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Regex to remove special characters in java

I have a string with a couple of special characters and need to remove only a few (~ and `). I have written the code below, but when I print the splitted strings, getting empty also with values.
String str = "ABC123-xyz`~`XYZ 1.7A";
String[] str1= varinaces.split("[\\~`]");
for(int i=0; i< str1.length ; i++){
System.out.println("str==="+ parts[i] );
}
Output:
str===ABC123-xyz
str===
str===
str===XYZ 1.7A
why empty strings also printing here ?

You’re splitting on one special char... split on 1 or more:
String[] str1= varinaces.split("[~`]+");
Note also that the tilda ~ doesn’t need escaping.

Its because when you use the .split() method it returns a String array of 4 items shown below:
String[4] { "ABC123-xyz", "", "", "XYZ 1.7A" }
And then in your for loop you printing all items of that array. You can use the following to resolve it:
for(int i=0; i< str1.length ; i++){
if(parts[i].compareTo("") > 0) {
System.out.println("str==="+ parts[i] );
}
}

The split method returns the stuff around every match of the regex. Your regex, [~`], matches to a single character that is either "~" or "`".
The parts of the string separated by matches to that regex are determined as follows:
The string "ABC123-xyz" is returned because it is split off the given string at the character: "`".
In between that character and the next match, "~", is the empty string, and so on.
If you want it to match to more, use [~`]+

Splitting characters

My characters is "!,;,%,#,**,**,(,)" which get from XML. when I split it with ',', I lost the ','.
How can I do to avoid it.
I have already tried to change the comma to '&#002C', but it does not work.
Thre result I want is "!,;,%,#,,,(,)", but not "!,;,%,#,,(,)"

String::split use regex so you can split with this regex ((?<!,),|,(?!,)) like this :
String string = "!,;,%,#,,,(,)";
String[] split = string.split("((?<!,),|,(?!,))");
Details
(?<!,), match a comma if not preceded by a comma
| or
,(?!,) match a comma if not followed by a comma
Outputs
!
;
%
#
,
(
)

If you are trying to extract all characters from string, you can do so by using String.toCharArray()[1] :
String str = "sample string here";
char[] char_array = s.toCharArray();
If you just want to iterate over the characters in the string, you can use the character array obtained from above method or do so by using a for loop and str.charAt(i)[2] to access the character at position i.
[1] https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#toCharArray()
[2]https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#charAt(int)

try this, this could be help full. First I replaced the ',' with other string and do split. After complete other string replace with ','
public static void main(String[] args) {
String str = "!,;,%,#,**,**,(,)";
System.out.println(str);
str = str.replace("**,**","**/!/**");
String[] array = str.split(",");
System.out.println(Arrays.stream(array).map(s -> s.replace("**/!/**", ",")).collect(Collectors.toList()));
}
out put
!,;,%,#,**,**,(,)
[!, ;, %, #, ,, (, )]

First, we need to define when the comma is an actual delimiter, and when it is part of a character sequence.
We need to assume that a sequence of commas surrounded by commas is an actual character sequence we want to capture. It can be done with lookarounds:
String s = "!,;,,,%,#,**,**,,,,(,)";
List<String> list = Arrays.asList(s.split(",(?!,)|(?<!,),"));
This regular expression splits by a comma that is either preceded by something that is not a comma, or followed by something that is not a comma.
Note that your formatting string, that is, every character sequence separated by a comma, is a bad design, since you require both the possibility to use a comma as sequence, and the possibility to use multiple characters to be used. That means you can combine them too!
What, for example, if I want to use these two character sequences:
,
,,,,
Then I construct the formatting string like this: ,,,,,,. It is now unclear whether , and ,,,, should be character sequences, or ,, and ,,,.

String.replace() not replacing all occurrences

I have a very long string which looks similar to this.
355,356,357,358,359,360,361,382,363,364,365,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,368,369,313,370,371,372,373,374,375,376,377,378,379,380,381,382,382,382,382,382,382,383,384,385,380,381,382,382,382,382,382,386,387,388,389,380,381,382,382,382,382,382,382,390,391,380,381,382,382,382,382,382,392,393,394,395,396,397,398,399,....
When I tried using the following code to remove the number 382 from the string.
String str = "355,356,357,358,359,360,361,382,363,364,365,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,368,369,313,370,371,372,373,374,375,376,377,378,379,380,381,382,382,382,382,382,382,383,384,385,380,381,382,382,382,382,382,386,387,388,389,380,381,382,382,382,382,382,382,390,391,380,381,382,382,382,382,382,392,393,394,395,396,397,398,399,...."
str = str.replace(",382,", ",");
But it seems that not all occurrences are being replaced. The string which originally had above 3000 occurrences still was left with about 630 occurrences after replacing.
Is the capability of String.replace() limited? If so, is there a possible way of achieving what I need?

