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How to match two string using java Regex

String 1= abc/{ID}/plan/{ID}/planID
String 2=abc/1234/plan/456/planID
How can I match these two strings using Java regex so that it returns true? Basically {ID} can contain anything. Java regex should match abc/{anything here}/plan/{anything here}/planID

If your "{anything here}" includes nothing, you can use .*. . matches any letter, and * means that match the string with any length with the letter before, including 0 length. So .* means that "match the string with any length, composed with any letter". If {anything here} should include at least one letter, you can use +, instead of *, which means almost the same, but should match at least one letter.
My suggestion: abc/.+/plan/.+/planID

If {ID} can contain anything I assume it can also be empty.
So this regex should work :
str.matches("^abc.*plan.*planID$");
^abc at the beginning
.* Zero or more of any Character
planID$ at the end

I am just writing a small code, just check it and start making changes as per you requirement. This is working, check for your other test cases, if there is any issue please comment that test case. Specifically I am using regex, because you want to match using java regex.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
class MatchUsingRejex
{
public static void main(String args[])
{
// Create a pattern to be searched
Pattern pattern = Pattern.compile("abc/.+/plan/.+/planID");
// checking, Is pattern match or not
Matcher isMatch = pattern.matcher("abc/1234/plan/456/planID");
if (isMatch.find())
System.out.println("Yes");
else
System.out.println("No");
}
}

If line always starts with 'abc' and ends with 'planid' then following way will work:
String s1 = "abc/{ID}/plan/{ID}/planID";
String s2 = "abc/1234/plan/456/planID";
String pattern = "(?i)abc(?:/\\S+)+planID$";
boolean b1 = s1.matches(pattern);
boolean b2 = s2.matches(pattern);

Java String- How to get a part of package name in android?

Its basically about getting string value between two characters. SO has many questions related to this. Like:
How to get a part of a string in java?
How to get a string between two characters?
Extract string between two strings in java
and more.
But I felt it quiet confusing while dealing with multiple dots in the string and getting the value between certain two dots.
I have got the package name as :
au.com.newline.myact
I need to get the value between "com." and the next "dot(.)". In this case "newline". I tried
Pattern pattern = Pattern.compile("com.(.*).");
Matcher matcher = pattern.matcher(beforeTask);
while (matcher.find()) {
int ct = matcher.group();
I tried using substrings and IndexOf also. But couldn't get the intended answer. Because the package name in android varies by different number of dots and characters, I cannot use fixed index. Please suggest any idea.

As you probably know (based on .* part in your regex) dot . is special character in regular expressions representing any character (except line separators). So to actually make dot represent only dot you need to escape it. To do so you can place \ before it, or place it inside character class [.].
Also to get only part from parenthesis (.*) you need to select it with proper group index which in your case is 1.
So try with
String beforeTask = "au.com.newline.myact";
Pattern pattern = Pattern.compile("com[.](.*)[.]");
Matcher matcher = pattern.matcher(beforeTask);
while (matcher.find()) {
String ct = matcher.group(1);//remember that regex finds Strings, not int
System.out.println(ct);
}
Output: newline
If you want to get only one element before next . then you need to change greedy behaviour of * quantifier in .* to reluctant by adding ? after it like
Pattern pattern = Pattern.compile("com[.](.*?)[.]");
// ^
Another approach is instead of .* accepting only non-dot characters. They can be represented by negated character class: [^.]*
Pattern pattern = Pattern.compile("com[.]([^.]*)[.]");
If you don't want to use regex you can simply use indexOf method to locate positions of com. and next . after it. Then you can simply substring what you want.
String beforeTask = "au.com.newline.myact.modelact";
int start = beforeTask.indexOf("com.") + 4; // +4 since we also want to skip 'com.' part
int end = beforeTask.indexOf(".", start); //find next `.` after start index
String resutl = beforeTask.substring(start, end);
System.out.println(resutl);

