C# how to split a string on the basis of <> character - c#

I want to split my text by <> characters.
Example suppose I have a string
string Name="this <link> is my <name>";
Now I want to split this so that I have a array of string like
ar[0]="this "
ar[1]="<link>"
ar[2]=" is my "
ar[3]="<name>"
I was trying with split function like
string[] ar=Name.Split('<');
I have also tried
string[] nameArray = Regex.Split(name, "<[^<]+>");
But this is not giving me
"<link>"
and "<name>"
But it is not a good approach.
Can I use regular expression here.

This
Regex r = new Regex(#"(?<=.)(?=<)|(?<=>)(?=.)");
foreach (var s in r.Split("this_<link>_is_my_<name>"))
{
Console.WriteLine(s);
}
gives
this_
<link>
_is_my_
<name>
(underscores used for clarity)
The regex splits on a zero-width point (so it doesn't remove anything) which is either:
preceeded by something and followed by <
preceeded by > and followed by something
The "something" checks are necessary to avoid empty strings at the start or end if your string starts or ends with something in brackets.
Note something like "<link<link>>" will give you { "<link", "<link>", ">" } so try to make your angle brackets balance.
If you want empty strings if the string starts with < or ends with > you can use (?=<)|(?<=>). If you want empty strings in the middle when you encounter ><, I think you need to first split on (?=<) and then split all the results on (?<=>) - I don't think you can do it in one go.

Related

C# Regex split() without removing the split condition character

I am splitting a string with regex using its Split() method.
var splitRegex = new Regex(#"[\s|{]");
string input = "/Tests/ShowMessage { 'Text': 'foo' }";
//second version of the input:
//string input = "/Tests/ShowMessage{ 'Text': 'foo' }";
string[] splittedText = splitRegex.Split(input, 2);
The string is just a sample pattern of the input. There are two different structures of input, once with a space before the { or without the space. I want to split the input on the { bracket in order to get the following result:
/Tests/ShowMessage
{ 'Text': 'foo' }
If there is a space, the string gets splitted there (space gets removed) and i get my desired result. But if there isnt a space i split the string on the {, so the { gets removed, what i dont want though. How can i use Regex.Split() without removing the split condition character?
The square brackets create a character set, so you want it to match exactly one of those inner characters. For your desire start off by removing them.
So to match it a random count of whitespaces you have to add *, the result is this one\s*.
\s is a whitespace
* means zero-or-more
That you don't remove the split condition character, you can use lookahead assertion (?=...).
(?=...) or (?!...) is a lookahead assertion
The combined Regex looks like this: \s*(?={)
This is a really good and detailed documentation of all the different Regex parts, you might have a look at it. Furthermore you can test your Regex easy and for free here.
In order to not include the curly brace in the match you can put it into a look ahead
\s*(?={)
That will match any number of white spaces up to the position before a open curly brace.
You can use regular string split, on "{" and trim the spaces off:
var bits = "/Tests/ShowMessage { 'Text': 'foo' }".Split("{", StringSplitOptions.RemoveEmptyEntries);
bits[0] = bits[0].TrimEnd();
bits[1] = "{" + bits[1];
If you want to use the RegEx route, you can add the { back if you change the regex a bit:
var splitRegex = new Regex(#"\s*{");
string input = "/Tests/ShowMessage { 'Text': 'foo' }";
//second version of the input:
//string input = "/Tests/ShowMessage{ 'Text': 'foo' }";
string[] splittedText = splitRegex.Split(input, 2);
splittedText[1] = "{" + splittedText[1];
It means "split at occurrence of (zero or more whitespace followed by {)" - so the split operation nukes your spaces (you want), and your { (you don't want) but you can put the { back with certainty that it will mean you get what you want
var splitedList = srt.Text.Replace(".", ".#").Replace("?", "?#").Replace("!", "!#").Split(new[] { "#"}, StringSplitOptions.RemoveEmptyEntries).ToList();
This will split text for .!? and will not remove condition chars. For better result just replace # with some uniq char. Like this one for example '®' That is all. Simple as it is. No regex.split which is slow and difficult due to many different task criterias, etc...
passing-> "Hello. I'am dev!"
result (split condition character exist )
"Hello."
"I'am dev!"

Split a string based on spaces (unless in quotes) and remove the quotes afterwards [duplicate]

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.

Html.Decoded ­ is problematic in string functions

string local= HttpUtility.HtmlDecode(GetLocalizedSupportPhone()).Replace("-", "").Replace(" ", "");
I am getting a string :
"0­12­4 41­481­73"
from the GetLocalizedSupportPhone() method. The Html Decode method returns:
"0-12-4 41-481-73"
I have a list of phone numbers like:- "01244148173", "01244148173", etc which are plain integers without any space character or html character.
Problem scenario:- All i want to do is to get decoded local string ("0-12-4 41-481-73"), replace the ­ as well as " " with empty string character and compare the resultant local string with the list items. If a similar list item exists, then remove that particular list item.
But strangely, the .Replace() method replaces space character with blank string but is unable to replace "-" with empty string.
I am just curious why is it happening? Why ANY OF THE STRING METHODS (like I tried with .split() ) can not detect "-"?
There are different types of hyphens. ­ is a soft hyphen. Specifically the soft hyphen is 173 and the hyphen on your keyboard is 45.
Try this instead.
var r = HttpUtility.HtmlDecode("0­12­4 41­481­73")
.Replace((char)173, ' ')
.Replace(" ", "");
That will replace the soft hyphen with a space and then your second replace will get rid of that.
Another option would be to use a regular expression to remove all non-numeric values.
Regex nonNumeric = new Regex(#"\D");
var r = nonNumeric.Replace(
HttpUtility.HtmlDecode("0­12­4 41­481­73"),
string.Empty);
This might help if you're just looking to strip spaces and soft hypens from a string without having to deal with HTML decoding:
var regex = new Regex(#"\u00ad| ");
var result = regex.Replace(stringWithSoftHyphens, string.Empty);
I tried doing this with Trim((char)173) but it (and methods like Split) do not seem to be able to handle the soft hyphen character like the Regex class can.