You need to replace the trailing comma as well (if one exists, which it won't if last in the list):
str = str.replaceAll("\\b382,?", "");
Note \b word boundary to prevent matching "-,1382,-".
The above will convert:
382,111,382,1382,222,382
to:
111,1382,222

I think the issue is your first argument to replace(), in particular the comma (,) before and after 382. If you have "382,382,383", you will only match the inner ",382," and leave the initial one behind. Try:
str.replace("382,", "");
Although this will fail to match "382" at the very end as it does not have a comma after it.
A full solution might entail two method calls thus:
str = str.replace("382", ""); // Remove all instances of 382
str.replaceAll(",,+", ","); // Compress all duplicates, triplicates, etc. of commas
This combines the two approaches:
str.replaceAll("382,?", ""); // Remove 382 and an optional comma after it.
Note: both of the last two approaches leave a trailing comma if 382 is at the end.

try this
str = str.replaceAll(",382,", ",");

Firstly, remove the preceding comma in your matching string. Then, remove duplicated commas by replacing commas with a single comma using java regular expression.
String input = "355,356,357,358,359,360,361,382,363,364,365,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,360,361,363,366,368,369,313,370,371,372,373,374,375,376,377,378,379,380,381,382,382,382,382,382,382,383,384,385,380,381,382,382,382,382,382,386,387,388,389,380,381,382,382,382,382,382,382,390,391,380,381,382,382,382,382,382,392,393,394,395,396,397,398,399";
String result = input.replace("382,", ","); // remove the preceding comma
String result2 = result.replaceAll("[,]+", ","); // replace duplicate commas
System.out.println(result2);

As dave already said, the problem is that your pattern overlaps. In the string "...,382,382,..." there are two occurrences of ",382,":
"...,382,382,..."
----- first occurrence
----- second occurrence
These two occurrences overlap at the comma, and thus Java can only replace one of them. When finding occurrences, it does not see yet what you replace the pattern with, and thus it doesn't see that new occurrence of ",382," is generated when replacing the first occurrence is replaced by the comma.
If your data is known not to contain numbers with more than 3 digits, then you might do:
str.replace("382,", "");
and then handle occurrences at the end as a special case. But if your data can contain big numbers, then "...,1382,..." will be replaced by "...,1,..." which probably is not what you want.
Here are two solutions that do not have the above problem:
First, simply repeat the replacement until no changes occur anymore:
String oldString = str;
str = str.replace(",382,", ",");
while (!str.equals(oldString)) {
oldString = str;
str = str.replace(",382,", ",");
}
After that, you will have to handle possible occurrences at the end of the string.
Second, if you have Java 8, you can do a little more work yourself and use Java streams:
str = Arrays.stream(str.split(","))
.filter(s -> !s.equals("382"))
.collect(Collectors.joining(","));
This first splits the string at ",", then filters out all strings which are equal to "382", and then concatenates the remaining strings again with "," in between.
(Both code snippets are untested.)

Traditional way:
String str = ",abc,null,null,0,0,7,8,9,10,11,12,13,14";
String newStr = "", word = "";
for (int i=0; i<str.length(); i++) {
if (str.charAt(i) == ',') {
if (word.equals("null") || word.equals("0"))
word = "";
newStr += word+",";
word = "";
} else {
word += str.charAt(i);
if (i == str.length()-1)
newStr += word;
}
}
System.out.println(newStr);
Output:
,abc,,,,,7,8,9,10,11,12,13,14

Parse and remove special characters in java regex

So we were looking at some of the other regex posts and we are having trouble removing a special case in one instance; the special character is in the beginning of the word.
We have the following line in our code:
String k = s.replaceAll("([a-z]+)[()?:!.,;]*", "$1");
where s is a singular word. For example, when parsing the sentence "(hi hi hi)" by tokenizing it, and then performing the replaceAll function on each token, we get an output of:
(hi
hi
hi
What are we missing in our regex?

You can use an easier approach - replace the characters that you do not want with spaces:
String k = s.replaceAll("[()?:!.,;]+", " ");

Position matters so you would need to match the excluded charcters before the capturing group also:
String k = s.replaceAll("[()?:!.,;]*([a-z]+)[()?:!.,;]*", "$1");

your replace just removed the "special chars" after the [a-z]+, that's why the ( before hi is left there.
If you know s is a single word
you could either:
String k = s.replaceAll("\\W*(\\w+)\\W*", "$1");
or
String k = s.replaceAll("\\W*", "");

This can be more simple
try this :
String oldString = "Hi There ##$ What is %#your name?##$##$ 0123$$";
System.out.println(oldString.replaceAll("[\\p{Punct}\\s\\d]+", " ");
output :
Hi There What is your name 0123
So it also accepts numeric.
.replaceAll("[\p{Punct}\s\d]+", " ");
will replace alll the Punctuations used which includes almost all the special characters.

Split on unescaped spaces

I've got strings in the form of:
/path/ /path\ with\ space/ /another/
I need to split this so I end up with an array containing:
[ /path/, /path with space/, /another/ ]
Is there an easy regex that would take care of this? Previously I was using \s+ but obviously that doesn't work here.

Use a negative lookbehind.
String s = "/path/ /path\\ with\\ space/ /another/";
String[] parts = s.split("(?<!\\\\)\\s+");
System.out.println(Arrays.toString(parts));
// prints [/path/, /path\ with\ space/, /another/]
Note that the second element still contains \s, which you'll need to strip out yourself.
for (int i=0; i<parts.length; i++)
{
parts[i] = parts[i].replaceAll("\\\\ ", " ");
}
System.out.println(Arrays.toString(parts));
// prints [/path/, /path with space/, /another/]
Yes, that's four \s just to match a single one in the string.

What if you split it on the and "/ " (note the space) and append a forward slash after each match? I'm not sure if split will work with a lookbehind.

Did you try with [/](w|s)+[/] ?