You can use reflections to get the name of any class. For example:
If I have a class Runner in com.some.package and I can run
Runner.class.toString() // string is "com.some.package.Runner"
to get the full name of the class which happens to have a package name inside.
TO get something after 'com' you can use Runner.class.toString().split(".") and then iterate over the returned array with boolean flag

All you have to do is split the strings by "." and then iterate through them until you find one that equals "com". The next string in the array will be what you want.
So your code would look something like:
String[] parts = packageName.split("\\.");
int i = 0;
for(String part : parts) {
if(part.equals("com")
break;
}
++i;
}
String result = parts[i+1];

private String getStringAfterComDot(String packageName) {
String strArr[] = packageName.split("\\.");
for(int i=0; i<strArr.length; i++){
if(strArr[i].equals("com"))
return strArr[i+1];
}
return "";
}

I have done heaps of projects before dealing with websites scraping and I
just have to create my own function/utils to get the job done. Regex might
be an overkill sometimes if you just want to extract a substring from
a given string like the one you have. Below is the function I normally
use to do this kind of task.
private String GetValueFromText(String sText, String sBefore, String sAfter)
{
String sRetValue = "";
int nPos = sText.indexOf(sBefore);
if ( nPos > -1 )
{
int nLast = sText.indexOf(sAfter,nPos+sBefore.length()+1);
if ( nLast > -1)
{
sRetValue = sText.substring(nPos+sBefore.length(),nLast);
}
}
return sRetValue;
}
To use it just do the following:
String sValue = GetValueFromText("au.com.newline.myact", ".com.", ".");

How to check if a string contains only digits in Java

In Java for String class there is a method called matches, how to use this method to check if my string is having only digits using regular expression. I tried with below examples, but both of them returned me false as result.
String regex = "[0-9]";
String data = "23343453";
System.out.println(data.matches(regex));
String regex = "^[0-9]";
String data = "23343453";
System.out.println(data.matches(regex));

Try
String regex = "[0-9]+";
or
String regex = "\\d+";
As per Java regular expressions, the + means "one or more times" and \d means "a digit".
Note: the "double backslash" is an escape sequence to get a single backslash - therefore, \\d in a java String gives you the actual result: \d
References:
Java Regular Expressions
Java Character Escape Sequences
Edit: due to some confusion in other answers, I am writing a test case and will explain some more things in detail.
Firstly, if you are in doubt about the correctness of this solution (or others), please run this test case:
String regex = "\\d+";
// positive test cases, should all be "true"
System.out.println("1".matches(regex));
System.out.println("12345".matches(regex));
System.out.println("123456789".matches(regex));
// negative test cases, should all be "false"
System.out.println("".matches(regex));
System.out.println("foo".matches(regex));
System.out.println("aa123bb".matches(regex));
Question 1:
Isn't it necessary to add ^ and $ to the regex, so it won't match "aa123bb" ?
No. In java, the matches method (which was specified in the question) matches a complete string, not fragments. In other words, it is not necessary to use ^\\d+$ (even though it is also correct). Please see the last negative test case.
Please note that if you use an online "regex checker" then this may behave differently. To match fragments of a string in Java, you can use the find method instead, described in detail here:
Difference between matches() and find() in Java Regex
Question 2:
Won't this regex also match the empty string, "" ?*
No. A regex \\d* would match the empty string, but \\d+ does not. The star * means zero or more, whereas the plus + means one or more. Please see the first negative test case.
Question 3
Isn't it faster to compile a regex Pattern?
Yes. It is indeed faster to compile a regex Pattern once, rather than on every invocation of matches, and so if performance implications are important then a Pattern can be compiled and used like this:
Pattern pattern = Pattern.compile(regex);
System.out.println(pattern.matcher("1").matches());
System.out.println(pattern.matcher("12345").matches());
System.out.println(pattern.matcher("123456789").matches());

