I have a string containing a number. Something like "Incident #492 - The Title Description".
I need to extract the number from this string.
Tried
Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(theString);
String substring =m.group();
By getting an error
java.lang.IllegalStateException: No match found
What am I doing wrong?
What is the correct expression?
I'm sorry for such a simple question, but I searched a lot and still not found how to do this (maybe because it's too late here...)
You are getting this exception because you need to call find() on the matcher before accessing groups:
Matcher m = p.matcher(theString);
while (m.find()) {
String substring =m.group();
System.out.println(substring);
}
Demo.
There are two things wrong here:
The pattern you're using is not the most ideal for your scenario, it's only checking if a string only contains numbers. Also, since it doesn't contain a group expression, a call to group() is equivalent to calling group(0), which returns the entire string.
You need to be certain that the matcher has a match before you go calling a group.
Let's start with the regex. Here's what it looks like now.
Debuggex Demo
That will only ever match a string that contains all numbers in it. What you care about is specifically the number in that string, so you want an expression that:
Doesn't care about what's in front of it
Doesn't care about what's after it
Only matches on one occurrence of numbers, and captures it in a group
To that, you'd use this expression:
.*?(\\d+).*
Debuggex Demo
The last part is to ensure that the matcher can find a match, and that it gets the correct group. That's accomplished by this:
if (m.matches()) {
String substring = m.group(1);
System.out.println(substring);
}
All together now:
Pattern p = Pattern.compile(".*?(\\d+).*");
final String theString = "Incident #492 - The Title Description";
Matcher m = p.matcher(theString);
if (m.matches()) {
String substring = m.group(1);
System.out.println(substring);
}
You need to invoke one of the Matcher methods, like find, matches or lookingAt to actually run the match.
Related
I have a String which looks like "<name><address> and <Phone_1>". I have get to get the result like
1) <name>
2) <address>
3) <Phone_1>
I have tried using regex "<(.*)>" but it returns just one result.
The regex you want is
<([^<>]+?)><([^<>]+?)> and <([^<>]+?)>
Which will then spit out the stuff you want in the 3 capture groups. The full code would then look something like this:
Matcher m = Pattern.compile("<([^<>]+?)><([^<>]+?)> and <([^<>]+?)>").matcher(string);
if (m.find()) {
String name = m.group(1);
String address = m.group(2);
String phone = m.group(3);
}
The pattern .* in a regex is greedy. It will match as many characters as possible between the first < it finds and the last possible > it can find. In the case of your string it finds the first <, then looks for as much text as possible until a >, which it will find at the very end of the string.
You want a non-greedy or "lazy" pattern, which will match as few characters as possible. Simply <(.+?)>. The question mark is the syntax for non-greedy. See also this question.
This will work if you have dynamic number of groups.
Pattern p = Pattern.compile("(<\\w+>)");
Matcher m = p.matcher("<name><address> and <Phone_1>");
while (m.find()) {
System.out.println(m.group());
}
This is the follow up to the question that i asked here
The given regex is perfect i.e., (?:[^\/]*\/){4}([A-Za-z]{3}[0-9]{3}). However, when i do it in java, The java matches the string upto the matching group rather just giving me that string.
String defaultRegex = "(?:[^\\/]*\\/){4}([A-Za-z]{3}[0-9]{3})";
String stringToMatch = "unknown/relevant/nonrelevant:2.2.2/random/ABC123:random/morerandom";
Pattern p = Pattern.compile(defaultRegex);
Matcher m = p.matcher (stringToMatch);
if (m.find()){
System.out.println(m.group());
}
The above thing is printing unknown/relevant/nonrelevant:2.2.2/random/ABC123 when I want regex just to give me ABC123
matcher.group() as well as matcher.group(0) always return the whole matched string.
To get the first capturing group, use matcher.group(1),
The second capturing group goes with matcher.group(2), and so on.
I need to print the simple bind variable names in the SQL query.
I need to print the words starting with : character But NOT ending with dot . character.
in this sample I need to print pOrg, pBusinessId but NOT the parameter.
The regular expression ="(:)(\\w+)^\\." is not working.
Could you help in correcting the regular expression.
Thanks
Peddi
public void testMethod(){
String regEx="(:)(\\w+)([^\\.])";
String input= "(origin_table like 'I%' or (origin_table like 'S%' and process_status =5))and header_id = NVL( :parameter.number1:NULL, header_id) and (orginization = :pOrg) and (businsess_unit = :pBusinessId";
Pattern pattern;
Matcher matcher;
pattern = Pattern.compile(regEx);
matcher = pattern.matcher(input);
String grp = null;
while(matcher.find()){
grp = matcher.group(2);
System.out.println(grp);
}
}
You can try with something like
String regEx = "(:)(\\w+)\\b(?![.])";
(:)(\\w+)\\b will make sure that you are matching only entire words starting with :
(?![.]) is look behind mechanism which makes sure that after found word there is no .
This regex will also allow :NULL so if there is some reason why it shouldn't be matched share it with us.
