I have this code snippet of a code base I am supposed to maintain.
String number = "1";
String value = "test";
String output = "";
output = value.replaceAll("\\Q{#}", number);
The value of output stays as "test" and I can only guess what this code is supposed to do: the value of numbershould be appended to whatever is in value. Maybe something like this: test1 or replace the value with the number entirely.
I found out that \\Q is the regex option to quote everything until \\E but there is no \\E. Anyway it is not doing anything at all and I am wondering if I oversee something?
Your regex just matches a literal {#}. It is true that after \Q the pattern is considered to have literal symbols (all the symbols after \Q get "quoted" or "escaped"), and \E stops this escaping/quoting, and if it is missing, the whole pattern will get quoted/escaped.
If your value variable holds test{#} value, the {#} will get replaced with the number.
See this demo:
String number = "1";
String value = "test{#}";
String output = "";
output = value.replaceAll("\\Q{#}", number);
System.out.println(output); // => test1
Note that without \Q, your regex ({#}) would throw a java.util.regex.PatternSyntaxException: Illegal repetition error because Java regex engine is not "smart" enough to disambiguate the braces (PCRE, JS, .NET can easily guess that since there is no number inside, it is not a limiting/bound quantifier).
As i haven't much worked on regex, can someone help me out in getting the answer for below thing:
(1)I want to remove a text say Element
(2)It may of may not followed by delimiter say pipe(||)
I tried below thing, but it is not working in the way i want:
String str = "String:abc||Element:abc||Value:abc"; // Sample text 1
String str1 = "String:abc||Element:abc"; // Sample text 2
System.out.println(str.replaceFirst("Element.*\\||", ""));
System.out.println(str1.replaceFirst("Element.*\\||", ""));
Required output in above cases:
String:abc||Value:abc //for the first case
String:abc //for the second case
Assuming that you can decide to give another value to the original pattern which is Element in this case, you can use Pattern.quote to escape it as below:
String str = "String:abc||Element:abc||Value:abc"; // Sample text 1
String str1 = "String:abc||Element:abc"; // Sample text 2
String originalPattern = "Element";
String pattern = String.format("\\|{2}%s[^\\|]+", Pattern.quote(originalPattern));
System.out.println(str.replaceFirst(pattern, ""));
System.out.println(str1.replaceFirst(pattern, ""));
Your patter is then generic and its value is String.format("\\|{2}%s[^\\|]+", Pattern.quote(originalPattern))
Output:
String:abc||Value:abc
String:abc
You put the escape wrong. It should be:
Element(.*?\|\||.*$)
Put the escape on each pipe, and use ? for non greedy Regex so you only replace just enough string, not everything.
String text = "String:abc||Element:abc||Value:abc";
text = text.replaceAll("\\belement\\b", "");
you might need to use replace all this will replace all element from your string here i am using '\b' word boundary in java regular expression in between the words
I'm new to regular expressions and would appreciate your help. I'm trying to put together an expression that will split the example string using all spaces that are not surrounded by single or double quotes. My last attempt looks like this: (?!") and isn't quite working. It's splitting on the space before the quote.
Example input:
This is a string that "will be" highlighted when your 'regular expression' matches something.
Desired output:
This
is
a
string
that
will be
highlighted
when
your
regular expression
matches
something.
Note that "will be" and 'regular expression' retain the space between the words.
I don't understand why all the others are proposing such complex regular expressions or such long code. Essentially, you want to grab two kinds of things from your string: sequences of characters that aren't spaces or quotes, and sequences of characters that begin and end with a quote, with no quotes in between, for two kinds of quotes. You can easily match those things with this regular expression:
[^\s"']+|"([^"]*)"|'([^']*)'
I added the capturing groups because you don't want the quotes in the list.
This Java code builds the list, adding the capturing group if it matched to exclude the quotes, and adding the overall regex match if the capturing group didn't match (an unquoted word was matched).
List<String> matchList = new ArrayList<String>();
Pattern regex = Pattern.compile("[^\\s\"']+|\"([^\"]*)\"|'([^']*)'");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
if (regexMatcher.group(1) != null) {
// Add double-quoted string without the quotes
matchList.add(regexMatcher.group(1));
} else if (regexMatcher.group(2) != null) {
// Add single-quoted string without the quotes
matchList.add(regexMatcher.group(2));
} else {
// Add unquoted word
matchList.add(regexMatcher.group());
}
}
If you don't mind having the quotes in the returned list, you can use much simpler code:
List<String> matchList = new ArrayList<String>();
Pattern regex = Pattern.compile("[^\\s\"']+|\"[^\"]*\"|'[^']*'");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
matchList.add(regexMatcher.group());
}
There are several questions on StackOverflow that cover this same question in various contexts using regular expressions. For instance:
parsings strings: extracting words and phrases
Best way to parse Space Separated Text
UPDATE: Sample regex to handle single and double quoted strings. Ref: How can I split on a string except when inside quotes?
m/('.*?'|".*?"|\S+)/g
Tested this with a quick Perl snippet and the output was as reproduced below. Also works for empty strings or whitespace-only strings if they are between quotes (not sure if that's desired or not).
