Related
I have the following string:
string testString = ",,,The,,,boy,,,kicked,,,the,,ball";
I want to remove the unwanted commas and have the sentence as this (simply printed to the console):
The boy kicked the ball
I tried the below code:
string testString = ",,,The,,,boy,,,kicked,,,the,,ball";
string manipulatedString = testString.Replace(",", " "); //line 2
Console.WriteLine(manipulatedString.Trim() + "\n");
string result = Regex.Replace(manipulatedString, " ", " ");
Console.WriteLine(result.TrimStart());
However, I end up with a result with double whitespaces as so:
The boy kicked the ball
It kind of makes sense why I am getting such an anomalous output because in line 2 I am saying that for every comma (,) character, replace that with whitespace and it will do that for every occurrence.
What's the best way to solve this?
This is a simple solution using Split and Join
string testString = ",,,The,,,boy,,,kicked,,,the,,ball";
var splitted = testString.Split(new char[] {','}, StringSplitOptions.RemoveEmptyEntries);
string result = string.Join(" ", splitted);
Console.WriteLine(result);
You could use regex to replace the pattern ,+ (one or more occurrences of a comma) by a whitespace.
var replacedString = Regex.Replace(testString, ",+", " ").Trim();
Added Trim to remove white spaces at beginning/end as I assume you want to remove them.
I'm trying to parse a a bunch of file with Replace method(string) while is doing what I expect: I feels is not practical. for instance I will process 10K files but in the First 72 I found like 30 values that need to be replace And this is the rule :
My Goal :"
My goal is to replace all Instance of the ':' Dont follows this Rules :
1- the 2nd or 3rd Character foward is Not Another ':'
2-the 3rd or 2nd Chacarcter backward is Not Another ':'
All other should be Replaced
1- Any time that I found this character (:) and this character is not preceded by two char or three characters like :00: or :12A: I should replace it with an (*).
This is the method that I have so far.....
private static string cleanMesage(string str)
{
string result = String.Empty;
try
{
result = str.Replace("BNF:", "BNF*").Replace("B/O:", "B/O*").Replace("O/B:", "O/B*");
result = result.Replace("Epsas:", "Epsas*").Replace("2017:", "2017*").Replace("BANK:", "BANK*");
result = result.Replace("CDT:", "CDT*").Replace("ENT:", "").Replace("GB22:", "GB22*");
result = result.Replace("A / C:", "A/C*").Replace("ORD:", "ORD*").Replace("A/C:", "A/C*");
result = result.Replace("REF:", "REF*").Replace("ISIN:", "ISIN*").Replace("PAY:", "PAY*");
result = result.Replace("DEPOSITO:", "DEPOSITO*").Replace("WITH:", "WITH*");
result = result.Replace("Operaciones:", "Operaciones*").Replace("INST:", "INST*");
result = result.Replace("DETAIL:", "DETAIL*").Replace("WITH:", "WITH*").Replace("BO:", "BO*");
result = result.Replace("CUST:", "CUST*").Replace("ISIN:", "ISIN*").Replace("SEDL:", "SEDL*");
result = result.Replace("Enero:", "Enero*").Replace("enero:", "Enero*");
result = result.Replace("agosto:", "agosto*").Replace("febrero:", "febrero*");
result = result.Replace("marzo:", "marzo*").Replace("abril:", "abril*");
result = result.Replace("mayo:", "mayo*").Replace("junio:", "junio*").Replace("RE:", "RE:*");
result = result.Replace("julio:", "julio*").Replace("septiembre:", "septiembre*");
result = result.Replace("NIF:", "NIF*").Replace("INST:", "INST*").Replace("SHS:", "SHS*")
.Replace("SK:", "");
result = result.Replace("PARTY:", "PARTY*").Replace("SEDOL:", "SEDOL*").Replace("PD:", "PD*");
}
catch (Exception e)
{
}
return result;
}
And this is some sample data :"
:13: <-- keep /ISIN/XS SVUNSK UXPORTKRUDIT ZX PZY DZTU:<- replace UX DZ
TU:<- replace02ZUG12 RZTU:<- replace W/H TZX RZTU:<- replace0.00000 SHZRUS PZID:<- replace
0.000000 IDDSIN:<- replace
:31: <-- keep 1201000100CD05302,24NSUC20523531001//00520023531014
:13: <-- keep /ISIN/XS0153242003 SVUNSK UXPORTKRUDIT ZX PZY DZTU:<- replace00ZUG12 UX DZ
TU:02ZUG12 RZTU:0.30241 W/H TZX RZTU:<- replace0.00000 SHZRUS PZID:<- replace
0.000000 ISIN:XS0153242003
:31: <-- keep 1201000100DD121253,25S202IMSSMSZUX534C//S0322211DF4301
S F/O 0150001400
:13: <-- keep XNF:<- replace this
If your goal is to replace all instances of the ':' character where it is not followed by 2 or 3 other characters. You could indeed try the System.Text.RegularExpressions library. You could then simplify your cleanMessage function in the following way.