Regex Replacing only whole matches

I am trying to replace a bunch of strings in files. The strings are stored in a datatable along with the new string value.
string contents = File.ReadAllText(file);
foreach (DataRow dr in FolderRenames.Rows)
{
contents = Regex.Replace(contents, dr["find"].ToString(), dr["replace"].ToString());
File.SetAttributes(file, FileAttributes.Normal);
File.WriteAllText(file, contents);
}
The strings look like this _-uUa, -_uU, _-Ha etc.
The problem that I am having is when for example this string "_uU" will also overwrite "_-uUa" so the replacement would look like "newvaluea"
Is there a way to tell regex to look at the next character after the found string and make sure it is not an alphanumeric character?
I hope it is clear what I am trying to do here.
Here is some sample data:
private function _-0iX(arg1:flash.events.Event):void
{
if (arg1.type == flash.events.Event.RESIZE)
{
if (this._-2GU)
{
this._-yu(this._-2GU);
}
}
return;
}
The next characters could be ;, (, ), dot, comma, space, :, etc.
First of all, you should use Regex.Escape.
You can use then
contents = Regex.Replace(
contents,
Regex.Escape(dr["find"].ToString()) + #"(?![a-zA-Z])",
Regex.Escape(dr["replace"].ToString()));
or even better
contents = Regex.Replace(
contents,
#"\b" + Regex.Escape(dr["find"].ToString()) + #"\b",
Regex.Escape(dr["replace"].ToString()));
I think this is what you're looking for:
contents = Regex.Replace(
contents,
string.Format(#"(?<!\w){0}(?!\w)", Regex.Escape(dr["find"].ToString())),
dr["replace"].ToString().Replace("$", "$$")
);
You can't use \b because your search strings don't always start and end with word characters. Instead, I used (?<!\w) and (?!\w) to make sure the matched substring is not immediately preceded or followed by a word character (i.e., a letter, a digit, or an underscore). I don't know the complete specs for your search strings, so this pattern might need some tweaking.
None of the sample patterns you provided contain regex metacharacters, but like the other responders, I used Regex.Escape() to render it safe anyway. In the replacement string the only character you have to watch out for is the dollar sign (ref), and the way to escape that is with another dollar sign. Notice that I used String.Replace() for that instead of Regex.Replace().
There are two tricks that can help you here:
Order all the search string by length, and replace the longest ones first, that way you won't accidentally replace the shorter ones.
Use a MatchEvaluator and instead of looping through all your rows, search fro all replacement patterns in the string and look them up in your dataset.
Option one is simple, option two would look like this:
Regex.Replace(contents", "_-\\w+", ReplaceIdentifier)
public string ReplaceIdentifier(Match m)
{
DataRow row = FolderRenames.Rows.FindRow("find"); // Requires a primary key on "find"
if (row != null) return row["replace"];
else return m.Value;
}

Regex not working in .NET

So I'm trying to match up a regex and I'm fairly new at this. I used a validator and it works when I paste the code but not when it's placed in the codebehind of a .NET2.0 C# page.
The offending code is supposed to be able to split on a single semi-colon but not on a double semi-colon. However, when I used the string
"entry;entry2;entry3;entry4;"
I get a nonsense array that contains empty values, the last letter of the previous entry, and the semi-colons themselves. The online javascript validator splits it correctly. Please help!
My regex:
((;;|[^;])+)
Split on the following regular expression:
(?<!;);(?!;)
It means match semicolons that are neither preceded nor succeeded by another semicolon.
For example, this code
var input = "entry;entry2;entry3;entry4;";
foreach (var s in Regex.Split(input, #"(?<!;);(?!;)"))
Console.WriteLine("[{0}]", s);
produces the following output:
[entry]
[entry2]
[entry3]
[entry4]
[]
The final empty field is a result of the semicolon on the end of the input.
If the semicolon is a terminator at the end of each field rather than a separator between consecutive fields, then use Regex.Matches instead
foreach (Match m in Regex.Matches(input, #"(.+?)(?<!;);(?!;)"))
Console.WriteLine("[{0}]", m.Groups[1].Value);
to get
[entry]
[entry2]
[entry3]
[entry4]
Why not use String.Split on the semicolon?
string sInput = "Entry1;entry2;entry3;entry4";
string[] sEntries = sInput.Split(';');
// Do what you have to do with the entries in the array...
Hope this helps,
Best regards,
Tom.
As tommieb75 wrote, you can use String.Split with StringSplitOptions Enumeration so you can control your output of newly created splitting array
string input = "entry1;;entry2;;;entry3;entry4;;";
char[] charSeparators = new char[] {';'};
// Split a string delimited by characters and return all non-empty elements.
result = input.Split(charSeparators, StringSplitOptions.RemoveEmptyEntries);
The result would contain only 4 elements like this:
<entry1><entry2><entry3><entry4>

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