Regex to match a list - java

I have the following text: some unknown random stuff here: (ben, tim, sam toben, suzei)
I need to use regex only to pull out each of the items from the (unknown sized) comma separated list: ben, tim, sam toben, suzei into the matched groups. Leading/trailing whitespace doesn't matter.
I tried the following: (?:\(([^,]+)) but it will only pull out ben as a group.
Any ideas?

This isn't something you can do in the way you want with regexes only. It is, however, possible to do it using regexes and String.split().
Something like this:
String[] getList(String text) {
Pattern p = Pattern.compile(".+\\((.+)\\)"); // Note the doubled backslashes
Matcher m = p.matcher(text);
if (m.find()) {
return m.group(1).split(",");
} else {
return new String[0];
}
}
Demo of the regex on Regex101
What this does is grab the contents of the brackets, then split those contents into an array on the comma character.

Related

Java Pattern matcher not matching for HTTP response code [duplicate]

I have this small piece of code
String[] words = {"{apf","hum_","dkoe","12f"};
for(String s:words)
{
if(s.matches("[a-z]"))
{
System.out.println(s);
}
}
Supposed to print
dkoe
but it prints nothing!!
Welcome to Java's misnamed .matches() method... It tries and matches ALL the input. Unfortunately, other languages have followed suit :(
If you want to see if the regex matches an input text, use a Pattern, a Matcher and the .find() method of the matcher:
Pattern p = Pattern.compile("[a-z]");
Matcher m = p.matcher(inputstring);
if (m.find())
// match
If what you want is indeed to see if an input only has lowercase letters, you can use .matches(), but you need to match one or more characters: append a + to your character class, as in [a-z]+. Or use ^[a-z]+$ and .find().
[a-z] matches a single char between a and z. So, if your string was just "d", for example, then it would have matched and been printed out.
You need to change your regex to [a-z]+ to match one or more chars.
String.matches returns whether the whole string matches the regex, not just any substring.
java's implementation of regexes try to match the whole string
that's different from perl regexes, which try to find a matching part
if you want to find a string with nothing but lower case characters, use the pattern [a-z]+
if you want to find a string containing at least one lower case character, use the pattern .*[a-z].*
Used
String[] words = {"{apf","hum_","dkoe","12f"};
for(String s:words)
{
if(s.matches("[a-z]+"))
{
System.out.println(s);
}
}
I have faced the same problem once:
Pattern ptr = Pattern.compile("^[a-zA-Z][\\']?[a-zA-Z\\s]+$");
The above failed!
Pattern ptr = Pattern.compile("(^[a-zA-Z][\\']?[a-zA-Z\\s]+$)");
The above worked with pattern within ( and ).
Your regular expression [a-z] doesn't match dkoe since it only matches Strings of lenght 1. Use something like [a-z]+.
you must put at least a capture () in the pattern to match, and correct pattern like this:
String[] words = {"{apf","hum_","dkoe","12f"};
for(String s:words)
{
if(s.matches("(^[a-z]+$)"))
{
System.out.println(s);
}
}
You can make your pattern case insensitive by doing:
Pattern p = Pattern.compile("[a-z]+", Pattern.CASE_INSENSITIVE);

Pattern in java regEx does not match [duplicate]

I have this small piece of code
String[] words = {"{apf","hum_","dkoe","12f"};
for(String s:words)
{
if(s.matches("[a-z]"))
{
System.out.println(s);
}
}
Supposed to print
dkoe
but it prints nothing!!
Welcome to Java's misnamed .matches() method... It tries and matches ALL the input. Unfortunately, other languages have followed suit :(
If you want to see if the regex matches an input text, use a Pattern, a Matcher and the .find() method of the matcher:
Pattern p = Pattern.compile("[a-z]");
Matcher m = p.matcher(inputstring);
if (m.find())
// match
If what you want is indeed to see if an input only has lowercase letters, you can use .matches(), but you need to match one or more characters: append a + to your character class, as in [a-z]+. Or use ^[a-z]+$ and .find().
[a-z] matches a single char between a and z. So, if your string was just "d", for example, then it would have matched and been printed out.
You need to change your regex to [a-z]+ to match one or more chars.
String.matches returns whether the whole string matches the regex, not just any substring.
java's implementation of regexes try to match the whole string
that's different from perl regexes, which try to find a matching part
if you want to find a string with nothing but lower case characters, use the pattern [a-z]+
if you want to find a string containing at least one lower case character, use the pattern .*[a-z].*
Used
String[] words = {"{apf","hum_","dkoe","12f"};
for(String s:words)
{
if(s.matches("[a-z]+"))
{
System.out.println(s);
}
}
I have faced the same problem once:
Pattern ptr = Pattern.compile("^[a-zA-Z][\\']?[a-zA-Z\\s]+$");
The above failed!
Pattern ptr = Pattern.compile("(^[a-zA-Z][\\']?[a-zA-Z\\s]+$)");
The above worked with pattern within ( and ).
Your regular expression [a-z] doesn't match dkoe since it only matches Strings of lenght 1. Use something like [a-z]+.
you must put at least a capture () in the pattern to match, and correct pattern like this:
String[] words = {"{apf","hum_","dkoe","12f"};
for(String s:words)
{
if(s.matches("(^[a-z]+$)"))
{
System.out.println(s);
}
}
You can make your pattern case insensitive by doing:
Pattern p = Pattern.compile("[a-z]+", Pattern.CASE_INSENSITIVE);