using System.Text.RegularExpressions;
function string cleanMessage(string str)
{
string pattern = ":(\s)"; //This will be a ':' followed by a space
Regex rgx = new Regex(pattern);
string replaceResult = rgx.Replace(str,"*$1") //this will replace the pattern with a '*' followed by a space.
return replaceResult;
}
If your goal is to replace all instances of the ':' character where it is not followed by 2 or 3 other characters and the 2nd or 3rd character forward or backward is not another ':'. You could change your cleanMessage to the following instead.
using System.Text.RegularExpressions;
function string cleanMessage(string str)
{
string pattern = "([^;]{2}.):(\s[^:]{2})";
//This will be 2 characters that cannot be ':' followed by anything then a ':' followed by a space and 2 more characters that cannot by ':'
//For instance, "BNF: :F" would FAIL and not get replaced but "BNF: HH" would pass and become "BNF* HH"
Regex rgx = new Regex(pattern);
string replaceResult = rgx.Replace(str,"$1*$2") //this will replace the : with a *
return replaceResult;
}
More information on the System.Text.RegularExpressions library replace can be found at
https://msdn.microsoft.com/en-us/library/xwewhkd1(v=vs.110).aspx
As #dymanoid mentioned, regular expressions are a way to handle this. By using the following you'd get what you want:
result = Regex.Replace(str, "([a-zA-Z0-9]{2,3})\:", "$1*");
However for large datasets this won't perform well. In that case I'd look at walking through str character by character using a for-loop. If the current character is not a colon, add it to the result string and to a temporary string. When the current character is a colon (:) and the temporary string has a length of 2 or 3, write an asterisk to the result and clear the temporary string.
In this case you don't do any string replacement, you just select what to write to a new string.
See here for a speed comparison between string replacement and regex replacement.
I am trying to remove some text and keep only small text from the string.
Actually I am very new to regex, I have read an article and did not get it very well.
Here is an example of my text (every line in separate string object)
2015-03-08 10:30:00 /user841/column-width
2015-03-08 10:30:01 /user849/connect
2015-03-08 10:30:01 /user262/open-level2-price/some other text
2015-03-08 10:30:01 /user839/open-detailed-quotes
I want to process them using regex in c# and have the following output:
column-width
connect
open-level2-price/some other text
open-detailed-quotes
I have used the following line to do that but it throws an exception:
Match match = Regex.Match(line, #"*./user\d+/*.");
The Exception:
System.ArgumentException: 'parsing "*./user\d+/*." - Quantifier {x,y} following nothing.'
could anyone help please!
The error you get is caused by the fact that you try to quantify the start of the pattern, which is considered an error in a .NET regex. Perhaps, you meant to use .* instead of the *. (to match any 0+ chars greedily, as many as possible), but it is certainly not what you need judging by the expected results.
