I need to extract first three octet from a IP address(class C) and I can do it by splitting on "//.". But is there a way to do it using REGEX.
Input : 192.168.1.1 Output : 192.168.1
Something like this:
/^[0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3}/
Use match and it is done..
More precisely for Java:
Pattern p = Pattern.compile("([0-9]{1,3})\\.([0-9]{1,3})\\.([0-9]{1,3}).*");
Matcher m = p.matcher("127.0.2.13");
if (m.matches()) {
String s0 = m.group(1); // contains "127"
String s1 = m.group(2); // contains "0"
String s2 = m.group(3); // contains "2"
System.out.println("s0 + "." + s1 + "." + s2);
}
This slightly more simple pattern also works:
Pattern p = Pattern.compile("(\\d{1,3})\\.(\\d{1,3})\\.(\\d{1,3}).*");
Really good regex tutorial here.
Related
I have strings in this format:
object[i].base.base_x[i] and I get lists like List(0,1).
I want to use regular expressions in scala to find the match [i] in the given string and replace the first occurance with 0 and the second with 1. Hence getting something like object[0].base.base_x[1].
I have the following code:
val stringWithoutIndex = "object[i].base.base_x[i]" // basically this string is generated dynamically
val indexReplacePattern = raw"\[i\]".r
val indexValues = List(0,1) // list generated dynamically
if(indexValues.nonEmpty){
indexValues.map(row => {
indexReplacePattern.replaceFirstIn(stringWithoutIndex , "[" + row + "]")
})
else stringWithoutIndex
Since String is immutable, I cannot update stringWithoutIndex resulting into an output like List("object[0].base.base_x[i]", "object[1].base.base_x[i]").
I tried looking into StringBuilder but I am not sure how to update it. Also, is there a better way to do this? Suggestions other than regex are also welcome.
You couldloop through the integers in indexValues using foldLeft and pass the string stringWithoutIndex as the start value.
Then use replaceFirst to replace the first match with the current value of indexValues.
If you want to use a regex, you might use a positive lookahead (?=]) and a positive lookbehind (?<=\[) to assert the i is between opening and square brackets.
(?<=\[)i(?=])
For example:
val strRegex = """(?<=\[)i(?=])"""
val res = indexValues.foldLeft(stringWithoutIndex) { (s, row) =>
s.replaceFirst(strRegex, row.toString)
}
See the regex demo | Scala demo
How about this:
scala> val str = "object[i].base.base_x[i]"
str: String = object[i].base.base_x[i]
scala> str.replace('i', '0').replace("base_x[0]", "base_x[1]")
res0: String = object[0].base.base_x[1]
This sounds like a job for foldLeft. No need for the if (indexValues.nonEmpty) check.
indexValues.foldLeft(stringWithoutIndex) { (s, row) =>
indexReplacePattern.replaceFirstIn(s, "[" + row + "]")
}
I am trying to do regex on a number based on the below conditions, however its returning an empty string
import java.util.regex.Matcher
import java.util.regex.Pattern
object clean extends App {
val ALPHANUMERIC: Pattern = Pattern.compile("^[a-zA-Z0-9]*$")
val SPECIALCHAR: Pattern = Pattern.compile("[a-zA-Z0-9\\-#\\.\\(\\)\\/%&\\s]")
val LEADINGZEROES: Pattern = Pattern.compile("^[0]+(?!$)")
val TRAILINGZEROES: Pattern = Pattern.compile("\\.0*$|(\\.\\d*?)0+$")
def evaluate(codes: String): String = {
var str2: String = codes.toString
var text:Matcher = LEADINGZEROES.matcher(str2)
str2 = text.replaceAll("")
text = ALPHANUMERIC.matcher(str2)
str2 = text.replaceAll("")
text = SPECIALCHAR.matcher(str2)
str2 = text.replaceAll("")
text = TRAILINGZEROES.matcher(str2)
str2 = text.replaceAll("")
}
}
the code is returning empty string for LEADINGZEROES match.
scala> println("cleaned value :" + evaluate("0001234"))
cleaned value :
What change should I do to make the code work as I expect. Basically i am trying to remove leading/trailing zeroes and if the numbers has special characters/alphanumeric values than entire value should be returned null
Your LEADINGZEROES pattern is working correct as
val LEADINGZEROES: Pattern = Pattern.compile("^[0]+(?!$)")
println(LEADINGZEROES.matcher("0001234").replaceAll(""))
gives
//1234
But then there is a pattern matching
text = ALPHANUMERIC.matcher(str2)
which replaces all alphanumeric to "" and this made str as empty ("")
As when you do
val ALPHANUMERIC: Pattern = Pattern.compile("^[a-zA-Z0-9]*$")
val LEADINGZEROES: Pattern = Pattern.compile("^[0]+(?!$)")
println(ALPHANUMERIC.matcher(LEADINGZEROES.matcher("0001234").replaceAll("")).replaceAll(""))
it will print empty
Updated
As you have commented
if there is a code that is alphanumeric i want to make that value NULL
but in case of leading or trailing zeroes its pure number, which should return me the value after removing zeroes
but its also returning null for trailing and leading zeroes matches
and also how can I skip a match , suppose i want the regex to not match the number 0999 for trimming leading zeroes
You can write your evaluate function and regexes as below
val LEADINGTRAILINGZEROES = """(0*)(\d{4})(0*)""".r
val ALPHANUMERIC = """[a-zA-Z]""".r
def evaluate(codes: String): String = {
val LEADINGTRAILINGZEROES(first, second, third) = if(ALPHANUMERIC.findAllIn(codes).length != 0) "0010" else codes
if(second.equalsIgnoreCase("0010")) "NULL" else second
}
which should give you
println("cleaned value : " + evaluate("000123400"))
// cleaned value : 1234
println("alphanumeric : " + evaluate("0001A234"))
// alphanumeric : NULL
println("skipping : " + evaluate("0999"))
// skipping : 0999
I hope the answer is helpful
If I am doing this it is working fine:
val string = "somestring;userid=someidT;otherstuffs"
var pattern = """[;?&]userid=([^;&]+)?(;|&|$)""".r
val result = pattern.findFirstMatchIn(string).get;
But I am getting an error when I am doing this
val string = "somestring;userid=someidT;otherstuffs"
val id_name = "userid"
var pattern = """[;?&]""" + id_name + """=([^;&]+)?(;|&|$)""".r
val result = pattern.findFirstMatchIn(string).get;
This is the error:
error: value findFirstMatchIn is not a member of String
You may use an interpolated string literal and use a bit simpler regex:
val string = "somestring;userid=someidT;otherstuffs"
val id_name = "userid"
var pattern = s"[;?&]${id_name}=([^;&]*)".r
val result = pattern.findFirstMatchIn(string).get.group(1)
println(result)
// => someidT
See the Scala demo.
