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How to remove everything after specific character in string using Java

I have a string that looks like this:
analitics#gmail.com#5
And it represents my userId.
I have to send that userId as parameter to the function and send it in the way that I remove number 5 after second # and append new number.
I started with something like this:
userService.getUser(user.userId.substring(0, userAfterMigration.userId.indexOf("#") + 1) + 3
What is the best way of removing everything that comes after the second # character in string above using Java?

Here is a splitting option:
String input = "analitics#gmail.com#5";
String output = String.join("#", input.split("#")[0], input.split("#")[1]) + "#";
System.out.println(output); // analitics#gmail.com#
Assuming your input would only have two at symbols, you could use a regex replacement here:
String input = "analitics#gmail.com#5";
String output = input.replaceAll("#[^#]*$", "#");
System.out.println(output); // analitics#gmail.com#

You can capture in group 1 what you want to keep, and match what comes after it to be removed.
In the replacement use capture group 1 denoted by $1
^((?:[^#\s]+#){2}).+
^ Start of string
( Capture group 1
(?:[^#\s]+#){2} Repeat 2 times matching 1+ chars other than #, and then match the #
) Close group 1
.+ Match 1 or more characters that you want to remove
Regex demo | Java demo
String s = "analitics#gmail.com#5";
System.out.println(s.replaceAll("^((?:[^#\\s]+#){2}).+", "$1"));
Output
analitics#gmail.com#
If the string can also start with ##1 and you want to keep ## then you might also use:
^((?:[^#]*#){2}).+
Regex demo

The simplest way that would seem to work for you:
str = str.replaceAll("#[^.]*$", "");
See live demo.
This matches (and replaces with blank to delete) # and any non-dot chars to the end.

String replacement when regex reverse group is null in java

I want to convert a software version number into a github tag name by regular expression.
For example, the version of ognl is usually 3.2.1. What I want is the tag name OGNL_3_2_1
So we can use String::replaceAll(String regex, String replacement) method like this
"3.2.1".replaceAll("(\d+).(\d+).(\d+)", "OGNL_$1_$2_$3")
And we can get the tag name OGNL_3_2_1 easily.
But when it comes to 3.2, I want the regex still working so I change it into (\d+).(\d+)(?:.(\d+))?.
Execute the code again, what I get is OGNL_3_2_ rather than OGNL_3_2. The underline _ at the tail is not what I want. It is resulted by the null group for $3
So how can I write a suitable replacement to solve this case?
When the group for $3 is null, the underline _ should disappear
Thanks for your help !!!

You can make the last . + digits part optional by enclosing it with an optional non-capturing group and use a lambda as a replacement argument with Matcher.replaceAll in the latest Java versions:
String regex = "(\\d+)\\.(\\d+)(?:\\.(\\d+))?";
Pattern p = Pattern.compile(regex);
String s="3.2.1";
Matcher m = p.matcher(s);
String result = m.replaceAll(x ->
x.group(3) != null ? "OGNL_" + x.group(1) + "_" + x.group(2) + "_" + x.group(3) :
"OGNL_" + x.group(1) + "_" + x.group(2) );
System.out.println(result);
See the Java demo.
The (\d+)\.(\d+)(?:\.(\d+))? pattern (note that literal . are escaped) matches and captures into Group 1 any one or more digits, then matches a dot, then captures one or more digits into Group 2 and then optionally matches a dot and digits (captured into Group 3). If Group 3 is not null, add the _ and Group 3 value, else, omit this part when building the final replacement value.

How to match java regexp between some '#'?

I am facing an issue with the String.replaceFirst method.
I have the following String :
String content = "select * from queries
where update_date >= to_timestamp('#date|Date debut|dd/MM/yyyy# 00:00:00','DD/MM/YYYY HH24:MI:SS')
and update_date <= to_timestamp('#date|Date fin|dd/MM/yyyy# 23:59:59','DD/MM/YYYY HH24:MI:SS')";
(The two expressions between '#' are dynamically defined).
And I have 2 dates too :
String begin = "28/05/2018";
String end = "29/05/2018";
Then I would to replace the first expression with begin, and the second with end.
I use :
content = content.replaceFirst("#(date)\\|(.*)\\|(.*)#", begin);
content = content.replaceFirst("#(date)\\|(.*)\\|(.*)#", end);
Although, replaceFirst takes the last '#' of entire String and I am obtaining:
select * from queries where update_date >= to_timestamp('28/05/2018 23:59:59','DD/MM/YYYY HH24:MI:SS');
I understand the error but I ask you to help me to find a solution.
Thank you a lot ! Axel.

If looking for a generic regex for both replacements as your question's code seems to want, this is how to make it work:
the regex for .* that captures all characters is greedy by default, it means that it will try to capture as many characters as it can. This is why your first replacement replaces all.
You can use the lazy quantifier ? to precise that you want to capture the less characters possible instead of the most.
try:
#(date)\|(.*?)\|(.*?)#
(or escaped version for your code: "#(date)\\|(.*?)\\|(.*?)#")
see regex in regex101

When reading your question, I was not sure whether the text between #s (here I mean "date|Date debut|dd/MM/yyyy" and "date|Date fin|dd/MM/yyyy") were dynamically defined or if you were just explaining that you wanted to dynamically replace the fix contents above with your dynamically defined dates.
So I will give you two answers (and both should work).
If the text is fix:
#date\|Date debut\|dd/MM/yyyy# - for the first range
#date\|Date fin\|dd/MM/yyyy# - for the second range
If the text between # is not fix:
#[^#]*#
The regex above means find a range of chars that start with a #, than contains any chars except a #, this is what [^#] means, 0 or several times (the *) and ends with a #
I hope it helps!