You need
/user\d+/(.*)
See the regex demo
Details:
/user - a literal substring /user
\d+ - 1 or more digits (use RegexOptions.ECMAScript option to only match ASCII digits with \d in a .NET regex)
/ - a literal /
(.*) - A capturing group #1 that matches any 0+ chars other than a newline (replace * with + to match at least 1 char).
C#:
var results = Regex.Matches(s, #"/user\d+/(.*)")
.Cast<Match>()
.Select(m => m.Groups[1].Value)
.ToList();
Instead of using Regex, just split on the '/' character and use the last index of the array (using LINQ):
string inputString = "2015-03-08 10:30:01 /user262/open-level2-price";
inputString.Split('/').Last();
Split returns an array of strings, in your case with the sample input above the string array would look like:
array[0] = "2015-03-08 10:30:01 "
array[1] = "user262"
array[2] = "open-level2-price"
You indicate you always want the last part so just use LINQ to take the .Last() index of the array.
Fiddle here
Here's a simple example of how to use the Regex.Replace static method.
https://dotnetfiddle.net/JuUF9E
using System;
using System.Text.RegularExpressions;
public class Program
{
public static void Main()
{
string[] lines = new string[] {
"2015-03-08 10:30:00 /user841/column-width",
"2015-03-08 10:30:01 /user849/connect",
"2015-03-08 10:30:01 /user262/open-level2-price",
"2015-03-08 10:30:01 /user839/open-detailed-quotes"
};
string pattern = #"(.*/.*/)(.*)";
string replacement = "$2";
foreach(var line in lines)
{
Console.WriteLine(Regex.Replace(line, pattern, replacement));
}
}
}
I don't know why you're trying to do this simple thing with regex, you just have to read the lines and split by the '\', them select the last index and that's it. For example, if you have that data in a file you can use something like this:
string newString = "";
StreamReader sr = new StreamReader('log.txt');
while(!sr.ReadLine)
{
string[] splitted = sr.ReadLine().Split('/');
if(splitted.Length > 0)
newString += splitted[splitted.Length - 1];
}
sr.Close();
At the end, the newString variable will contains what you want. Otherwise you can add every line in a list if you will do some with the data.
How about using Look around
var line = "2015-03-08 10:30:01 /user839/open-detailed otes/dsada/dsa/das/dsadsa";
// dsadsa
var match = Regex.Match(line, #"(?!.*/).*").Value;
In C# what's the best way to remove blank lines i.e., lines that contain only whitespace from a string? I'm happy to use a Regex if that's the best solution.
EDIT: I should add I'm using .NET 2.0.
Bounty update: I'll roll this back after the bounty is awarded, but I wanted to clarify a few things.
First, any Perl 5 compat regex will work. This is not limited to .NET developers. The title and tags have been edited to reflect this.
Second, while I gave a quick example in the bounty details, it isn't the only test you must satisfy. Your solution must remove all lines which consist of nothing but whitespace, as well as the last newline. If there is a string which, after running through your regex, ends with "/r/n" or any whitespace characters, it fails.
If you want to remove lines containing any whitespace (tabs, spaces), try:
string fix = Regex.Replace(original, #"^\s*$\n", string.Empty, RegexOptions.Multiline);
Edit (for #Will): The simplest solution to trim trailing newlines would be to use TrimEnd on the resulting string, e.g.:
string fix =
Regex.Replace(original, #"^\s*$\n", string.Empty, RegexOptions.Multiline)
.TrimEnd();
string outputString;
using (StringReader reader = new StringReader(originalString)
using (StringWriter writer = new StringWriter())
{
string line;
while((line = reader.ReadLine()) != null)
{
if (line.Trim().Length > 0)
writer.WriteLine(line);
}
outputString = writer.ToString();
}
off the top of my head...
string fixed = Regex.Replace(input, "\s*(\n)","$1");
turns this:
fdasdf
asdf
[tabs]
[spaces]
asdf
into this:
fdasdf
asdf
asdf
Using LINQ:
var result = string.Join("\r\n",
multilineString.Split(new string[] { "\r\n" }, ...None)
.Where(s => !string.IsNullOrWhitespace(s)));
If you're dealing with large inputs and/or inconsistent line endings you should use a StringReader and do the above old-school with a foreach loop instead.
