I have sample content string repeated in a file which I wanna to retrieve its double value from it.the string content is "(AIC)|234.654 |" which I wanna retrieve the 234.654 from that...the "(AIC)|" is always fixed but the numbers change in other occasions so I am using regular expression as follow..but it says there is no match using below expression..any help would be appreciated
String contents="(AIC)|234.654 |";
Pattern p = Pattern.compile("AIC\\u0029{1}\\u007C{1}\\d+u002E{1}\\d+");
Matcher m = p.matcher(contents);
boolean b = m.find();
String t=m.group();
The above expression doest find any match and throw exception..
Thanks for any help
Your code has several typos, but beside them, you say you need to match the number inside the brackets, but you are referring to the whole match with .group(). You need to set a capturing group to access that number with .group(1).
Here is a fixed code:
String content="(AIC)|234.654 |";
Pattern p = Pattern.compile("AIC\\)\\|(\\d+\\.\\d+)");
Matcher m = p.matcher(content);
if (m.find())
{
System.out.println(m.group(1));
}
See IDEONE demo
If the number can be integer, just use an optional non-capturing group around the decimal part: Pattern.compile("AIC\\)\\|(\\d+(?:\\.\\d+)?)");
I think this regex should do the work:
(?<=\|)[\d\.]*(?=\s*\|)
It will only match digits and dots after a | and before an optional space and another |
And the complete code:
String content="(AIC)|234.654 |";
Pattern p = Pattern.compile("(?<=\\|)[\\d\\.]*(?=\\s*\\|)");
Matcher m = p.matcher(content);
boolean b = m.find();
String t=m.group();
Related
I have used this (?:#\d{7}) regex for extracting only 7 digit after '#'.
For example I have string something like "#1234567890". After using the above patterrn I will get 7 digit after '#'.
Now the problem is : I have string something like that "Referenc number #1234567890"
where "Referenc number #" fixed.
Now I am finding for regex which can return the 1234567 number from the above string.
I have a one file which contains above string and there are also other data available.
You can try something like this:
String ref_no = "Referenc number #123456789";
Pattern p = Pattern.compile("Referenc number #([0-9]{7})");
Matcher m = p.matcher(ref_no);
while (m.find())
{
System.out.println(m.group(1));
}
The ?: should make your group "non-capturing", so if you add that separately around the hash sign, it should used for matching but excluded from capture.
(?:#)(\d{7})
If the String always starts with Referenc number # you could just use the following code:
String text = "Referenc number #1234567890";
Pattern pattern = Pattern.compile("\\d{7}");
Matcher matcher = pattern.matcher(text);
while(matcher.find()){
System.out.println(matcher.group());
}
I need a regular expression that matches a substring in string /*exa*/mple*/ ,
the matched string must be /*exa*/ not /*exa*/mple*/.
It also must not contain "*/" in it.
I have tried these regex:
"/\\*[.*&&[^*/]]\\*/" ,
"/\\*.*&&(?!^*/$)\\*/"
but im not able to get the exact solution.
I understand you want to pick out comments from a text.
Pattern p = Pattern.compile("/\\*.*?\\*/");
Matcher m = p.matcher("/*ex*a*/mple*/and/*more*/ther*/");
while (m.find()){
System.out.println(m.group());
}
you can try this:
/\*[^\*\/\*]+\*/ --> anything that is in between (including) "/*" and "*/"
Here is a sample:
Pattern p = Pattern.compile("/\\*[^\\*\\/\\*]+\\*/");
Matcher m = p.matcher("/*exa*/mple*/");
while (m.find()){
System.out.println(m.group());
}
OUTPUT:
/*exa*/
I have this code to find this pattern: 201409250200131738007947036000 - 1 ,inside the text
final String patternStr = "(\\d{30} - \\d{1})";
final Pattern p = Pattern.compile(patternStr);
final Matcher m = p.matcher(page);
if (m.matches()) {
System.out.println("SUCCESS");
}
But for any strange reasson in Java did't work, Can somebody help me where is the error please?
The reason is that the matches method checks for the entire given string to match the regex.
