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Decode and replace hex values in a string in Java

I have the a string in Java which contains hex values beneath normal characters. It looks something like this:
String s = "Hello\xF6\xE4\xFC\xD6\xC4\xDC\xDF"
What I want is to convert the hex values to the characters they represent, so it will look like this:
"HelloöäüÖÄÜß"
Is there a way to replace all hex values with the actual character they represent?
I can achieve what I want with this, but I have to do one line for every character and it does not cover unexcepted characters:
indexRequest = indexRequest.replace("\\xF6", "ö");
indexRequest = indexRequest.replace("\\xE4", "ä");
indexRequest = indexRequest.replace("\\xFC", "ü");
indexRequest = indexRequest.replace("\\xD6", "Ö");
indexRequest = indexRequest.replace("\\xC4", "Ä");
indexRequest = indexRequest.replace("\\xDC", "Ü");
indexRequest = indexRequest.replace("\\xDF", "ß");

public static void main(String[] args) {
String s = "Hello\\xF6\\xE4\\xFC\\xD6\\xC4\\xDC\\xDF\\xFF ";
StringBuffer sb = new StringBuffer();
Pattern p = Pattern.compile("\\\\x[0-9A-F]+");
Matcher m = p.matcher(s);
while(m.find()){
String hex = m.group(); //find hex values
int num = Integer.parseInt(hex.replace("\\x", ""), 16); //parse to int
char bin = (char)num; // cast int to char
m.appendReplacement(sb, bin+""); // replace hex with char
}
m.appendTail(sb);
System.out.println(sb.toString());
}

I would loop through every chacter to find the '\' and than skip one char and start a methode with the next two chars.
And than just use the code by Michael Berry
here:
Convert a String of Hex into ASCII in Java

You can use a regex [xX][0-9a-fA-F]+ to identify all the hex code in your string, convert them to there corresponding character using Integer.parseInt(matcher.group().substring(1), 16) and replace them in string. Below is a sample code for it
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class HexToCharacter {
public static void main(String[] args) {
String s = "HelloxF6xE4xFCxD6xC4xDCxDF";
StringBuilder sb = new StringBuilder(s);
Pattern pattern = Pattern.compile("[xX][0-9a-fA-F]+");
Matcher matcher = pattern.matcher(s);
while(matcher.find()) {
int indexOfHexCode = sb.indexOf(matcher.group());
sb.replace(indexOfHexCode, indexOfHexCode+matcher.group().length(), Character.toString((char)Integer.parseInt(matcher.group().substring(1), 16)));
}
System.out.println(sb.toString());
}
}
I have tested this regex pattern using your string. If there are other test-cases that you have in mind, then you might need to change regex accordingly

How to make control characters visible?

I have to display string with visible control characters like \n, \t etc.
I have tried quotations like here, also I have tried to do something like
Pattern pattern = Pattern.compile("\\p{Cntrl}");
Matcher matcher = pattern.matcher(str);
String controlChar = matcher.group();
String replace = "\\" + controlChar;
result = result.replace(controlChar, replace);
but I have failed

Alternative: Use visible characters instead of escape sequences.
To make control characters "visible", use the characters from the Unicode Control Pictures Block, i.e. map \u0000-\u001F to \u2400-\u241F, and \u007F to \u2421.
Note that this requires output to be Unicode, e.g. UTF-8, not a single-byte code page like ISO-8859-1.
private static String showControlChars(String input) {
StringBuffer buf = new StringBuffer();
Matcher m = Pattern.compile("[\u0000-\u001F\u007F]").matcher(input);
while (m.find()) {
char c = m.group().charAt(0);
m.appendReplacement(buf, Character.toString(c == '\u007F' ? '\u2421' : (char) (c + 0x2400)));
if (c == '\n') // Let's preserve newlines
buf.append(System.lineSeparator());
}
return m.appendTail(buf).toString();
}
Output using method above as input text:
␉private static String showControlChars(String input) {␍␊
␉␉StringBuffer buf = new StringBuffer();␍␊
␉␉Matcher m = Pattern.compile("[\u0000-\u001F\u007F]").matcher(input);␍␊
␉␉while (m.find()) {␍␊
␉␉␉char c = m.group().charAt(0);␍␊
␉␉␉m.appendReplacement(buf, Character.toString(c == '\u007F' ? '\u2421' : (char) (c + 0x2400)));␍␊
␉␉␉if (c == '\n')␍␊
␉␉␉␉buf.append(System.lineSeparator());␍␊
␉␉}␍␊
␉␉return m.appendTail(buf).toString();␍␊
␉}␍␊