Alright this answer is in accordance to the clarified requirements specified in the bounty:
I also need to remove any trailing newlines, and my Regex-fu is
failing. My bounty goes to anyone who can give me a regex which passes
this test: StripWhitespace("test\r\n \r\nthis\r\n\r\n") ==
"test\r\nthis"
So Here's the answer:
(?<=\r?\n)(\s*$\r?\n)+|(?<=\r?\n)(\r?\n)+|(\r?\n)+\z
Or in the C# code provided by #Chris Schmich:
string fix = Regex.Replace("test\r\n \r\nthis\r\n\r\n", #"(?<=\r?\n)(\s*$\r?\n)+|(?<=\r?\n)(\r?\n)+|(\r?\n)+\z", string.Empty, RegexOptions.Multiline);
Now let's try to understand it. There are three optional patterns in here which I am willing to replace with string.empty.
(?<=\r?\n)(\s*$\r?\n)+ - matches one to unlimited lines containing only white space and preceeded by a line break (but does not match the first preceeding line breaks).
(?<=\r?\n)(\r?\n)+ - matches one to unlimited empty lines with no content that are preceeded by a line break (but does not match the first preceeding line breaks).
(\r?\n)+\z - matches one to unlimited line breaks at the end of the tested string (trailing line breaks as you called them)
That satisfies your test perfectly! But also satisfies both \r\n and \n line break styles! Test it out! I believe this will be the most correct answer, although simpler expression would pass your specified bounty test, this regex passes more complex conditions.
EDIT: #Will pointed out a potential flaw in the last pattern match of the above regex in that it won't match multiple line breaks containing white space at the end of the test string. So let's change that last pattern to this:
\b\s+\z The \b is a word boundry (beginning or END of a word), the \s+ is one or more white space characters, the \z is the end of the test string (end of "file"). So now it will match any assortment of whitespace at the end of the file including tabs and spaces in addition to carriage returns and line breaks. I tested both of #Will's provided test cases.
So all together now, it should be:
(?<=\r?\n)(\s*$\r?\n)+|(?<=\r?\n)(\r?\n)+|\b\s+\z
EDIT #2: Alright there is one more possible case #Wil found that the last regex doesn't cover. That case is inputs that have line breaks at the beginning of the file before any content. So lets add one more pattern to match the beginning of the file.
\A\s+ - The \A match the beginning of the file, the \s+ match one or more white space characters.
So now we've got:
\A\s+|(?<=\r?\n)(\s*$\r?\n)+|(?<=\r?\n)(\r?\n)+|\b\s+\z
So now we have four patterns for matching:
whitespace at the beginning of the file,
redundant line breaks containing white space, (ex: \r\n \r\n\t\r\n)
redundant line breaks with no content, (ex: \r\n\r\n)
whitespace at the end of the file
not good. I would use this one using JSON.net:
var o = JsonConvert.DeserializeObject(prettyJson);
new minifiedJson = JsonConvert.SerializeObject(o, Formatting.None);
In response to Will's bounty, which expects a solution that takes "test\r\n \r\nthis\r\n\r\n" and outputs "test\r\nthis", I've come up with a solution that makes use of atomic grouping (aka Nonbacktracking Subexpressions on MSDN). I recommend reading those articles for a better understanding of what's happening. Ultimately the atomic group helped match the trailing newline characters that were otherwise left behind.