So i.e. if your string is 123456123412345612341234561234 - 8 it will match, if it is my number 123456123412345612341234561234 - 8 is inside other text it won't.
Use the find method to accomplish your task:
if (m.find()) {
System.out.println("SUCCESS");
}
It will search inside the given string instead of attempting to match the entire string.
From the documentation for Matcher, matches:
Attempts to match the entire region against the pattern.
As opposed to find which:
Attempts to find the next subsequence of the input sequence that matches the pattern.
So use matches to match an entire String against a pattern, use find to locate a pattern inside a String.
Try:
final String patternStr = "\\d{30}+\\s-\\s\\d";
final Pattern p = Pattern.compile(patternStr);
final Matcher m = p.matcher(page);
while (m.find()) {
System.out.printf("FOUND A MATCH: %s%n", matcher.group());
}
I edited your pattern slightly to make it more robust. This will print each match that it finds.
Pattern is:
private static Pattern r = Pattern.compile("(.*\\..*\\..*)\\..*");
String is:
sentVersion = "1.1.38.24.7";
I do:
Matcher m = r.matcher(sentVersion);
if (m.find()) {
guessedClientVersion = m.group(1);
}
I expect 1.1.38 but the pattern match fails. If I change to Pattern.compile("(.*\\..*\\..*)\\.*");
// notice I remove the "." before the last *
then 1.1.38.XXX fails
My goal is to find (x.x.x) in any incoming string.
Where am I wrong?
Problem is probably due to greedy-ness of your regex. Try this negation based regex pattern:
private static Pattern r = Pattern.compile("([^.]*\\.[^.]*\\.[^.]*)\\..*");
Online Demo: http://regex101.com/r/sJ5rD4
Make your .* matches reluctant with ?
Pattern r = Pattern.compile("(.*?\\..*?\\..*?)\\..*");
otherwise .* matches the whole String value.
See here: http://regex101.com/r/lM2lD5
How do I build a regular expression for a long data type in Java, I currently have a regex expression for 3 double values as my pattern:
String pattern = "(max=[0-9]+\\.?[0-9]*) *(total=[0-9]+\\.?[0-9]*) *(free=[0-9]+\\.?[0-9]*)";
I am constructing the pattern using the line:
Pattern a = Pattern.compile("control.avgo:", Pattern.CASE_INSENSITIVE);
I want to match the numbers following the equals signs in the example text below, from the file control.avgo.
max=259522560, total=39325696, free=17979640
What do I need to do to correct my code to match them?
Could it be that you actually need
Pattern a = Pattern.compile(pattern, Pattern.CASE_INSENSITIVE);
instead of
Pattern a = Pattern.compile("control.avgo:", Pattern.CASE_INSENSITIVE);
because your current code uses "control.avgo:" as the regex, and not the pattern you have defined.
You need to address several errors, including:
Your pattern specifies real numbers, but your question asks for long integers.
Your pattern omits the commas in the string being searched.
The first argument to Pattern.compile() is the regular expression, not the string being searched.
This will work:
String sPattern = "max=([0-9]+), total=([0-9]+), free=([0-9]+)";
Pattern pattern = Pattern.compile( sPattern, Pattern.CASE_INSENSITIVE );
String source = "control.avgo: max=259522560, total=39325696, free=17979640";
Matcher matcher = pattern.matcher( source );
if ( matcher.find()) {
System.out.println("max=" + matcher.group(1));
System.out.println("total=" + matcher.group(2));
System.out.println("free=" + matcher.group(3));
}
If you want to convert the numbers you find to a numeric type, use Long.valueOf( String ).
In case you only need to find any numerical preceded by "="...
String test = "3.control.avgo: max=259522560, total=39325696, free=17979640";
// looks for the "=" sign preceding any numerical sequence of any length
Pattern pattern = Pattern.compile("(?<=\\=)\\d+");
Matcher matcher = pattern.matcher(test);
// keeps on searching until cannot find anymore
while (matcher.find()) {
// prints out whatever found
System.out.println(matcher.group());
}
Output:
259522560
39325696
17979640