The [;?&]$id_name=([^;&]*) pattern finds ;, ? or & and then userId (since ${id_name} is interpolated) and then = is matched and then any 0+ chars other than ; and & are captured into Group 1 that is returned.
NOTE: if you want to use a $ as an end of string anchor in the interpolated string literal use $$.
Also, remember to Regex.quote("pattern") if the variable may contain special regex operators like (, ), [, etc. See Scala: regex, escape string.
Add parenthesis around the string so that regex is made after the string has been constructed instead of the other way around:
var pattern = ("[;?&]" + id_name + "=([^;&]+)?(;|&|$)").r
// pattern: scala.util.matching.Regex = [;?&]userid=([^;&]+)?(;|&|$)
val result = pattern.findFirstMatchIn(string).get;
// result: scala.util.matching.Regex.Match = ;userid=someidT;
I want to split the following string around +, but I couldn't succeed in getting the correct regex for this.
String input = "SOP3a'+bEOP3'+SOP3b'+aEOP3'";
I want to have a result like this
[SOP3a'+bEOP3', SOP3b'+aEOP3']
In some cases I may have the following string
c+SOP2SOP3a'+bEOP3'+SOP3b'+aEOP3'EOP2
which should be split as
[c, SOP2SOP3a'+bEOP3'+SOP3b'+aEOP3'EOP2]
I have tried the following regex but it doesn't work.
input.split("(SOP[0-9](.*)EOP[0-9])*\\+((SOP)[0-9](.*)(EOP)[0-9])*");
Any help or suggestions are appreciated.
Thanks
You can use the following regex to match the string and by replacing it using captured group you can get the expected result :
(?m)(.*?)\+(SOP.*?$)
see demo / explanation
Following is the code in Java that would work for you:
public static void main(String[] args) {
String input = "SOP3a'+bEOP3'+SOP3b'+aEOP3'";
String pattern = "(?m)(.*?)\\+(SOP.*?$)";
Pattern regex = Pattern.compile(pattern);
Matcher m = regex.matcher(input);
if (m.find()) {
System.out.println("Found value: " + m.group(0));
System.out.println("Found value: " + m.group(1));
System.out.println("Found value: " + m.group(2));
} else {
System.out.println("NO MATCH");
}
}
The m.group(1) and m.group(2) are the values that you are looking for.
Do you really need to use split method?
And what are the rules? They are unclear to me.
Anyway, considering the regex you provided, I've only removed some unnecessary groups and I've found what you are looking for, however, instead of split, I just joined the matches as splitting it would generate some empty elements.
const str = "SOP1a+bEOP1+SOP2SOP3a'+bEOP3'+SOP3b'+aEOP3'EOP2";
const regex = RegExp(/(SOP[0-9].*EOP[0-9])*\+(SOP[0-9].*EOP[0-9])*/)
const matches = str.match(regex);
console.log('Matches ', matches);
console.log([matches[1],matches[2]]);
The pattern I want to match is a sequence of length n where n is right before the sequence.
For example, when the input is "1aaaaa", I want to match the single character "a", as the first number specifies only 1 character is matched.
Similar, when the input is "2aaaaa", I want to match the first two characters "aa", but not the rest, as the number 2 specifies two characters will be matched.
I understand a{1} and a{2} will match "a" one or two times. But how to match a{n} in which n is not fixed?
Is it possible to do this type of match using regular expressions?
This will work for repeating numbers.
import re
a="1aaa2bbbbb1cccccccc4dddddddddddd"
for b in re.findall(r'\d[a-z]+', a):
print b[int(b[0])+1:int(b[0])+1+int(b[0])]
Output:
a
bb
c
dddd
Though I have done in Java, it will help you get going in your program.
Here you can select the first letter as sub-string from the given input string and use it in your regex to match the string accordingly.
public class DynamicRegex {
public static void main(String args[]){
Scanner scan = new Scanner(System.in);
System.out.println("Enter a string: ");
String str = scan.nextLine();
String testStr = str.substring(0, 1); //Get the first character from the string using sub-string.
String pattern = "a{"+ testStr +"}"; //Use the sub-string in your regex as length of the string to match.
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(str);
if(m.find()){
System.out.println(m.group());
}
}
}