Use RegexOptions.Multiline with this pattern:
^\s+(?!\B)|\s*(?>[\r\n]+)$
Here is an example with some test cases, including some I gathered from Will's comments on other posts, as well as my own.
string[] inputs =
{
"one\r\n \r\ntwo\r\n\t\r\n \r\n",
"test\r\n \r\nthis\r\n\r\n",
"\r\n\r\ntest!",
"\r\ntest\r\n ! test",
"\r\ntest \r\n ! "
};
string[] outputs =
{
"one\r\ntwo",
"test\r\nthis",
"test!",
"test\r\n ! test",
"test \r\n ! "
};
string pattern = #"^\s+(?!\B)|\s*(?>[\r\n]+)$";
for (int i = 0; i < inputs.Length; i++)
{
string result = Regex.Replace(inputs[i], pattern, "",
RegexOptions.Multiline);
Console.WriteLine(result == outputs[i]);
}
EDIT: To address the issue of the pattern failing to clean up text with a mix of whitespace and newlines, I added \s* to the last alternation portion of the regex. My previous pattern was redundant and I realized \s* would handle both cases.
string corrected =
System.Text.RegularExpressions.Regex.Replace(input, #"\n+", "\n");
I'll go with:
public static string RemoveEmptyLines(string value) {
using (StringReader reader = new StringReader(yourstring)) {
StringBuilder builder = new StringBuilder();
string line;
while ((line = reader.ReadLine()) != null) {
if (line.Trim().Length > 0)
builder.AppendLine(line);
}
return builder.ToString();
}
}
Here's another option: use the StringReader class. Advantages: one pass over the string, creates no intermediate arrays.
public static string RemoveEmptyLines(this string text) {
var builder = new StringBuilder();
using (var reader = new StringReader(text)) {
while (reader.Peek() != -1) {
string line = reader.ReadLine();
if (!string.IsNullOrWhiteSpace(line))
builder.AppendLine(line);
}
}
return builder.ToString();
}
Note: the IsNullOrWhiteSpace method is new in .NET 4.0. If you don't have that, it's trivial to write on your own:
public static bool IsNullOrWhiteSpace(string text) {
return string.IsNullOrEmpty(text) || text.Trim().Length < 1;
}
In response to Will's bounty here is a Perl sub that gives correct response to the test case:
sub StripWhitespace {
my $str = shift;
print "'",$str,"'\n";
$str =~ s/(?:\R+\s+(\R)+)|(?:()\R+)$/$1/g;
print "'",$str,"'\n";
return $str;
}
StripWhitespace("test\r\n \r\nthis\r\n\r\n");
output:
'test
this
'
'test
this'
In order to not use \R, replace it with [\r\n] and inverse the alternative. This one produces the same result:
$str =~ s/(?:(\S)[\r\n]+)|(?:[\r\n]+\s+([\r\n])+)/$1/g;
There're no needs for special configuration neither multi line support. Nevertheless you can add s flag if it's mandatory.
$str =~ s/(?:(\S)[\r\n]+)|(?:[\r\n]+\s+([\r\n])+)/$1/sg;
if its only White spaces why don't you use the C# string method
string yourstring = "A O P V 1.5";
yourstring.Replace(" ", string.empty);
result will be "AOPV1.5"
char[] delimiters = new char[] { '\r', '\n' };
string[] lines = value.Split(delimiters, StringSplitOptions.RemoveEmptyEntries);
string result = string.Join(Environment.NewLine, lines)
Here is something simple if working against each individual line...
(^\s+|\s+|^)$
Eh. Well, after all that, I couldn't find one that would hit all the corner cases I could figure out. The following is my latest incantation of a regex that strips
All empty lines from the start of a string
Not including any spaces at the beginning of the first non-whitespace line
All empty lines after the first non-whitespace line and before the last non-whitespace line
Again, preserving all whitespace at the beginning of any non-whitespace line
All empty lines after the last non-whitespace line, including the last newline
(?<=(\r\n)|^)\s*\r\n|\r\n\s*$
which essentially says:
Immediately after
The beginning of the string OR
The end of the last line
Match as much contiguous whitespace as possible that ends in a newline*
OR
Match a newline and as much contiguous whitespace as possible that ends at the end of the string
The first half catches all whitespace at the start of the string until the first non-whitespace line, or all whitespace between non-whitespace lines. The second half snags the remaining whitespace in the string, including the last non-whitespace line's newline.
