringBuilder(Character.MAX_VALUE);
for(char c=0; c<Character.MAX_VALUE; c++) sb.append(c);
String s=sb.toString();
// if it should behave like a compile-time constant:
s=s.intern();
or
// all unicode characters (aka code points)
StringBuilder sb=new StringBuilder(2162686);
for(int c=0; c<Character.MAX_CODE_POINT; c++) sb.appendCodePoint(c);
String s=sb.toString();
// if it should behave like a compile-time constant:
s=s.intern();
If you wan’t the String to contain valid unicode characters only you can use if(Character.isDefined(c)) … inside the loop. But that’s a moving target— newer JRE’s will most probably know more defined characters.
Smply use Apache classes, org.apache.commons.lang.RandomStringUtils (commons-lang) can solve your purpose.
http://commons.apache.org/proper/commons-lang/javadocs/api-3.1/org/apache/commons/lang3/RandomStringUtils.html
Also please refer to below code for api usage,
import org.apache.commons.lang3.RandomStringUtils;
public class RandomString {
public static void main(String[] args) {
// Random string only with numbers
String string = RandomStringUtils.random(64, false, true);
System.out.println("Random 0 = " + string);
// Random alphabetic string
string = RandomStringUtils.randomAlphabetic(64);
System.out.println("Random 1 = " + string);
// Random ASCII string
string = RandomStringUtils.randomAscii(32);
System.out.println("Random 2 = " + string);
// Create a random string with indexes from the given array of chars
string = RandomStringUtils.random(32, 0, 20, true, true, "bj81G5RDED3DC6142kasok".toCharArray());
System.out.println("Random 3 = " + string);
}
}
I want to display a Unicode character in Java. If I do this, it works just fine:
String symbol = "\u2202";
symbol is equal to "∂". That's what I want.
The problem is that I know the Unicode number and need to create the Unicode symbol from that. I tried (to me) the obvious thing:
int c = 2202;
String symbol = "\\u" + c;
However, in this case, symbol is equal to "\u2202". That's not what I want.
How can I construct the symbol if I know its Unicode number (but only at run-time---I can't hard-code it in like the first example)?
If you want to get a UTF-16 encoded code unit as a char, you can parse the integer and cast to it as others have suggested.
If you want to support all code points, use Character.toChars(int). This will handle cases where code points cannot fit in a single char value.
Doc says:
Converts the specified character (Unicode code point) to its UTF-16 representation stored in a char array. If the specified code point is a BMP (Basic Multilingual Plane or Plane 0) value, the resulting char array has the same value as codePoint. If the specified code point is a supplementary code point, the resulting char array has the corresponding surrogate pair.
Just cast your int to a char. You can convert that to a String using Character.toString():
String s = Character.toString((char)c);
EDIT:
Just remember that the escape sequences in Java source code (the \u bits) are in HEX, so if you're trying to reproduce an escape sequence, you'll need something like int c = 0x2202.
The other answers here either only support unicode up to U+FFFF (the answers dealing with just one instance of char) or don't tell how to get to the actual symbol (the answers stopping at Character.toChars() or using incorrect method after that), so adding my answer here, too.
To support supplementary code points also, this is what needs to be done:
// this character:
// http://www.isthisthingon.org/unicode/index.php?page=1F&subpage=4&glyph=1F495
// using code points here, not U+n notation
// for equivalence with U+n, below would be 0xnnnn
int codePoint = 128149;
// converting to char[] pair
char[] charPair = Character.toChars(codePoint);
// and to String, containing the character we want
String symbol = new String(charPair);
// we now have str with the desired character as the first item
// confirm that we indeed have character with code point 128149
System.out.println("First code point: " + symbol.codePointAt(0));
I also did a quick test as to which conversion methods work and which don't
int codePoint = 128149;
char[] charPair = Character.toChars(codePoint);
System.out.println(new String(charPair, 0, 2).codePointAt(0)); // 128149, worked
System.out.println(charPair.toString().codePointAt(0)); // 91, didn't work
System.out.println(new String(charPair).codePointAt(0)); // 128149, worked
System.out.println(String.valueOf(codePoint).codePointAt(0)); // 49, didn't work
System.out.println(new String(new int[] {codePoint}, 0, 1).codePointAt(0));
// 128149, worked
--
Note: as #Axel mentioned in the comments, with java 11 there is Character.toString(int codePoint) which would arguably be best suited for the job.