This
is
a
string
that
"will be"
highlighted
when
your
'regular expression'
matches
something.
Note that this does include the quote characters themselves in the matched values, though you can remove that with a string replace, or modify the regex to not include them. I'll leave that as an exercise for the reader or another poster for now, as 2am is way too late to be messing with regular expressions anymore ;)
If you want to allow escaped quotes inside the string, you can use something like this:
(?:(['"])(.*?)(?<!\\)(?>\\\\)*\1|([^\s]+))
Quoted strings will be group 2, single unquoted words will be group 3.
You can try it on various strings here: http://www.fileformat.info/tool/regex.htm or http://gskinner.com/RegExr/
The regex from Jan Goyvaerts is the best solution I found so far, but creates also empty (null) matches, which he excludes in his program. These empty matches also appear from regex testers (e.g. rubular.com).
If you turn the searches arround (first look for the quoted parts and than the space separed words) then you might do it in once with:
("[^"]*"|'[^']*'|[\S]+)+
(?<!\G".{0,99999})\s|(?<=\G".{0,99999}")\s
This will match the spaces not surrounded by double quotes.
I have to use min,max {0,99999} because Java doesn't support * and + in lookbehind.
It'll probably be easier to search the string, grabbing each part, vs. split it.
Reason being, you can have it split at the spaces before and after "will be". But, I can't think of any way to specify ignoring the space between inside a split.
(not actual Java)
string = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
regex = "\"(\\\"|(?!\\\").)+\"|[^ ]+"; // search for a quoted or non-spaced group
final = new Array();
while (string.length > 0) {
string = string.trim();
if (Regex(regex).test(string)) {
final.push(Regex(regex).match(string)[0]);
string = string.replace(regex, ""); // progress to next "word"
}
}
Also, capturing single quotes could lead to issues:
"Foo's Bar 'n Grill"
//=>
"Foo"
"s Bar "
"n"
"Grill"
String.split() is not helpful here because there is no way to distinguish between spaces within quotes (don't split) and those outside (split). Matcher.lookingAt() is probably what you need:
String str = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
str = str + " "; // add trailing space
int len = str.length();
Matcher m = Pattern.compile("((\"[^\"]+?\")|('[^']+?')|([^\\s]+?))\\s++").matcher(str);
for (int i = 0; i < len; i++)
{
m.region(i, len);
if (m.lookingAt())
{
String s = m.group(1);
if ((s.startsWith("\"") && s.endsWith("\"")) ||
(s.startsWith("'") && s.endsWith("'")))
{
s = s.substring(1, s.length() - 1);
}
System.out.println(i + ": \"" + s + "\"");
i += (m.group(0).length() - 1);
}
}
which produces the following output:
0: "This"
5: "is"
8: "a"
10: "string"
17: "that"
22: "will be"
32: "highlighted"
44: "when"
49: "your"
54: "regular expression"
75: "matches"
83: "something."
I liked Marcus's approach, however, I modified it so that I could allow text near the quotes, and support both " and ' quote characters. For example, I needed a="some value" to not split it into [a=, "some value"].
(?<!\\G\\S{0,99999}[\"'].{0,99999})\\s|(?<=\\G\\S{0,99999}\".{0,99999}\"\\S{0,99999})\\s|(?<=\\G\\S{0,99999}'.{0,99999}'\\S{0,99999})\\s"
Jan's approach is great but here's another one for the record.
If you actually wanted to split as mentioned in the title, keeping the quotes in "will be" and 'regular expression', then you could use this method which is straight out of Match (or replace) a pattern except in situations s1, s2, s3 etc
The regex:
'[^']*'|\"[^\"]*\"|( )
The two left alternations match complete 'quoted strings' and "double-quoted strings". We will ignore these matches. The right side matches and captures spaces to Group 1, and we know they are the right spaces because they were not matched by the expressions on the left. We replace those with SplitHere then split on SplitHere. Again, this is for a true split case where you want "will be", not will be.