You can also use NumberUtil.isNumber(String str) from Apache Commons

Using regular expressions is costly in terms of performance. Trying to parse string as a long value is inefficient and unreliable, and may be not what you need.
What I suggest is to simply check if each character is a digit, what can be efficiently done using Java 8 lambda expressions:
boolean isNumeric = someString.chars().allMatch(x -> Character.isDigit(x));

One more solution, that hasn't been posted, yet:
String regex = "\\p{Digit}+"; // uses POSIX character class

You must allow for more than a digit (the + sign) as in:
String regex = "[0-9]+";
String data = "23343453";
System.out.println(data.matches(regex));

Long.parseLong(data)
and catch exception, it handles minus sign.
Although the number of digits is limited this actually creates a variable of the data which can be used, which is, I would imagine, the most common use-case.

We can use either Pattern.compile("[0-9]+.[0-9]+") or Pattern.compile("\\d+.\\d+"). They have the same meaning.
the pattern [0-9] means digit. The same as '\d'.
'+' means it appears more times.
'.' for integer or float.
Try following code:
import java.util.regex.Pattern;
public class PatternSample {
public boolean containNumbersOnly(String source){
boolean result = false;
Pattern pattern = Pattern.compile("[0-9]+.[0-9]+"); //correct pattern for both float and integer.
pattern = Pattern.compile("\\d+.\\d+"); //correct pattern for both float and integer.
result = pattern.matcher(source).matches();
if(result){
System.out.println("\"" + source + "\"" + " is a number");
}else
System.out.println("\"" + source + "\"" + " is a String");
return result;
}
public static void main(String[] args){
PatternSample obj = new PatternSample();
obj.containNumbersOnly("123456.a");
obj.containNumbersOnly("123456 ");
obj.containNumbersOnly("123456");
obj.containNumbersOnly("0123456.0");
obj.containNumbersOnly("0123456a.0");
}
}
Output:
"123456.a" is a String
"123456 " is a String
"123456" is a number
"0123456.0" is a number
"0123456a.0" is a String

According to Oracle's Java Documentation:
private static final Pattern NUMBER_PATTERN = Pattern.compile(
"[\\x00-\\x20]*[+-]?(NaN|Infinity|((((\\p{Digit}+)(\\.)?((\\p{Digit}+)?)" +
"([eE][+-]?(\\p{Digit}+))?)|(\\.((\\p{Digit}+))([eE][+-]?(\\p{Digit}+))?)|" +
"(((0[xX](\\p{XDigit}+)(\\.)?)|(0[xX](\\p{XDigit}+)?(\\.)(\\p{XDigit}+)))" +
"[pP][+-]?(\\p{Digit}+)))[fFdD]?))[\\x00-\\x20]*");
boolean isNumber(String s){
return NUMBER_PATTERN.matcher(s).matches()
}

Refer to org.apache.commons.lang3.StringUtils
public static boolean isNumeric(CharSequence cs) {
if (cs == null || cs.length() == 0) {
return false;
} else {
int sz = cs.length();
for(int i = 0; i < sz; ++i) {
if (!Character.isDigit(cs.charAt(i))) {
return false;
}
}
return true;
}
}

In Java for String class, there is a method called matches(). With help of this method you can validate the regex expression along with your string.
String regex = "^[\\d]{4}$";
String value = "1234";
System.out.println(data.matches(value));
The Explanation for the above regex expression is:-
^ - Indicates the start of the regex expression.
[] - Inside this you have to describe your own conditions.
\\\d - Only allows digits. You can use '\\d'or 0-9 inside the bracket both are same.
{4} - This condition allows exactly 4 digits. You can change the number according to your need.
$ - Indicates the end of the regex expression.
Note: You can remove the {4} and specify + which means one or more times, or * which means zero or more times, or ? which means once or none.
For more reference please go through this website: https://www.rexegg.com/regex-quickstart.html

Offical regex way
I would use this regex for integers:
^[-1-9]\d*$
This will also work in other programming languages because it's more specific and doesn't make any assumptions about how different programming languages may interpret or handle regex.
Also works in Java
\\d+
Questions regarding ^ and $
As #vikingsteve has pointed out in java, the matches method matches a complete string, not parts of a string. In other words, it is unnecessary to use ^\d+$ (even though it is the official way of regex).
Online regex checkers are more strict and therefore they will behave differently than how Java handles regex.