Try this:
String content = "select * from queries " +
"where update_date >= to_timestamp('#date|Date debut|dd/MM/yyyy# 00:00:00','DD/MM/YYYY HH24:MI:SS') " +
"and update_date <= to_timestamp('#date|Date fin|dd/MM/yyyy# 23:59:59','DD/MM/YYYY HH24:MI:SS') ;";
String begin = "28/05/2018";
String end = "29/05/2018";
content = content.replaceFirst( "#date\\|[^\\|]*\\|[^#]*#", begin );
content = content.replaceFirst( "#date\\|[^\\|]*\\|[^#]*#", end );
System.out.println( content );
Here we don't need to use the () and we are matching until our character like | or # matched.

REGEX: GETTING VALUE OF href="" EXCEPT FOR PATICULAR STRING

Here's my regex code:
\\s*(?i)href\\s*=\\s*(\"(([^\"]*\")|'[^']*'|([^'\">\\s]+)))
Actually the real problem is like this. I want to change the value for each href that will match except for these two types <link href="foo.css"> and <link href="boo.ico">. I want to retain the value of these two Strings.
Pattern p = Pattern.compile(HTML_A_HREF);
Matcher m = p.matcher(getLine());
setNewLine(m.replaceAll((String.format("%-1s", sp))+"href=\"javascript:history.go(0)\"" + (String.format("%-1s", sp))));
getLine() is the html file itself.
String sp = "";

Your regex is off. To show you, let me explode it:
\\s*(?i)href\\s*=\\s*
(\"
(
([^\"]*\")
|
'[^']*'
|
([^'\">\\s]+)
)
)
The leading double-quote is outside the multi-choice block. It needs to be in the first choice section.
Also:
You should put (?i) first.
With \" inside first choice, one set of parenthesis goes away.
You don't need parenthesis in choice sections.
Parenthesis around choice block should be non-capturing.
So, that means:
(?i)\\s*href\\s*=\\s*
(?:
\"[^\"]*\"
|
'[^']*'
|
[^'\">\\s]+
)
Which is (?i)\\s*href\\s*=\\s*(?:\"[^\"]*\"|'[^']*'|[^'\">\\s]+).
As for the replacement code:
String sp = "";
m.replaceAll((String.format("%-1s", sp))
+
"href=\"javascript:history.go(0)\""
+
(String.format("%-1s", sp))
)
What is the purpose of (String.format("%-1s", sp)) when sp = ""??? An empty string, formatted to fit at least 1 space, left-aligned. That is a single space, i.e. " ", so why all that overhead?
m.replaceAll(" href=\"javascript:history.go(0)\" ")
Finally, you want to exclude foo.css and boo.ico.
One way to do that is with a negative lookahead. Since you have 3 choices, you need to repeat it 3 times:
(?i)\\s*href\\s*=\\s*
(?:
\"(?!foo\\.css|boo\\.ico)[^\"]*\"
|
'(?!foo\\.css|boo\\.ico)[^']*'
|
(?!foo\\.css|boo\\.ico)[^'\">\\s]+
)
I'll let you collapse that back to one line.
UPDATE
If you want to exclude all .css and .ico files, use a negative lookbehind instead.
Also, I forgot to escape the . before, sorry. Fixed that.
(?i)\\s*href\\s*=\\s*
(?:
\"[^\"]*(?<!\\.css|\\.ico)\"
|
'[^']*(?<!\\.css|\\.ico)'
|
[^'\">\\s]+(?<!\\.css|\\.ico)
)

Java String Replace Regex

I am doing some string replace in SQL on the fly.
MySQLString = " a.account=b.account ";
MySQLString = " a.accountnum=b.accountnum ";
Now if I do this
MySQLString.replaceAll("account", "account_enc");
the result will be
a.account_enc=b.account_enc
(This is good)
But look at 2nd result
a.account_enc_num=a.account_enc_num
(This is not good it should be a.accountnum_enc=b.accountnum_enc)
Please advise how can I achieve what I want with Java String Replace.
Many Thanks.

From your comment:
Is there anyway to tell in Regex only replace a.account=b.account or a.accountnum=b.accountnum. I do not want accountname to be replace with _enc
If I understand correctly you want to add _enc part only to account or accountnum. To do this you can use
MySQLString = MySQLString.replaceAll("\\baccount(num)?\\b", "$0_enc");
(num)? mean that num is optional so regex will accept account or accountnum
\\b at start mean that there can be no letters, numbers or "_" before account so it wont accept (affect) something like myaccount, or my_account.
\\b at the end will prevent other letters, numbers or "_" after account or accountnum.

It's hard to extrapolate from so few examples, but maybe what you want is:
MySQLString = MySQLString.replaceAll("account\\w*", "$0_enc");
which will append _enc to any sequence of letters, digits, and underscores that starts with account.

try
String s = " a.accountnum=b.accountnum ".replaceAll("(account[^ =]*)", "$1_enc");
it means replace any sequence characters which are not ' ' or '=' which starts the word "account" with the sequence found + "_enc".
$1 is a reference to group 1 in regex; group 1 is the expression in parenthesis (account[^ =]+), i.e. our sequence
See http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html for details