Simply replace occurences of '\n' with the escaped version (i.e. '\\n'), like this:
final String result = str.replace("\n", "\\n");
For example:
public static void main(final String args[]) {
final String str = "line1\nline2";
System.out.println(str);
final String result = str.replace("\n", "\\n");
System.out.println(result);
}
Will yield the output:
line1
newline
line1\nnewline

just doing
result = result.replace("\\", "\\\\");
will work!!

Replacing \\u by \u in java string

I have a string which contains normal text and Unicode in between, for example "abc\ue415abc".
I want to replace all occurrences of \\u with \u. How can I achieve this?
I used the following code but it's not working properly.
String s = "aaa\\u2022bbb\\u2014ccc";
StringBuffer buf = new StringBuffer();
Matcher m = Pattern.compile("\\\\u([0-9A-Fa-f]{4})").matcher(s);
while (m.find()) {
try {
int cp = Integer.parseInt(m.group(1), 16);
m.appendReplacement(buf, "");
buf.appendCodePoint(cp);
} catch (NumberFormatException e) {
}
}
m.appendTail(buf);
s = buf.toString();
Please help. Thanks in advance.

From API reference: http://developer.android.com/reference/java/lang/String.html#replace(java.lang.CharSequence, java.lang.CharSequence)
You can use public
public String replace (CharSequence target, CharSequence replacement)
string = string.replace("\\u", "\u");
or
String replacedString = string.replace("\\u", "\u");

Your initial string doesn't, in fact, have any double backslashes.
String s = "aaa\\u2022bbb\\u2014ccc";
yields a string that contains aaa\u2022bbb\u2014ccc, as \\ is just java string-literal escaping for \.
If you want unicode characters: (StackOverflow21028089.java)
import java.util.regex.*;
class StackOverflow21028089 {
public static void main(String[] args) {
String s = "aaa\\u2022bbb\\u2014ccc";
StringBuffer buf = new StringBuffer();
Matcher m = Pattern.compile("\\\\u([0-9A-Fa-f]{4})").matcher(s);
while (m.find()) {
try {
// see example:
// http://docs.oracle.com/javase/7/docs/api/java/util/regex/Matcher.html#appendReplacement%28java.lang.StringBuffer,%20java.lang.String%29
int cp = Integer.parseInt(m.group(1), 16);
char[] chars = Character.toChars(cp);
String rep = new String(chars);
System.err.printf("Found %d which means '%s'\n", cp, rep);
m.appendReplacement(buf, rep);
} catch (NumberFormatException e) {
System.err.println("Confused: " + e);
}
}
m.appendTail(buf);
s = buf.toString();
System.out.println(s);
}
}
=>
Found 8226 which means '•'
Found 8212 which means '—'
aaa•bbb—ccc
If you want aaa\u2022bbb\u2014ccc, that's what you started with. If you meant to start with a string literal with aaa\\u2022bbb\\u2014ccc, that's this:
String s = "aaa\\\\u2022bbb\\\\u2014ccc";
and converting it to the one with single slashes can be as simple as #Overv's code:
s = s.replaceAll("\\\\u", "\\u");
though since backslash has a special meaning in regex patterns and replacements (see Matcher's docs) (in addition to java parsing), this should probably be:
s = s.replaceAll("\\\\\\\\u", "\\\\u");
=>
aaa\u2022bbb\u2014ccc

Try this:
s = s.replace(s.indexOf("\\u"), "\u");

There is a contains method and a replace method in String. That being said
String hello = "hgjgu\udfgyud\\ushddsjn\hsdfds\\ubjn";
if(hello.contains("\\u"))
hello.replace("\\u","\u");
System.out.println(hello);
It will print :- hgjgu\udfgyud\ushddsjn\hsdfds\ubjn