Thanks to all who tried to help out; your answers helped me think through everything I needed to consider when matching.
*(This regex considers a newline to be \r\n, and so will have to be adjusted depending on the source of the string. No options need to be set in order to run the match.)
String Extension
public static string UnPrettyJson(this string s)
{
try
{
// var jsonObj = Json.Decode(s);
// var sObject = Json.Encode(value); dont work well with array of strings c:['a','b','c']
object jsonObj = JsonConvert.DeserializeObject(s);
return JsonConvert.SerializeObject(jsonObj, Formatting.None);
}
catch (Exception e)
{
throw new Exception(
s + " Is Not a valid JSON ! (please validate it in http://www.jsoneditoronline.org )", e);
}
}
Im not sure is it efficient but =)
List<string> strList = myString.Split(new string[] { "\n" }, StringSplitOptions.None).ToList<string>();
myString = string.Join("\n", strList.Where(s => !string.IsNullOrWhiteSpace(s)).Distinct().ToList());
Try this.
string s = "Test1" + Environment.NewLine + Environment.NewLine + "Test 2";
Console.WriteLine(s);
string result = s.Replace(Environment.NewLine, String.Empty);
Console.WriteLine(result);
s = Regex.Replace(s, #"^[^\n\S]*\n", "");
[^\n\S] matches any character that's not a linefeed or a non-whitespace character--so, any whitespace character except \n. But most likely the only characters you have to worry about are space, tab and carriage return, so this should work too:
s = Regex.Replace(s, #"^[ \t\r]*\n", "");
And if you want it to catch the last line, without a final linefeed:
s = Regex.Replace(s, #"^[ \t\r]*\n?", "");
I need to parse a string so the result should output like that:
"abc,def,ghi,klm,nop"
But the string I am receiving could looks more like that:
",,,abc,,def,ghi,,,,,,,,,klm,,,nop"
The point is, I don't know in advance how many commas separates the words.
Is there a regex I could use in C# that could help me resolve this problem?
You can use the ,{2,} expression to match any occurrences of 2 or more commas, and then replace them with a single comma.
You'll probably need a Trim call in there too, to remove any leading or trailing commas left over from the Regex.Replace call. (It's possible that there's some way to do this with just a regex replace, but nothing springs immediately to mind.)
string goodString = Regex.Replace(badString, ",{2,}", ",").Trim(',');
Search for ,,+ and replace all with ,.
So in C# that could look like
resultString = Regex.Replace(subjectString, ",,+", ",");
,,+ means "match all occurrences of two commas or more", so single commas won't be touched. This can also be written as ,{2,}.
a simple solution without regular expressions :
string items = inputString.Split(new[] { ',' }, StringSplitOptions.RemoveEmptyEntries);
string result = String.Join(",", items);
Actually, you can do it without any Trim calls.
text = Regex.Replace(text, "^,+|,+$|(?<=,),+", "");
should do the trick.
The idea behind the regex is to only match that, which we want to remove. The first part matches any string of consecutive commas at the start of the input string, the second matches any consecutive string of commas at the end, while the last matches any consecutive string of commas that follows a comma.
Here is my effort:
//Below is the test string
string test = "YK 002 10 23 30 5 TDP_XYZ "
private static string return_with_comma(string line)
{
line = line.TrimEnd();
line = line.Replace(" ", ",");
line = Regex.Replace(line, ",,+", ",");
string[] array;
array = line.Split(',');
for (int x = 0; x < array.Length; x++)
{
line += array[x].Trim();
}
line += "\r\n";
return line;
}
string result = return_with_comma(test);
//Output is
//YK,002,10,23,30,5,TDP_XYZ