My replacerRegex is
("schedulingCancelModal": \{\s*? "title": ")(.+?)(?=")
The right value is getting picked up, i.e. valueToBePicked:
But how do I get ("schedulingCancelModal": \{\s*? "title": ") not to be included in the result like positive lookbehind does?
My Java code so far:
Pattern replacerPattern = Pattern.compile(replacerRegex);
Matcher matcher = replacerPattern.matcher(value);
while (matcher.find()) {
String valueToBePicked = matcher.group();
}
You can simply select matcher.group(2) which will give you the contents of the second capture group. For example:
String replacerRegex = "(\"schedulingCancelModal\": \\{\\s*? \"title\": \")(.+?)(?=\")";
String value = "\"valueToBePicked\": \"schedulingCancelModal\": {\n \"title\": \"Are you sure you want to leave scheduling?\", ... }";
Pattern replacerPattern = Pattern.compile(replacerRegex);
Matcher matcher = replacerPattern.matcher(value);
while (matcher.find()) {
String valueToBePicked = matcher.group(2);
System.out.println(valueToBePicked);
}
Output:
Are you sure you want to leave scheduling?
Demo on rextester
Related
I would like to be able to find the first occurrence of m² and then numbers in front of it, could be integers or decimal numbers.
E.g.
"some text" 38 m² "some text" ,
"some text" 48,8 m² "some text",
"some text" 48 m² "some text", etc..
What I have so far is:
\d\d,\d\s*(\m\u00B2)|\d\d\s*(\m\u00B2)
This right now finds all occurrences, although I guess it could be fixed with findFirst(). Any ideas how to improve the Regex part?
To get the first match, you just need to use Matcher#find() inside an if block:
String rx = "\\d+(?:,\\d+)?\\s*m\\u00B2";
Pattern p = Pattern.compile(rx);
Matcher matcher = p.matcher("E.g. : 4668,68 m² some text, some text 48 m² etc");
if (matcher.find()){
System.out.println(matcher.group());
}
See IDEONE demo
Note that you can get rid of the alternation group using an optional non-capturing group (?:..)?
Pattern breakdown:
\d+ - 1+ digits
(?:,\d+)? - 0+ sequences of a comma followed with 1+ digits
\s* - 0+ whitespace symbols
m\u00B2 - m2.
This is what I came up with you help :) (work in progress, later it should return BigDecimal value), for now it seems to work:
public static String findArea(String description) {
String tempString = "";
Pattern p = Pattern.compile("\\d+(?:,\\d+)?\\s*m\\u00B2");
Matcher m = p.matcher(description);
if(m.find()) {
tempString = m.group();
}
//remove the m and /u00B2 to parse it to BigDecimal later
tempString = tempString.replaceAll("[^0-9|,]","");
System.out.println(tempString);
return tempString;
}
One simple way of doing it!
description.replaceFirst(#NotNull String regex,
#NotNull String replacement)
JAVADoc: Replaces the first substring of this string that matches the given regular expression with the given replacement.
To find only last one:
#Test
public void testFindFirstRegExp() {
String pattern = ".* (\\d+,\\d+) .*";
Pattern r = Pattern.compile(pattern);
String line = "some text 44,66 m² some 33,11 m² text 11,22 m² some text";
Matcher m = r.matcher(new StringBuilder(line).reverse().toString());
String expected = "44,66";
String actual = null;
if (m.find()) {
actual = new StringBuilder(m.group(1)).reverse().toString();
}
System.out.println("got first:" + actual);
Assert.assertEquals(expected, actual);
m = r.matcher(line);
expected = "11,22";
actual = null;
if (m.find()) {
actual = m.group(1);
}
System.out.println("got last:" + actual);
Assert.assertEquals(expected, actual);
}
prints:
got first:44,66
got last:11,22
Note: think that you need to reverse pattern when needed for ex:
pattern = ".* (\\d+,\\d+-?) .*"; //reverse for (-?\\d+,\\d+)
but this will work as waited:
pattern = " (\\-?\\d+,\\d+) ";
you get all of them in loop:
while (m.find()) {
actual = m.group(1);
System.out.println("got last:" + actual);
}
Will print:
got last:44,66
got last:33,11
got last:11,22
i have a string like this:
font-size:36pt;color:#ffffff;background-color:#ff0000;font-family:Times New Roman;
How can I get the value of the color and the value of background-color?
color:#ffffff;
background-color:#ff0000;
i have tried the following code but the result is not my expected.