Try this part of code:
void containsOnlyNumbers(String str)
{
try {
Integer num = Integer.valueOf(str);
System.out.println("is a number");
} catch (NumberFormatException e) {
// TODO: handle exception
System.out.println("is not a number");
}
}

Substring replacement in Java

I am trying to implement substring replacement, but I am not getting the desired results. Can someone comment on what I may missing here?
public class SubtringReplacement {
public static void main (String[] args){
String input = "xPIy";
if (input.contains("PI") || input.contains("pi") || input.contains("Pi")){
input.replace("PI", "3.14");
}
System.out.println(input);
}
}

Strings are immutable!!
input = input.replace("PI", "3.14");

One problem is that you need to capture the return value when you do the replacement. The other problem is that you will only replace upper case "PI", while it seems you want to replace mixed case instances. Try this instead:
input = input.replaceAll("(PI|pi|Pi)", "3.14");
replace looks for a literal match; replaceAll does a regular expression match, which is what you need.
By the way, you don't need the if condition -- if there is no match, there will be no replacement.
P.S. Look at the comment by #NullUserException if you also want to replace instances of "pI".

Obviously you were missing the Left hand assignment to make your code working with given conditions.
input.replace("PI", "3.14");
But it will only replace, if input contains PI while it will attempt for pi and Pi as well. To better handle this, I think you can use "[pP][iI]" or "[pP]{1}[iI]{1}" as replacement pattern, which will look for one occurrence of P or p followed by one occurrence of I or i e.g. below:
  String input = "xPIyPizpIapib";
  input = input.replaceAll("[pP][iI]", "3.14"); 
  System.out.println(input); //<- all "pi"s irrespective of case are replaced.
String input = "xPIyPizpIapib";
input = input.replaceAll("[pP]{1}[iI]{1}", "3.14");
System.out.println(input); //<- all "pi"s irrespective of case are replaced.
Please Note: This will also replace pI as wel, if found.

Regular Expression problem in Java

I am trying to create a regular expression for the replaceAll method in Java. The test string is abXYabcXYZ and the pattern is abc. I want to replace any symbol except the pattern with +. For example the string abXYabcXYZ and pattern [^(abc)] should return ++++abc+++, but in my case it returns ab++abc+++.
public static String plusOut(String str, String pattern) {
pattern= "[^("+pattern+")]" + "".toLowerCase();
return str.toLowerCase().replaceAll(pattern, "+");
}
public static void main(String[] args) {
String text = "abXYabcXYZ";
String pattern = "abc";
System.out.println(plusOut(text, pattern));
}
When I try to replace the pattern with + there is no problem - abXYabcXYZ with pattern (abc) returns abxy+xyz. Pattern (^(abc)) returns the string without replacement.
Is there any other way to write NOT(regex) or group symbols as a word?