Anyway to exclude NULL from results you can use
String regEx = "(:)(\\w+)\\b(?![.])(?<!:NULL)";
To make regex case insensitive so NULL could also match null compile this pattern with Pattern.CASE_INSENSITIVE flag like
Pattern pattern = Pattern.compile(regEx,Pattern.CASE_INSENSITIVE);
Since it looks like you're using camelcase, you can actually simplify things a bit when it comes to excluding :NULL:
:([a-z][\\w]+)\\b(?!\\.)
And $1 will return your variable names.
Alternative that doesn't rely on negative lookahead:
:([a-z][\\w]+)\\b(?:[^\\.]|$)
You can try:
Pattern regex = Pattern.compile("^:.*?[^.]$");
Demo
I have a string that begins with one or more occurrences of the sequence "Re:". This "Re:" can be of any combinations, for ex. Re<any number of spaces>:, re:, re<any number of spaces>:, RE:, RE<any number of spaces>:, etc.
Sample sequence of string : Re: Re : Re : re : RE: This is a Re: sample string.
I want to define a java regular expression that will identify and strip off all occurrences of Re:, but only the ones at the beginning of the string and not the ones occurring within the string.
So the output should look like This is a Re: sample string.
Here is what I have tried:
String REGEX = "^(Re*\\p{Z}*:?|re*\\p{Z}*:?|\\p{Z}Re*\\p{Z}*:?)";
String INPUT = title;
String REPLACE = "";
Pattern p = Pattern.compile(REGEX);
Matcher m = p.matcher(INPUT);
while(m.find()){
m.appendReplacement(sb,REPLACE);
}
m.appendTail(sb);
I am using p{Z} to match whitespaces(have found this somewhere in this forum, as Java regex does not identify \s).
The problem I am facing with this code is that the search stops at the first match, and escapes the while loop.
Try something like this replace statement:
yourString = yourString.replaceAll("(?i)^(\\s*re\\s*:\\s*)+", "");
Explanation of the regex:
(?i) make it case insensitive
^ anchor to start of string
( start a group (this is the "re:")
\\s* any amount of optional whitespace
re "re"
\\s* optional whitespace
: ":"
\\s* optional whitespace
) end the group (the "re:" string)
+ one or more times
in your regex:
String regex = "^(Re*\\p{Z}*:?|re*\\p{Z}*:?|\\p{Z}Re*\\p{Z}*:?)"
here is what it does:
see it live here
it matches strings like:
\p{Z}Reee\p{Z: or
R\p{Z}}}
which make no sense for what you try to do:
you'd better use a regex like the following:
yourString.replaceAll("(?i)^(\\s*re\\s*:\\s*)+", "");
or to make #Doorknob happy, here's another way to achieve this, using a Matcher:
Pattern p = Pattern.compile("(?i)^(\\s*re\\s*:\\s*)+");
Matcher m = p.matcher(yourString);
if (m.find())
yourString = m.replaceAll("");
(which is as the doc says the exact same thing as yourString.replaceAll())
Look it up here
(I had the same regex as #Doorknob, but thanks to #jlordo for the replaceAll and #Doorknob for thinking about the (?i) case insensitivity part ;-) )
I have a long string. I want to replace all the matches with part of the matching regex (group).
For example:
String = "This is a great day, is it not? If there is something, THIS IS it. <b>is</b>".
I want to replace all the words "is" by, let's say, "<h1>is</h1>". The case should remain the same as original. So the final string I want is:
This <h1>is</h1> a great day, <h1>is</h1> it not? If there <h1>is</h1> something,
THIS <h1>IS</h1> it. <b><h1>is</h1></b>.
The regex I was trying:
Pattern pattern = Pattern.compile("[.>, ](is)[.<, ]", Pattern.CASE_INSENSITIVE);
The Matcher class is commonly used in conjunction with Pattern. Use the Matcher.replaceAll() method to replace all matches in the string
String str = "This is a great day...";
Pattern p = Pattern.compile("\\bis\\b", Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(str);
String result = m.replaceAll("<h1>is</h1>");
Note: Using the \b regex command will match on a word boundary (like whitespace). This is helpful to use in order to ensure that only the word "is" is matched and not words that contain the letters "i" and "s" (like "island").
Like this:
str = str.replaceAll(yourRegex, "<h1>$1</h1>");
The $1 refers to the text captured by group #1 in your regex.
Michael's answer is better, but if you happen to specifically only want [.>, ] and [.<, ] as boundaries, you can do it like this:
String input = "This is a great day, is it not? If there is something, THIS IS it. <b>is</b>";
Pattern p = Pattern.compile("(?<=[.>, ])(is)(?=[.<, ])", Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(input);
String result = m.replaceAll("<h1>$1</h1>");
yourStr.replaceAll("(?i)([.>, ])(is)([.<, ])","$1<h1>$2</h1>$3")
(?i)to indicate ignoring case; wrap everything your want to reuse with brackets, reuse them with $1 $2 and $3, concatenate them into what you want.
Simply use a backreference for that.
"This is a great day, is it not? If there is something, THIS IS it. <b>is</b>".replaceAll("[.>, ](is)[.<, ]", "<h1>$2</h1>"); should do.
It may be a late addition, but if anyone is looking for this like
Searching for 'thing' and also he needs 'Something' too to be taken as result,
Pattern p = Pattern.compile("([^ ])is([^ \.])");
String result = m.replaceAll("<\h1>$1is$2</h1>");
will result <\h1>Something</h1> too