Here is a full working implementation (see the results on the online demo).
import java.util.*;
import java.io.*;
import java.util.regex.*;
import java.util.List;
class Program {
public static void main (String[] args) throws java.lang.Exception {
String subject = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
Pattern regex = Pattern.compile("\'[^']*'|\"[^\"]*\"|( )");
Matcher m = regex.matcher(subject);
StringBuffer b= new StringBuffer();
while (m.find()) {
if(m.group(1) != null) m.appendReplacement(b, "SplitHere");
else m.appendReplacement(b, m.group(0));
}
m.appendTail(b);
String replaced = b.toString();
String[] splits = replaced.split("SplitHere");
for (String split : splits) System.out.println(split);
} // end main
} // end Program
If you are using c#, you can use
string input= "This is a string that \"will be\" highlighted when your 'regular expression' matches <something random>";
List<string> list1 =
Regex.Matches(input, #"(?<match>\w+)|\""(?<match>[\w\s]*)""|'(?<match>[\w\s]*)'|<(?<match>[\w\s]*)>").Cast<Match>().Select(m => m.Groups["match"].Value).ToList();
foreach(var v in list1)
Console.WriteLine(v);
I have specifically added "|<(?[\w\s]*)>" to highlight that you can specify any char to group phrases. (In this case I am using < > to group.
Output is :
This
is
a
string
that
will be
highlighted
when
your
regular expression
matches
something random
1st one-liner using String.split()
String s = "This is a string that \"will be\" highlighted when your 'regular expression' matches something.";
String[] split = s.split( "(?<!(\"|').{0,255}) | (?!.*\\1.*)" );
[This, is, a, string, that, "will be", highlighted, when, your, 'regular expression', matches, something.]
don't split at the blank, if the blank is surrounded by single or double quotes
split at the blank when the 255 characters to the left and all characters to the right of the blank are neither single nor double quotes
adapted from original post (handles only double quotes)
I'm reasonably certain this is not possible using regular expressions alone. Checking whether something is contained inside some other tag is a parsing operation. This seems like the same problem as trying to parse XML with a regex -- it can't be done correctly. You may be able to get your desired outcome by repeatedly applying a non-greedy, non-global regex that matches the quoted strings, then once you can't find anything else, split it at the spaces... that has a number of problems, including keeping track of the original order of all the substrings. Your best bet is to just write a really simple function that iterates over the string and pulls out the tokens you want.
A couple hopefully helpful tweaks on Jan's accepted answer:
(['"])((?:\\\1|.)+?)\1|([^\s"']+)
Allows escaped quotes within quoted strings
Avoids repeating the pattern for the single and double quote; this also simplifies adding more quoting symbols if needed (at the expense of one more capturing group)
You can also try this:
String str = "This is a string that \"will be\" highlighted when your 'regular expression' matches something";
String ss[] = str.split("\"|\'");
for (int i = 0; i < ss.length; i++) {
if ((i % 2) == 0) {//even
String[] part1 = ss[i].split(" ");
for (String pp1 : part1) {
System.out.println("" + pp1);
}
} else {//odd
System.out.println("" + ss[i]);
}
}
The following returns an array of arguments. Arguments are the variable 'command' split on spaces, unless included in single or double quotes. The matches are then modified to remove the single and double quotes.
using System.Text.RegularExpressions;
var args = Regex.Matches(command, "[^\\s\"']+|\"([^\"]*)\"|'([^']*)'").Cast<Match>
().Select(iMatch => iMatch.Value.Replace("\"", "").Replace("'", "")).ToArray();
When you come across this pattern like this :
String str = "2022-11-10 08:35:00,470 RAV=REQ YIP=02.8.5.1 CMID=caonaustr CMN=\"Some Value Pyt Ltd\"";
//this helped
String[] str1= str.split("\\s(?=(([^\"]*\"){2})*[^\"]*$)\\s*");
System.out.println("Value of split string is "+ Arrays.toString(str1));
This results in :[2022-11-10, 08:35:00,470, PLV=REQ, YIP=02.8.5.1, CMID=caonaustr, CMN="Some Value Pyt Ltd"]
This regex matches spaces ONLY if it is followed by even number of double quotes.
I want to replace words in a string, but I am having little difficulties. Here is what I want to do. I have string:
String a = "I want to replace some words in this string";
It should work like some kind of a translator. I am doing this with String.replaceAll(), but it doesn't work completely because of this. Let's say I am translating from English to German, than this should be the output (Ich means I in German).
String toTranslate = "I";
String translated = "Ich";
a = a.replaceAll(toTranslate.toLowerCase(), translated.toLowerCase());
Now the output of the String a will be this:
"ich want to replace some words ich**n** **th**ich**s** **str**ich**ng**"
How to replace just the words, not the subwords in the words?
replaceAll uses regex, so you may add word boundaries or look-around mechanisms to check if there are no non-space characters surrounding word you want to replace.
