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Converting letters to alphabet position with two JLists

I'm trying to replace all words (alphabet letters) from JList1 to the number corresponding its place in the alphabet to JList2 with the press of the Run button. (ex. A to 01) And if it's not an English alphabet letter then leaving it as it is. Capitalization doesn't matter (a and A is still 01) and spaces should be kept.
For visual purposes:
"Apple!" should be converted to "0116161205!"
"stack Overflow" to "1920010311 1522051806121523"
"über" to "ü020518"
I have tried a few methods I found on here, but had zero clue how to add the extra 0 in front of the first 9 letters or keep the spaces. Any help is much appreciated.

Here is a solution :
//Create a Map of character and equivalent number
Map<Character, String> lettersToNumber = new HashMap<>();
int i = 1;
for(char c = 'a'; c <= 'z'; c++) {
lettersToNumber.put(c, String.format("%02d", i++));
}
//Loop over the characters of your input and the corresponding number
String result = "";
for(char c : "Apple!".toCharArray()) {
char x = Character.toLowerCase(c);
result+= lettersToNumber.containsKey(x) ? lettersToNumber.get(x) : c;
}
Input, Output
Apple! => 0116161205!
stack Overflow => 1920010311 1522051806121523
über => ü020518

So given...
(ex. A to 01) And if it's not an English alphabet letter then leaving it as it is. Capitalization doesn't matter (a and A is still 01) and spaces should be kept.
This raises some interesting points:
We don't care about non-english characters, so we can dispense with issues around UTF encoding
Capitalization doesn't matter
Spaces should be kept
The reason these points are interesting to me is it means we're only interested in a small subset of characters (1-26). This immediately screams "ASCII" to me!
This provides an immediate lookup table which doesn't require us to produce anything up front, it's immediately accessible.
A quick look at any ascii table provides us with all the information we need. A-Z is in the range of 65-90 (since we don't care about case, we don't need to worry about the lower case range.
But how does that help us!?
Well, this now means the primary question becomes, "How do we convert a char to an int?", which is amazingly simple! A char can be both a "character" and a "number" at the same time, because of the ASCII encoding support!
So if you were to print out (int)'A', it would print 65! And since all the characters are in order, we just need to subtract 64 from 65 to get 1!
That's basically your entire problem solved right there!
Oh, okay, you need to deal with the edge cases of characters not falling between A-Z, but that's just a simple if statement
A solution based on the above "might" look something like...
public static String convert(String text) {
int offset = 64;
StringBuilder sb = new StringBuilder(32);
for (char c : text.toCharArray()) {
char input = Character.toUpperCase(c);
int value = ((int) input) - offset;
if (value < 1 || value > 25) {
sb.append(c);
} else {
sb.append(String.format("%02d", value));
}
}
return sb.toString();
}
Now, there are a number of ways you might approach this, I've chosen a path based on my understanding of the problem and my experience.
And based on your example input...
String[] test = {"Apple!", "stack Overflow", "über"};
for (String value : test) {
System.out.println(value + " = " + convert(value));
}
would produce the following output...
Apple! = 0116161205!
stack Overflow = 1920010311 1522051806121523
über = ü020518

How to compare Chinese characters in Java using 'equals()'

I want to compare a string portion (i.e. character) against a Chinese character. I assume due to the Unicode encoding it counts as two characters, so I'm looping through the string with increments of two. Now I ran into a roadblock where I'm trying to detect the '兒' character, but equals() doesn't match it, so what am I missing ? This is the code snippet:
for (int CharIndex = 0; CharIndex < tmpChar.length(); CharIndex=CharIndex+2) {
// Account for 'r' like in dianr/huir
if (tmpChar.substring(CharIndex,CharIndex+2).equals("兒")) {
Also, feel free to suggest a more elegant way to parse this ...
[UPDATE] Some pics from the debugger, showing that it doesn't match, even though it should. I pasted the Chinese character from the spreadsheet I use as input, so I don't think it's a copy and paste issue (unless the unicode gets lost along the way)
oh, dang, apparently it does not work simply copy and pasting:

Use CharSequence.codePoints(), which returns a stream of the codepoints, rather than having to deal with chars:
tmpChar.codePoints().forEach(c -> {
if (c == '兒') {
// ...
}
});
(Of course, you could have used tmpChar.codePoints().filter(c -> c == '兒').forEach(c -> { /* ... */ })).

Either characters, accepting 兒 as substring.
String s = ...;
if (s.contains("兒")) { ... }
int position = s.indexOf("兒");
if (position != -1) {
int position2 = position + "兒".length();
s = s.substring(0, position) + "*" + s.substring(position2);
}
if (s.startsWith("兒", i)) {
// At position i there is a 兒.
}
Or code points where it would be one code point. As that is not really easier, variable substring seem fine.

if (tmpChar.substring(CharIndex,CharIndex+2).equals("兒")) {
Is your problem. 兒 is only one UTF-16 character. Many Chinese characters can be represented in UTF-16 in one code unit; Java uses UTF-16. However, other characters are two code units.
There are a variety of APIs on the String class for coping.
As offered in another answer, obtaining the IntStream from codepoints allows you to get a 32-bit code point for each character. You can compare that to the code point value for the character you are looking for.
Or, you can use the ICU4J library with a richer set of facilities for all of this.