What you are trying to achieve is pretty tough with regular expressions, since there is no way to express “replace strings not matching a pattern”. You will have to use a “positive” pattern, telling what to match instead of what not to match.
Furthermore, you want to replace every character with a replacement character, so you have to make sure that your pattern matches exactly one character. Otherwise, you will replace whole strings with a single character, returning a shorter string.
For your toy example, you can use negative lookaheads and lookbehinds to achieve the task, but this may be more difficult for real-world examples with longer or more complex strings, since you will have to consider each character of your string separately, along with its context.
Here is the pattern for “not ‘abc’”:
[^abc]|a(?!bc)|(?<!a)b|b(?!c)|(?<!ab)c
It consists of five sub-patterns, connected with “or” (|), each matching exactly one character:
[^abc] matches every character except a, b or c
a(?!bc) matches a if it is not followed by bc
(?<!a)b matches b if it is not preceded with a
b(?!c) matches b if it is not followed by c
(?<!ab)c matches c if it is not preceded with ab
The idea is to match every character that is not in your target word abc, plus every word character that, according to the context, is not part of your word. The context can be examined using negative lookaheads (?!...) and lookbehinds (?<!...).
You can imagine that this technique will fail once you have a target word containing one character more than once, like example. It is pretty hard to express “match e if it is not followed by x and not preceded by l”.
Especially for dynamic patterns, it is by far easier to do a positive search and then replace every character that did not match in a second pass, as others have suggested here.

[^ ... ] will match one character that is not any of ...
So your pattern "[^(abc)]" is saying "match one character that is not a, b, c or the left or right bracket"; and indeed that is what happens in your test.
It is hard to say "replace all characters that are not part of the string 'abc'" in a single trivial regular expression. What you might do instead to achieve what you want could be some nasty thing like
while the input string still contains "abc"
find the next occurrence of "abc"
append to the output a string containing as many "+"s as there are characters before the "abc"
append "abc" to the output string
skip, in the input string, to a position just after the "abc" found
append to the output a string containing as many "+"s as there are characters left in the input
or possibly if the input alphabet is restricted you could use regular expressions to do something like
replace all occurrences of "abc" with a single character that does not occur anywhere in the existing string
replace all other characters with "+"
replace all occurrences of the target character with "abc"
which will be more readable but may not perform as well

Negating regexps is usually troublesome. I think you might want to use negative lookahead. Something like this might work:
String pattern = "(?<!ab).(?!abc)";
I didn't test it, so it may not really work for degenerate cases. And the performance might be horrible too. It is probably better to use a multistep algorithm.
Edit: No I think this won't work for every case. You will probably spend more time debugging a regexp like this than doing it algorithmically with some extra code.

Try to solve it without regular expressions:
String out = "";
int i;
for(i=0; i<text.length() - pattern.length() + 1; ) {
if (text.substring(i, i + pattern.length()).equals(pattern)) {
out += pattern;
i += pattern.length();
}
else {
out += "+";
i++;
}
}
for(; i<text.length(); i++) {
out += "+";
}

Rather than a single replaceAll, you could always try something like:
#Test
public void testString() {
final String in = "abXYabcXYabcHIH";
final String expected = "xxxxabcxxabcxxx";
String result = replaceUnwanted(in);
assertEquals(expected, result);
}
private String replaceUnwanted(final String in) {
final Pattern p = Pattern.compile("(.*?)(abc)([^a]*)");
final Matcher m = p.matcher(in);
final StringBuilder out = new StringBuilder();
while (m.find()) {
out.append(m.group(1).replaceAll(".", "x"));
out.append(m.group(2));
out.append(m.group(3).replaceAll(".", "x"));
}
return out.toString();
}

Instead of using replaceAll(...), I'd go for a Pattern/Matcher approach:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static String plusOut(String str, String pattern) {
StringBuilder builder = new StringBuilder();
String regex = String.format("((?:(?!%s).)++)|%s", pattern, pattern);
Matcher m = Pattern.compile(regex).matcher(str.toLowerCase());
while(m.find()) {
builder.append(m.group(1) == null ? pattern : m.group().replaceAll(".", "+"));
}
return builder.toString();
}
public static void main(String[] args) {
String text = "abXYabcXYZ";
String pattern = "abc";
System.out.println(plusOut(text, pattern));
}
}
Note that you'll need to use Pattern.quote(...) if your String pattern contains regex meta-characters.
Edit: I didn't see a Pattern/Matcher approach was already suggested by toolkit (although slightly different)...