String toTranslate = "I";
String translated = "Ich";
a = a.replaceAll("(?<!\\S)"+toTranslate.toLowerCase()+"(?!\\S)", translated.toLowerCase());
You can also add quotation mechanism to escape any regex metacharacters like + * ( inside word you want to replace. BTW you don't need to change your string to lower case, simply add case-insensitive flag to regex (?i).
a = a.replaceAll("(?i)(?<!\\S)"+Pattern.quote(toTranslate)+"(?!\\S)", translated.toLowerCase());
Use split(" ") for getting each word in the sentence. And then use replaceAll on each word.
String a = "I want to replace some words in this string";
String toTranslate = "I";
String translated = "Ich";
String newString[]=a.split(" ");
for (String string : newString) {
string=string.replaceAll(toTranslate, toTranslate.toLowerCase());//Adding this line ensures you dont miss any uppercase toTranslate
string=string.replaceAll(toTranslate.toLowerCase(), translated.toLowerCase());
System.out.println("after translation ="+string);
}
String toTranslate = "I ";
String translated = "Ich ";
a = a.replaceAll(toTranslate.toLowerCase(), translated.toLowerCase());
If you add a space after the "I" it should replace it when it comes to the word "Ich" but if your word ends in a "I" then thats another problem
If you assume that I will always be capitalized in English as it should be then
a = a.replaceAll(toTranslate, translated);
will work, otherwise you need to replace both cases
a = a.replaceAll(toTranslate, translated);
a = a.replaceAll("([^a-zA-Z])("+toTranslate.toLowerCase()+")([^a-zA-Z])", "$1"+translated.toLowerCase()+"$3");
Here is a working example
Yes, the word boundaries are the solution. I just did this in the regex:
text.replaceAll("\\b" + parts1[i] + "\\b", map.element.value);
Don't be confused with the second argument it's string (from Hash table).
You can use RegEx's word bound, which is \b
String toTranslate = "\\bI\\b";
String translated = "Ich";
a = a.replaceAll(toTranslate.toLowerCase(), translated.toLowerCase());
This should ensure I is separated entirely into its own word
Edit: I misread the question and realized you want whole words. See above, as I have accounted for that
I am trying to split a string according to a certain set of delimiters.
My delimiters are: ,"():;.!? single spaces or multiple spaces.
This is the code i'm currently using,
String[] arrayOfWords= inputString.split("[\\s{2,}\\,\"\\(\\)\\:\\;\\.\\!\\?-]+");
which works fine for most cases but i'm have a problem when the the first word is surrounded by quotation marks. For example
String inputString = "\"Word\" some more text.";
Is giving me this output
arrayOfWords[0] = ""
arrayOfWords[0] = "Word"
arrayOfWords[1] = "some"
arrayOfWords[2] = "more"
arrayOfWords[3] = "text"
I want the output to give me an array with
arrayOfWords[0] = "Word"
arrayOfWords[1] = "some"
arrayOfWords[2] = "more"
arrayOfWords[3] = "text"
This code has been working fine when quotation marks are used in the middle of the sentence, I'm not sure what the trouble is when it's at the beginning.
EDIT: I just realized I have same problem when any of the delimiters are used as the first character of the string
Unfortunately you wont be able to remove this empty first element using only split. You should probably remove first elements from your string that match your delimiters and split after it. Also your regex seems to be incorrect because
by adding {2,} inside [...] you are in making { 2 , and } characters delimiters,
you don't need to escape rest of your delimiters (note that you don't have to escape - only because it is at end of character class [] so he cant be used as range operator).
Try maybe this way
String regexDelimiters = "[\\s,\"():;.!?\\-]+";
String inputString = "\"Word\" some more text.";
String[] arrayOfWords = inputString.replaceAll(
"^" + regexDelimiters,"").split(regexDelimiters);
for (String s : arrayOfWords)
System.out.println("'" + s + "'");
output:
'Word'
'some'
'more'
'text'
A delimiter is interpreted as separating the strings on either side of it, thus the empty string on its left is added to the result as well as the string to its right ("Word"). To prevent this, you should first strip any leading delimiters, as described here:
How to prevent java.lang.String.split() from creating a leading empty string?
So in short form you would have:
String delim = "[\\s,\"():;.!?\\-]+";
String[] arrayOfWords = inputString.replaceFirst("^" + delim, "").split(delim);
Edit: Looking at Pshemo's answer, I realize he is correct regarding your regex. Inside the brackets it's unnecessary to specify the number of space characters, as they will be caught be the + operator.