This one worked fine for me.
String cc2 = "2202";
String text2 = String.valueOf(Character.toChars(Integer.parseInt(cc2, 16)));
Now text2 will have ∂.
Remember that char is an integral type, and thus can be given an integer value, as well as a char constant.
char c = 0x2202;//aka 8706 in decimal. \u codepoints are in hex.
String s = String.valueOf(c);
String st="2202";
int cp=Integer.parseInt(st,16);// it convert st into hex number.
char c[]=Character.toChars(cp);
System.out.println(c);// its display the character corresponding to '\u2202'.
Although this is an old question, there is a very easy way to do this in Java 11 which was released today: you can use a new overload of Character.toString():
public static String toString​(int codePoint)
Returns a String object representing the specified character (Unicode code point). The result is a string of length 1 or 2, consisting solely of the specified codePoint.
Parameters:
codePoint - the codePoint to be converted
Returns:
the string representation of the specified codePoint
Throws:
IllegalArgumentException - if the specified codePoint is not a valid Unicode code point.
Since:
11
Since this method supports any Unicode code point, the length of the returned String is not necessarily 1.
The code needed for the example given in the question is simply:
int codePoint = '\u2202';
String s = Character.toString(codePoint); // <<< Requires JDK 11 !!!
System.out.println(s); // Prints ∂
This approach offers several advantages:
It works for any Unicode code point rather than just those that can be handled using a char.
It's concise, and it's easy to understand what the code is doing.
It returns the value as a string rather than a char[], which is often what you want. The answer posted by McDowell is appropriate if you want the code point returned as char[].
This is how you do it:
int cc = 0x2202;
char ccc = (char) Integer.parseInt(String.valueOf(cc), 16);
final String text = String.valueOf(ccc);
This solution is by Arne Vajhøj.
The code below will write the 4 unicode chars (represented by decimals) for the word "be" in Japanese. Yes, the verb "be" in Japanese has 4 chars!
The value of characters is in decimal and it has been read into an array of String[] -- using split for instance. If you have Octal or Hex, parseInt take a radix as well.
// pseudo code
// 1. init the String[] containing the 4 unicodes in decima :: intsInStrs
// 2. allocate the proper number of character pairs :: c2s
// 3. Using Integer.parseInt (... with radix or not) get the right int value
// 4. place it in the correct location of in the array of character pairs
// 5. convert c2s[] to String
// 6. print
String[] intsInStrs = {"12354", "12426", "12414", "12377"}; // 1.
char [] c2s = new char [intsInStrs.length * 2]; // 2. two chars per unicode
int ii = 0;
for (String intString : intsInStrs) {
// 3. NB ii*2 because the 16 bit value of Unicode is written in 2 chars
Character.toChars(Integer.parseInt(intsInStrs[ii]), c2s, ii * 2 ); // 3 + 4
++ii; // advance to the next char
}
String symbols = new String(c2s); // 5.
System.out.println("\nLooooonger code point: " + symbols); // 6.