Java String comparison issue with blank spaces

I invoke using java some web service which returns some values to me.
Those values are some attribute names which I successful receive. I put the returned values in database.
For example the values I retrieved are:
DSLAM port
Interfejs na ISP
So when I look in the database those values are stored in DB as DSLAM port and Interfejs na ISP.
That is how I received them and that is how they are stored in DB (so only with one blank space between the words).
So I'm receiving those values from a web service but when I try to do a comparison additional in the class:
if ( attribute.trim().equalsIgnoreCase("Interfejs na ISP") ) {
System.out.println("attr2");
}
or
if ( attribute.trim().equalsIgnoreCase("DSLAM port") ) {
System.out.println("attr3");
}
I am not having the System Print lines to my console, if is always false.
What can be the problem and how can I solve it?
This is a really strange behavior for me. Attributes are stored correctly and only when I try to compare it I get strange behavior. The if clause is never true. Can there be some issue with the language format?
Additionally if I try with single word:
if (attribute.trim().equalsIgnoreCase("Telefon"))
{ System.out.println("attr5");
}
Then it writes in System Out.
So with sinlge word it seems it does not have problems

Output a serialization of the byte array behind the values of attribute to see exactly what characters it contains. This will help you catch stuff like nbsp for whitespace etc. When this is done, you can change your string matching to what you actually get back from the database, or tune your persistence to produce straight forward values.
It is not enough to just println the variable, instead, you have to iterate over all the buckets in the array and output them in hex or dec - your target output is something like "[44, 53, 4C, .. ]", which would correspond to "DSL..". To convert the byte array to a hex representation, you can use this snippet, or as an exercise, try it on your own:
public static String convertToHexString(byte[] data) {
StringBuffer buf = new StringBuffer();
for (int i = 0; i < data.length; i++) {
int nibble = (data[i] >>> 4) & 0x0F;
int two_nibbles = 0;
do {
if ((0 <= nibble) && (nibble <= 9))
buf.append((char) ('0' + nibble));
else
buf.append((char) ('a' + (nibble - 10)));
nibble = data[i] & 0x0F;
} while (two_nibbles++ < 1);
}
return buf.toString();
}
Now when you have that output, take an ascii table to look up which values are contained in the string and change your ifs depending on what is actually contained in the Strings. Possibly by using a matching regex.
Chances are the whitespaces are some non-trivial blank characters like (Hex: A0), but it's also possible that you are having an encoding problem. Feel free to post the hex values if the character tables don't help.

XOR mask in java android

User enter code word and text to encrypt and program should put XOR mask - code on text - and turn back to normal , but it just put on mask and dont turn back to normal look , why ?
public void onClick(View arg0) {
code = etCode.getText().toString();
text = etText.getText().toString();
while(code.length()<text.length()){
code+=code;
}
char[] Ccode = code.toCharArray();
char[] Ctext = text.toCharArray();
for(i=0;i<Ctext.length;i++){
Ctext[i]^=Ccode[i];
}
rezult=Ctext.toString();
for(i=0;i<Ctext.length;i++){
Ctext[i]^=Ccode[i];
}
rezult+="\n";
rezult+=Ctext.toString();
tvMain.setText(rezult);
}
});
if I enter code : code , text : text
it shows:
[C#40527808
[C#40527808

You output the address of the array. You want the content.
Arrays have not useful toString() mwthod.
change
rezult=Ctext.toString();
to
rezult=new String(Ctext);
same for
rezult+=Ctext.toString();
=>
rezult+=new String(Ctext);

You cannot use toString() to convert a char array to String.
Use tvMain.setText(new String(Ctext));

In Java, the default toString() operation on arrays is to convert to an internal identifier, which is what you're seeing. Try using:
rezult = new String(Ctext);
...
rezult += new String(Ctext);
or, depending on what you want to display (since it's not clear to me that Ctext always contains displayable characters):
rezult = Arrays.toString(Ctext);
...
rezult += Arrays.toString(Ctext);
This will give you a comma-separated array of character values, surrounded by square brackets.

Convert foreign characthers (Greek) to English ones.

I am installing alarms, but they have an interface that I have to give name to various zones. The problem is that they can only accept greek characters that does not exist in english language else they have to be the English equivelant.
For example if I write "ΠΑΡΑΘΥΡΟ", the characthers 1,2,3,5,6,7 must enter in english because are the same with the greek ones in appearence. But chars 0 and 4 only must be in Greek.
I care only for capitals.
Any idea on how to do it with 2 simple jtextfields ?
Thank you!

Use a HashMap to translate characters. Since the problem domain is small and will probably never change, it's justifiable to hard-code the content of the map, like so:
private static final Map<Character, Character> GREEK_TO_ROMAN = new HashMap<>();
static {
GREEK_TO_ROMAN.put('\u0391', '\u0041'); // uppercase alpha
GREEK_TO_ROMAN.put('\u03A1', '\u0050'); // uppercase rho
// ...
}
Then get the input string's character array, translate characters as needed, and create a new String from the changed array:
String s = "ΠΑΡΑΘΥΡΟ";
char[] chars = s.toCharArray();
for (int i = 0; i < chars.length; i++) {
Character repl = GREEK_TO_ROMAN.get(chars[i]);
if (repl != null)
chars[i] = repl;
}
s = new String(chars);
How JTextField would come into play I don't quite see, but maybe if you want you can subclass it, overwrite the getText() method and make sure that any String it yields is already converted.