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How to remove spaces from string only if it occurs once between two words but not if it occurs thrice?

I am a beginner working on a diff and regenerate algorithm but for Strings. I store the patch in a file. To regenerate the new string from old I use that file. Although the code works, I face a problem when using space.
I use replaceAll(" ", ""); for removing spaces. This is fine when the string is [char][space][char], but creates problem when it is like [space][space][space]. Here, I want that the space be retained(only one).
I thought of doing replaceAll(" ", " ");. But this would leave spaces in type [char][space][char]. I am using scanner to scan through the string.
Is there a way to achieve this?
Input Output
c => c
cc => cc
c c => cc
c c => This is not possible. Since there will be padding of one space for each character
c c => c c

We can also split the string on where there are more than one white space, then join the resulting array by into a string using the Stream and Collector API.
Also we would replace the single spaces by using replaceAll() in a Stream#map operation:
String test = " this is a test of space in string ";
//using the pattern \\s{n,} for splitting at multi spaces
String[] arr = test.split("\\s{2,}");
String s = Arrays.stream(arr)
.map(str -> str.replaceAll(" ", ""))
.collect(Collectors.joining(" "));
System.out.println(s);
Output:
this isatestof spaceinstring

You could use lookarounds to do your replacement:
String newText = text
.replaceAll("(?<! ) (?! )", "")
.replaceAll(" +", " ");
The first replaceAll removes any space not surrounded by spaces; the second one replaces the remaining sequences of spaces by a single one.
Ideone example. Sequences of two or more spaces become a single space, and single spaces are removed.
Lookarounds
A lookaround in the context of regular expressions is a collective term for lookbehinds and lookaheads. These are so-called zero-width assertions, that means they match a certain pattern, but do not actually consume characters. There are positive and negative lookarounds.
A short example: the pattern Ira(?!q) matches the substring Ira, but only if it's not followed by a q. So if the input string is Iraq, it won't match, but if the input string is Iran, then the match is Ira.
More info:
https://www.regular-expressions.info/lookaround.html

If you want to replace any group of space by one you could use:
value.replaceAll("\\s+", " ")

I had to use two replacements:
String e = "a b c";
e = e.replaceAll("([A-Z|a-z])\\s([A-Z|a-z])", "$1$2");
e = e.replaceAll(" "," ");
System.out.println(e);
Which prints
ab c
The first one replaces any letter-space-letter combo with just the two letters, and then the second replaces any triple-space with a single space.
The first replacement is using backreferences. $1 refers to the part inside the first set of parenthesis that matches the first letter, and $2 refers to the part inside the second set of parenthesis.
If you have leading/trailing spaces on the input, you can call trim() before doing the replacements.
e = e.trim()

How not to match the first empty string in this regex?

(Disclaimer: the title of this question is probably too generic and not helpful to future readers having the same issue. Probably, it's just because I can't phrase it properly that I've not been able to find anything yet to solve my issue... I engage in modifying the title, or just close the question once someone will have helped me to figure out what the real problem is :) ).
High level description
I receive a string in input that contains two information of my interest:
A version name, which is 3.1.build and something else later
A build id, which is somenumbers-somenumbers-eitherwordsornumbers-somenumbers
I need to extract them separately.
More details about the inputs
I have an input which may come in 4 different ways:
Sample 1: v3.1.build.dev.12345.team 12345-12345-cici-12345 (the spaces in between are some \t first, and some whitespaces then).
Sample 2: v3.1.build.dev.12345.team 12345-12345-12345-12345 (this is very similar than the first example, except that in the second part, we only have numbers and -, no alphabetic characters).
Sample 3:
v3.1.build.dev.12345.team
12345-12345-cici-12345
(the above is very similar to sample 1, except that instead of \t and whitespaces, there's just a new line.
Sample 4:
v3.1.build.dev.12345.team
12345-12345-12345-12345
(same than above, with only digits and dashes in the second line).
Please note that in sample 3 and sample 4, there are some trailing spaces after both strings (not visible here).
To sum up, these are the 4 possible inputs:
String str1 = "v3.1.build.dev.12345.team\t\t\t\t\t 12345-12345-cici-12345";
String str2 = "v3.1.build.dev.12345.team\t\t\t\t\t 12345-12345-12345-12345";
String str3 = "v3.1.build.dev.12345.team \n12345-12345-cici-12345 ";
String str4 = "v3.1.build.dev.12345.team \n12345-12345-12345-12345 ";
My code currently
I have written the following code to extract the information I need (here reporting only relevant, please visit the fiddle link to have a complete and runnable example):
String versionPattern = "^.+[\\s]";
String buildIdPattern = "[\\s].+";
Pattern pVersion = Pattern.compile(versionPattern);
Pattern pBuildId = Pattern.compile(buildIdPattern);
for (String str : possibilities) {
Matcher mVersion = pVersion.matcher(str);
Matcher mBuildId = pBuildId.matcher(str);
while(mVersion.find()) {
System.out.println("Version found: \"" + mVersion.group(0).replaceAll("\\s", "") + "\"");
}
while (mBuildId.find()) {
System.out.println("Build-id found: \"" + mBuildId.group(0).replaceAll("\\s", "") + "\"");
}
}
The issue I'm facing
The above code works, pretty much. However, in the Sample 3 and Sample 4 (those where the build-id is separated by the version with a \n), I'm getting two matches: the first, is just a "", the second is the one I wish.
I don't feel this code is stable, and I think I'm doing something wrong with the regex pattern to match the build-id:
String buildIdPattern = "[\\s].+";
Does anyone have some ideas in order to exclude the first empty match on the build-id for sample 3 and 4, while keeping all the other matches?
Or some better way to write the regexs themselves (I'm open to improvements, not a big expert of regex)?