// I tested it in Eclipse and Java 7 and it works. Enjoy
Here is a block to print out unicode chars between \u00c0 to \u00ff:
char[] ca = {'\u00c0'};
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 16; j++) {
String sc = new String(ca);
System.out.print(sc + " ");
ca[0]++;
}
System.out.println();
}
Unfortunatelly, to remove one backlash as mentioned in first comment (newbiedoodle) don't lead to good result. Most (if not all) IDE issues syntax error. The reason is in this, that Java Escaped Unicode format expects syntax "\uXXXX", where XXXX are 4 hexadecimal digits, which are mandatory. Attempts to fold this string from pieces fails. Of course, "\u" is not the same as "\\u". The first syntax means escaped 'u', second means escaped backlash (which is backlash) followed by 'u'. It is strange, that on the Apache pages is presented utility, which doing exactly this behavior. But in reality, it is Escape mimic utility. Apache has some its own utilities (i didn't testet them), which do this work for you. May be, it is still not that, what you want to have. Apache Escape Unicode utilities But this utility 1 have good approach to the solution. With combination described above (MeraNaamJoker). My solution is create this Escaped mimic string and then convert it back to unicode (to avoid real Escaped Unicode restriction). I used it for copying text, so it is possible, that in uencode method will be better to use '\\u' except '\\\\u'. Try it.
/**
* Converts character to the mimic unicode format i.e. '\\u0020'.
*
* This format is the Java source code format.
*
* CharUtils.unicodeEscaped(' ') = "\\u0020"
* CharUtils.unicodeEscaped('A') = "\\u0041"
*
* #param ch the character to convert
* #return is in the mimic of escaped unicode string,
*/
public static String unicodeEscaped(char ch) {
String returnStr;
//String uniTemplate = "\u0000";
final static String charEsc = "\\u";
if (ch < 0x10) {
returnStr = "000" + Integer.toHexString(ch);
}
else if (ch < 0x100) {
returnStr = "00" + Integer.toHexString(ch);
}
else if (ch < 0x1000) {
returnStr = "0" + Integer.toHexString(ch);
}
else
returnStr = "" + Integer.toHexString(ch);
return charEsc + returnStr;
}
/**
* Converts the string from UTF8 to mimic unicode format i.e. '\\u0020'.
* notice: i cannot use real unicode format, because this is immediately translated
* to the character in time of compiling and editor (i.e. netbeans) checking it
* instead reaal unicode format i.e. '\u0020' i using mimic unicode format '\\u0020'
* as a string, but it doesn't gives the same results, of course
*
* This format is the Java source code format.
*
* CharUtils.unicodeEscaped(' ') = "\\u0020"
* CharUtils.unicodeEscaped('A') = "\\u0041"
*
* #param String - nationalString in the UTF8 string to convert
* #return is the string in JAVA unicode mimic escaped
*/
public String encodeStr(String nationalString) throws UnsupportedEncodingException {
String convertedString = "";
for (int i = 0; i < nationalString.length(); i++) {
Character chs = nationalString.charAt(i);
convertedString += unicodeEscaped(chs);
}
return convertedString;
}
/**
* Converts the string from mimic unicode format i.e. '\\u0020' back to UTF8.
*
* This format is the Java source code format.
*
* CharUtils.unicodeEscaped(' ') = "\\u0020"
* CharUtils.unicodeEscaped('A') = "\\u0041"
*
* #param String - nationalString in the JAVA unicode mimic escaped
* #return is the string in UTF8 string
*/
public String uencodeStr(String escapedString) throws UnsupportedEncodingException {
String convertedString = "";
String[] arrStr = escapedString.split("\\\\u");
String str, istr;
for (int i = 1; i < arrStr.length; i++) {
str = arrStr[i];
if (!str.isEmpty()) {
Integer iI = Integer.parseInt(str, 16);
char[] chaCha = Character.toChars(iI);
convertedString += String.valueOf(chaCha);
}
}
return convertedString;
}
char c=(char)0x2202;
String s=""+c;
(ANSWER IS IN DOT NET 4.5 and in java, there must be a similar approach exist)
I am from West Bengal in INDIA.
As I understand your problem is ...
You want to produce similar to ' অ ' (It is a letter in Bengali language)
which has Unicode HEX : 0X0985.
Now if you know this value in respect of your language then how will you produce that language specific Unicode symbol right ?
In Dot Net it is as simple as this :
int c = 0X0985;
string x = Char.ConvertFromUtf32(c);
Now x is your answer.
But this is HEX by HEX convert and sentence to sentence conversion is a work for researchers :P