Java Regex BBCode not replacing but

I am going to use the following method to replace special BB Codes for html links
public String replace(String text , String bbcode , String imageLocation ){
StringBuffer imageBuffer = new StringBuffer ("");
Pattern pattern = Pattern.compile("\\"+bbcode );
Matcher matcher = pattern.matcher(text);
StringBuilder builder = new StringBuilder();
int i = 0;
while (matcher.find()) {
//String orginal = replacements.get(matcher.group(1));
imageBuffer.append("<img src=\"" + imageLocation + "\" />");
String replacement = imageBuffer.toString();
builder.append(text.substring(i, matcher.start()));
if (replacement == null) {
builder.append(matcher.group(0));
} else {
builder.append(replacement);
}
i = matcher.end();
}
builder.append(text.substring(i, text.length()));
return builder.toString();
}
but when it comes to replacing the following bbcodes ,
:D
O:-)
:-[
:o)
:~(
:xx(
:-]
:-(
^3^
#_#
:O
:)
:P
;-)
???
?_?
Z_Z
It turns out to be not closing bracket and : <-- not recognized
How should I override the regex functional code and replace the abovementioned list of icons as html image links?
I am currently using this string array but
It comes out with the following error
error: Error: No resource type specified (at '^index_6' with value '#_#').
<string-array name="hkgicon_array">
<item>[369]</item>
<item>#adore#</item>
<item>#yup#</item>
<item>#ass#</item>
<item>:-(</item>
<item>^3^</item>
<item>#_#</item>
</string-array>

USE QUOTE
You can use Pattern pattern = Pattern.compile(Pattern.quote(bbcode ));
in your code instead of Pattern.compile("\\"+bbcode );.
Try this code:
public static String replace(String text , String bbcode , String imageLocation ){
StringBuffer imageBuffer = new StringBuffer ("");
Pattern pattern = Pattern.compile(Pattern.quote(bbcode ));
Matcher matcher = pattern.matcher(text);
StringBuilder builder = new StringBuilder();
int i = 0;
while (matcher.find()) {
//String orginal = replacements.get(matcher.group(1));
imageBuffer.append("<img src=\"" + imageLocation + "\" />");
String replacement = imageBuffer.toString();
builder.append(text.substring(i, matcher.start()));
if (replacement == null) {
builder.append(matcher.group(0));
} else {
builder.append(replacement);
}
i = matcher.end();
}
builder.append(text.substring(i, text.length()));
return builder.toString();
}
Refer stackoverflow for more details .

Processing a String and replacing al URL's with working links via Java

How would I transform a text like the following or any other text containing an URL (http ftp etc)
Go to this link http://www.google.com (ofc stack overflow already does this, on my website this is just plain text);
Into this
Go to this link www.google.com
I've come up with this method
public String transformURLIntoLinks(String text){
String urlValidationRegex = "(https?|ftp)://(www\\d?|[a-zA-Z0-9]+)?.[a-zA-Z0-9-]+(\\:|.)([a-zA-Z0-9.]+|(\\d+)?)([/?:].*)?";
Pattern p = Pattern.compile(urlValidationRegex);
Matcher m = p.matcher(text);
StringBuffer sb = new StringBuffer();
while(m.find()){
String found =m.group(1); //this String is only made of the http or ftp (just the first part of the link)
m.appendReplacement(sb, ""http"
}
m.appendTail(sb);
return sb.toString();
}
The problem is the regexes i've tried match only the first part("http" or "ftp").
My output becomes: Go to this link <a href='http'>http</a>
It should be this
Go to this link <a href='http://www.google.com'>http://www.google.com</a>

public String transformURLIntoLinks(String text){
String urlValidationRegex = "(https?|ftp)://(www\\d?|[a-zA-Z0-9]+)?.[a-zA-Z0-9-]+(\\:|.)([a-zA-Z0-9.]+|(\\d+)?)([/?:].*)?";
Pattern p = Pattern.compile(urlValidationRegex);
Matcher m = p.matcher(text);
StringBuffer sb = new StringBuffer();
while(m.find()){
String found =m.group(0);
m.appendReplacement(sb, "<a href='"+found+"'>"+found+"</a>");
}
m.appendTail(sb);
return sb.toString();
}
m.group(1) was the mistake. m.group(0) works.
This will transform any URL found in the text into an anchor.