Pattern pattern = Pattern.compile("^.*(color:|background-color:).*;$");
The result will display:
font-size:36pt; color:#ffffff; background-color:#ff0000; font-family:Times New Roman;
If you want to have multiple matches in a string, don't assert ^ and $ because if those matches, then the whole string matches, which means that you can't match it again.
Also, use a lazy quantifier like *?. This will stop matching as soon as it finds some string that matches the pattern after it.
This is the regex you should use:
(color:|background-color:)(.*?);
Group 1 is either color: or background-color:, group 2 is the color code.
Demo
To do this you should use the (?!abc) expression in regex. This finds a match but doesn't select it. After that you can simply select the hexcode, like this:
String s = "font-size:36pt;color:#ffffff;background-color:#ff0000;font-family:Times New Roman";
Pattern pattern = Pattern.compile("(?!color:)#.{6}");
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
System.out.println(matcher.group());
}
Pattern pattern = Pattern.compile("color\\s*:\\s*([^;]+)\\s*;\\s*background-color\\s*:\\s*([^;]+)\\s*;");
Matcher matcher = pattern.matcher("font-size:36pt; color:#ffffff; background-color:#ff0000; font-family:Times New Roman;");
if (matcher.find()) {
System.out.println("color:" + matcher.group(1));
System.out.println("background-color:" + matcher.group(2));
}
No need to describe the whole input, only the relevant part(s) that you're looking to extract.
The regex color:(#[\\w\\d]+); does the trick for me:
String input = "font-size:36pt;color:#ffffff;background-color:#ff0000;font-family:Times New Roman;";
String regex = "color:(#[\\w\\d]+);";
Matcher m = Pattern.compile(regex).matcher(input);
while (m.find()) {
System.out.println(m.group(1));
}
Notice that m.group(1) returns the matching group which is inside the parenthesis in the regex. So the regex actually matches the whole color:#ffffff; and color:#ff0000; parts, but the print only handles the number itself.
Use a CSS parser like ph-css
String input = "font-size:36pt; color:#ffffff; background-color:#ff0000; font-family:Times New Roman;";
final CSSDeclarationList cssPropertyList =
CSSReaderDeclarationList.readFromString(input, ECSSVersion.CSS30);
System.out.println(cssPropertyList.get(1).getProperty() + " , "
+ cssPropertyList.get(1).getExpressionAsCSSString());
System.out.println(cssPropertyList.get(2).getProperty() + " , "
+ cssPropertyList.get(2).getExpressionAsCSSString());
Prints:
color , #ffffff
background-color , #ff0000
Find more about ph-css on github
I am trying to match a keyword with following string
"abc,pqr(1),xyz"
It will be succesfull match if the whole one word matched for e.g. "par" or "abc" or "xyz"
Can anyone please help me in creating regex for this match ?
String text = "hello, hellos(1),bye";
String keyword = "account";
String patternString = "["+ keyword + "]";
Pattern pattern = Pattern.compile(patternString, Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(text);
boolean matches = matcher.matches();
System.out.println("matches = " + matches);
This Should Work.
([a-zA-Z]+)
Input:
"abc,pqr(1),xyz"
Output:
abc
pqr
xyz
See: https://regex101.com/r/Us6G3X/2
I would like to be able to find the first occurrence of m² and then numbers in front of it, could be integers or decimal numbers.
E.g.
"some text" 38 m² "some text" ,
"some text" 48,8 m² "some text",
"some text" 48 m² "some text", etc..
What I have so far is:
\d\d,\d\s*(\m\u00B2)|\d\d\s*(\m\u00B2)
This right now finds all occurrences, although I guess it could be fixed with findFirst(). Any ideas how to improve the Regex part?