Based on your description it looks like your data is in form
NonWhiteSpaces whiteSpaces NonWhiteSpaces (optionalWhiteSpaces)
and you want to get only NonWhiteSpaces parts.
This can be achieved in numerous ways. One of them would be to trim() your string to get rid of potential trailing whitespaces and then split on the whitespaces (there should now only be in the middle of string). Something like
String[] arr = data.trim().split("\\s+");// \s also represents line separators like \n \r
String version = arr[0];
String buildID = arr[1];

(^v\w.+)\s+(\d+-\d+-\w+-\d+)\s*
It will capture 2 groups. One will capture the first section (v3.1.build.dev.12345.team), the second gets the last section (12345-12345-cici-12345)
It breaks down like: (^v\w.+) ensures that the string starts with a v, then captures all characters that are a number or letter (stopping on white space tabs etc.) \s+ matches any white space or tabs/newlines etc. as many times as it can. (\d+-\d+-\w+-\d+) this reads it in, ensuring that it conforms to your specified formatting. Note that this will still read in the dashes, making it easier for you to split the string after to get the information you need. If you want you could even make these their own capture groups making it even easier to get your info.
Then it ends with \s* just to make sure it doesn't get messed up by trailing white space. It uses * instead of + because we don't want it to break if there's no trailing white space.

I think this would be strong for production (aside from the fact that the strings cannot begin with any white-space - which is fixable, but I wasn't sure if it's what you're going for).
public class Other {
static String patternStr = "^([\\S]{1,})([\\s]{1,})(.*)";
static String str1 = "v3.1.build.dev.12345.team\t\t\t\t\t 12345-12345-cici-12345";
static String str2 = "v3.1.build.dev.12345.team\t\t\t\t\t 12345-12345-12345-12345";
static String str3 = "v3.1.build.dev.12345.team \n12345-12345-cici-12345 ";
static String str4 = "v3.1.build.dev.12345.team \n12345-12345-12345-12345 ";
static Pattern pattern = Pattern.compile(patternStr);
public static void main(String[] args) {
List<String> possibilities = Arrays.asList(str1, str2, str3, str4);
for (String str : possibilities) {
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
System.out.println("Version found: \"" + matcher.group(1).replaceAll("\\s", "") + "\"");
System.out.println("Some whitespace found: \"" + matcher.group(2).replaceAll("\\s", "") + "\"");
System.out.println("Build-id found: \"" + matcher.group(3).replaceAll("\\s", "") + "\"");
} else {
System.out.println("Pattern NOT found");
}
System.out.println();
}
}
}
Imo, it looks very similar to your original code. In case the regex doesn't look familiar to you, I'll explain what's going on.
Capital S in [\\S] basically means match everything except for [\\s]. .+ worked well in your case, but all it is really saying is match anything that isn't empty - even a whitespace. This is not necessarily bad, but would be troublesome if you ever had to modify the regex.
{1,} simple means one or more occurrences. {1,2}, to give another example, would be 1 or 2 occurrences. FYI, + usually means 0 or 1 occurrences (maybe not in Java) and * means one or more occurrences.
The parentheses denote groups. The entire match is group 0. When you add parentheses, the order from left to right represent group 1 .. group N. So what I did was combine your patterns using groups, separated by one or more occurrences of whitespace. (.*) is used for group 2, since that group can have both whitespace and non-whitespace, as long as it doesn't begin with whitespace.
If you have any questions feel free to ask. For the record, your current code is fine if you just add '+' to the buildId pattern: [\\s]+.+.
Without that, your regex is saying: match the whitespace that is followed by no characters or a single character. Since all of your whitespace is followed by more whitespace, you matching just a single whitespace.