Check only string and only digits with regex in Java [duplicate]

I have this small piece of code
String[] words = {"{apf","hum_","dkoe","12f"};
for(String s:words)
{
if(s.matches("[a-z]"))
{
System.out.println(s);
}
}
Supposed to print
dkoe
but it prints nothing!!
Welcome to Java's misnamed .matches() method... It tries and matches ALL the input. Unfortunately, other languages have followed suit :(
If you want to see if the regex matches an input text, use a Pattern, a Matcher and the .find() method of the matcher:
Pattern p = Pattern.compile("[a-z]");
Matcher m = p.matcher(inputstring);
if (m.find())
// match
If what you want is indeed to see if an input only has lowercase letters, you can use .matches(), but you need to match one or more characters: append a + to your character class, as in [a-z]+. Or use ^[a-z]+$ and .find().
[a-z] matches a single char between a and z. So, if your string was just "d", for example, then it would have matched and been printed out.
You need to change your regex to [a-z]+ to match one or more chars.
String.matches returns whether the whole string matches the regex, not just any substring.
java's implementation of regexes try to match the whole string
that's different from perl regexes, which try to find a matching part
if you want to find a string with nothing but lower case characters, use the pattern [a-z]+
if you want to find a string containing at least one lower case character, use the pattern .*[a-z].*
Used
String[] words = {"{apf","hum_","dkoe","12f"};
for(String s:words)
{
if(s.matches("[a-z]+"))
{
System.out.println(s);
}
}
I have faced the same problem once:
Pattern ptr = Pattern.compile("^[a-zA-Z][\\']?[a-zA-Z\\s]+$");
The above failed!
Pattern ptr = Pattern.compile("(^[a-zA-Z][\\']?[a-zA-Z\\s]+$)");
The above worked with pattern within ( and ).
Your regular expression [a-z] doesn't match dkoe since it only matches Strings of lenght 1. Use something like [a-z]+.
you must put at least a capture () in the pattern to match, and correct pattern like this:
String[] words = {"{apf","hum_","dkoe","12f"};
for(String s:words)
{
if(s.matches("(^[a-z]+$)"))
{
System.out.println(s);
}
}
You can make your pattern case insensitive by doing:
Pattern p = Pattern.compile("[a-z]+", Pattern.CASE_INSENSITIVE);

Split a sentence in array of string with special characters or spaces intact

I want to split a sentence having spaces or any special character into an array of words with spaces or special character also an element of array.
Sentence like:
aman,amit and sumit went to top-up
should be split into an array of String:
{"aman",",","amit"," ","and"," ","sumit"," ","went"," ","to"," ","top","-","up")
Please suggest any regex or logic to split the same using java.
I missed one thing in my question. I also need to split on numeric character as well.. But using split("\b") does not split a string having something like
abc12def
into
{ "abc", "12","def") or {"abc","1","2","def")
It seems all you need is to match either word characters (\w+) or non-word ones (\W+). Combine these with an alternation operator and - perhaps - add a Pattern.UNICODE_CHARACTER_CLASS (or its inline/embedded version (?U)) to make the pattern Unicode-aware:
String value = "aman,amit and sumit went to top-up";
String pattern = "(?U)\\w+|\\W+";
List<String> lst = new ArrayList<>();
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(value);
while (m.find())
lst.add(m.group(0));
System.out.println(lst);
See the Java demo
I hope the below code snippet helps you solve this.
public static void main(final String[] args) {
String message = "aman,amit and sumit went to top-up";
String[] messages = message.split("\\b");
for(String string : messages) {
System.out.println(string);
}
}

regex pattern - extract a string only if separated by a hyphen

I've looked at other questions, but they didn't lead me to an answer.
I've got this code:
Pattern p = Pattern.compile("exp_(\\d{1}-\\d)-(\\d+)");
The string I want to be matched is: exp_5-22-718
I would like to extract 5-22 and 718. I'm not too sure why it's not working What am I missing? Many thanks
Try this one:
Pattern p = Pattern.compile("exp_(\\d-\\d+)-(\\d+)");
In your original pattern you specified that second number should contain exactly one digit, so I put \d+ to match as more digits as we can.
Also I removed {1} from the first number definition as it does not add value to regexp.
If the string is always prefixed with exp_ I wouldn't use a regular expression.
I would:
replaceFirst() exp_
split() the resulting string on -
Note: This answer is based on the assumptions. I offer it as a more robust if you have multiple hyphens. However, if you need to validate the format of the digits then a regular expression may be better.
In your regexp you missed required quantifier for second digit \\d. This quantifier is + or {2}.
String yourString = "exp_5-22-718";
Matcher matcher = Pattern.compile("exp_(\\d-\\d+)-(\\d+)").matcher(yourString);
if (matcher.find()) {
System.out.println(matcher.group(1)); //prints 5-22
System.out.println(matcher.group(2)); //prints 718
}
You can use the string.split methods to do this. Check the following code.
I assume that your strings starts with "exp_".
String str = "exp_5-22-718";
if (str.contains("-")){
String newStr = str.substring(4, str.length());
String[] strings = newStr.split("-");
for (String string : strings) {
System.out.println(string);
}
}

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