To get the first match, you just need to use Matcher#find() inside an if block:
String rx = "\\d+(?:,\\d+)?\\s*m\\u00B2";
Pattern p = Pattern.compile(rx);
Matcher matcher = p.matcher("E.g. : 4668,68 m² some text, some text 48 m² etc");
if (matcher.find()){
System.out.println(matcher.group());
}
See IDEONE demo
Note that you can get rid of the alternation group using an optional non-capturing group (?:..)?
Pattern breakdown:
\d+ - 1+ digits
(?:,\d+)? - 0+ sequences of a comma followed with 1+ digits
\s* - 0+ whitespace symbols
m\u00B2 - m2.
This is what I came up with you help :) (work in progress, later it should return BigDecimal value), for now it seems to work:
public static String findArea(String description) {
String tempString = "";
Pattern p = Pattern.compile("\\d+(?:,\\d+)?\\s*m\\u00B2");
Matcher m = p.matcher(description);
if(m.find()) {
tempString = m.group();
}
//remove the m and /u00B2 to parse it to BigDecimal later
tempString = tempString.replaceAll("[^0-9|,]","");
System.out.println(tempString);
return tempString;
}
One simple way of doing it!
description.replaceFirst(#NotNull String regex,
#NotNull String replacement)
JAVADoc: Replaces the first substring of this string that matches the given regular expression with the given replacement.
To find only last one:
#Test
public void testFindFirstRegExp() {
String pattern = ".* (\\d+,\\d+) .*";
Pattern r = Pattern.compile(pattern);
String line = "some text 44,66 m² some 33,11 m² text 11,22 m² some text";
Matcher m = r.matcher(new StringBuilder(line).reverse().toString());
String expected = "44,66";
String actual = null;
if (m.find()) {
actual = new StringBuilder(m.group(1)).reverse().toString();
}
System.out.println("got first:" + actual);
Assert.assertEquals(expected, actual);
m = r.matcher(line);
expected = "11,22";
actual = null;
if (m.find()) {
actual = m.group(1);
}
System.out.println("got last:" + actual);
Assert.assertEquals(expected, actual);
}
prints:
got first:44,66
got last:11,22
Note: think that you need to reverse pattern when needed for ex:
pattern = ".* (\\d+,\\d+-?) .*"; //reverse for (-?\\d+,\\d+)
but this will work as waited:
pattern = " (\\-?\\d+,\\d+) ";
you get all of them in loop:
while (m.find()) {
actual = m.group(1);
System.out.println("got last:" + actual);
}
Will print:
got last:44,66
got last:33,11
got last:11,22
One common usage for regex is the replacement of the matches with something that is based on the matches.
For example a commit-text with ticket numbers ABC-1234: some text (ABC-1234) has to be replaced with <ABC-1234>: some text (<ABC-1234>) (<> as example for some surroundings.)
This is very simple in Java
String message = "ABC-9913 - Bugfix: Some text. (ABC-9913)";
String finalMessage = message;
Matcher matcher = Pattern.compile("ABC-\\d+").matcher(message);
if (matcher.find()) {
String ticket = matcher.group();
finalMessage = finalMessage.replace(ticket, "<" + ticket + ">");
}
System.out.println(finalMessage);
results in<ABC-9913> - Bugfix: Some text. (<ABC-9913>).
But if there are different matches in the input String, this is different. I tried a slightly different code replacing if (matcher.find()) { with while (matcher.find()) {. The result is messed up with doubled replacements (<<ABC-9913>>).
How can I replace all matching values in an elegant way?
You can simply use replaceAll:
String input = "ABC-1234: some text (ABC-1234)";
System.out.println(input.replaceAll("ABC-\\d+", "<$0>"));
prints:
<ABC-1234>: some text (<ABC-1234>)
$0 is a reference to the matched string.
Java regex reference (see "Groups and capturing").
The problem is that the replace() method transforms the string over and over again.
A better way is to replace one match at a time. The matcher class has an appendReplacement-method for this.
String message = "ABC-9913, ABC-9915 - Bugfix: Some text. (ABC-9913,ABC-9915)";
Matcher matcher = Pattern.compile("ABC-\\d+").matcher(message);
StringBuffer sb = new StringBuffer();
while (matcher.find()) {
String ticket = matcher.group();
matcher.appendReplacement(sb, "<" + ticket + ">");
}
matcher.appendTail(sb);
System.out.println(sb);