TLDR;
Use the pattern ^(v\\S+)\\s+(\\S+), where the capture-groups capture the version and build respectively, here's the complete snippet:
String unitPattern ="^(v\\S+)\\s+(\\S+)";
Pattern pattern = Pattern.compile(unitPattern);
for (String str : possibilities) {
System.out.println("Analyzing \"" + str + "\"");
Matcher matcher = pattern.matcher(str);
while(matcher.find()) {
System.out.println("Version found: \"" + matcher.group(1) + "\"");
System.out.println("Build-id found: \"" + matcher.group(2) + "\"");
}
}
Fiddle to try it.
Nitty Gritties
Reason for the empty lines in the output
It's because of how the Matcher class interprets the .; The . DOES NOT match newlines, it stops matching just before the \n. For that you need to add the flag Pattern.DOTALL using Pattern.compile(String pattern, int flags).
An attempt
But even with Pattern.DOTALL, you'll still not be able to match, because of the way you have defined the pattern. A better approach is to match the full build and version as a unit and then extract the necessary parts.
^(v\\S+)\\s+(\\S+)
This does trick where :
^(v\\S+) defines the starting of the unit and also captures version information
\\s+ matches the tabs, new line, spaces etc
(\\S+) captures the final contiguous build id

Regex pattern doesn't work when ending without a space

I want to remove strings that contain either http or https. I have the following code segment:
String line="abc http://someurl something https://someurl";
if (line.contains("https") || line.contains("http")) {
System.out.println(line);
String x = line.replaceAll("https?://.*?\\s+", " ");
System.out.println(x);
}
The output is: abc something https://someurl (doesn't remove the ending url)
Desired output is: abc something
I'm guessing its a simple change to the regex...
Edit: Sorry, the previous example didn't contain an actual url after the http.

Your regex is
https?://.*?\\s+
That final token \s+ means one or more space characters. If you want to remove substrings that don't necessarily end in spaces, you can repeat with * instead of + - * means to repeat the preceding token zero or more times:
String x = line.replaceAll("https?://.*?\\s*", " ");
That said, if the URLs you have are valid and don't contain any space characters, it would probably make more sense to match non-space characters with \S and replace with the empty string, rather than look for space characters, match them, and then replace with another space:
String x = line.replaceAll("https?://\\S*", "");

How to replace last letter to another letter in java using regular expression

i have seen to replace "," to "." by using ".$"|",$", but this logic is not working with alphabets.
i need to replace last letter of a word to another letter for all word in string containing EXAMPLE_TEST using java
this is my code
Pattern replace = Pattern.compile("n$");//here got the real problem
matcher2 = replace.matcher(EXAMPLE_TEST);
EXAMPLE_TEST=matcher2.replaceAll("k");
i also tried "//n$" ,"\n$" etc
Please help me to get the solution
input text=>njan ayman
output text=> njak aymak

Instead of the end of string $ anchor, use a word boundary \b
String s = "njan ayman";
s = s.replaceAll("n\\b", "k");
System.out.println(s); //=> "njak aymak"

You can use lookahead and group matching:
String EXAMPLE_TEST = "njan ayman";
s = EXAMPLE_TEST.replaceAll("(n)(?=\\s|$)", "k");
System.out.println("s = " + s); // prints: s = njak aymak
Explanation:
(n) - the matched word character
(?=\\s|$) - which is followed by a space or at the end of the line (lookahead)
The above is only an example! if you want to switch every comma with a period the middle line should be changed to:
s = s.replaceAll("(,)(?=\\s|$)", "\\.");

Here's how I would set it up:
(?=.\b)\w
Which in Java would need to be escaped as following:
(?=.\\b)\\w
It translates to something like "a character (\w) after (?=) any single character (.) at the end of a word (\b)".
String s = "njan ayman aowkdwo wdonwan. wadawd,.. wadwdawd;";
s = s.replaceAll("(?=.\\b)\\w", "");
System.out.println(s); //nja ayma aowkdw wdonwa. wadaw,.. wadwdaw;
This removes the last character of all words, but leaves following non-alphanumeric characters. You can specify only specific characters to remove/replace by changing the . to something else.
However, the other answers are perfectly good and might achieve exactly what you are looking for.

if (word.endsWith("char oldletter")) {
name = name.substring(0, name.length() - 1 "char newletter");
}

returning online numeric and the + sign

I have the following string : +996a34fdsc112
I want to return only the numbers with the plus sign : +99634112
How to do it using the replaceAll regex function ?

You can use:
String repl = "+996a34fdsc112".replaceAll("[^+\\d]+", "");
//=> +99634112

It seems that you want to remove all characters in range a-z. If that is the case then you can use
String noAlpha = "+996a34fdsc112".replaceAll("[a-z]+","");
+ in this case represents {1,} which means "at least once", it doesn't represent + character from your text.