Related
Say I have this string:
["teads.tv, 15429, reseller, 15a9c44f6d26cbe1 ","video.unrulymedia.com,367782854,reseller","google.com, pub-8173359565788166, direct, f08c47fec0942fa0","google.com, pub-8804303781641925, reseller, f08c47fec0942fa0 "]
I am trying to extract all the text strings like teads.tv, google.com and etc.
Each text string is placed in the following way "text.text,, but there are aslo combinations of ", without any character in between.
I tried this Regex expression:
"(.*?)\,
but I also capture the empty combinations, you can check it out here.
How can I modify the Regex expression, so it would capture only the combination with a string between ",?
Cheers,
If there should be at least a single non whitespace char present other than " , [ ] you can match optional whitespace chars and use a negated character class listing all the characters that should not be matched and repeat that 1 or more times.
"(\s*[^\][\s",]+),
Regex demo
The more broad variation is to repeat 1+ times any char except a comma:
"([^,]+),
Regex demo
How about using + (one or more) instead of * (zero or more) as quantifier:
"(.+?),
Additionally, you may not need to escape , with backslash.
Reading the question as retrieving the string with a dotted notation such as domain names means that we are looking for the first string after a ".
This string will grab strings with dots within them, but avoid the quote characters.
const regEx = /(?:\")([\w\d\.\-]+)/g;
const input = '["teads.tv, 15429, reseller, 15a9c44f6d26cbe1 ","video.unrulymedia.com,367782854,reseller","google.com, pub-8173359565788166, direct, f08c47fec0942fa0","google.com, pub-8804303781641925, reseller, f08c47fec0942fa0 "]';
const regMatch = Array.from(input.matchAll(regEx), m => m[1]);
console.log(regMatch)
I have the following two lines of code:
json_str = _cases.to_json
path += " #{USER} #{PASS} #{json_str}"
When I use the debugger, I noticed that json_str appears to be formatted as JSON:
"[["FMCE","Wiltone","Wiltone","04/10/2018","Marriage + - DOM"]]"
However, when I interpolate it into another string, the quotes are removed:
"node superuser 123456 [["FMCE","Wiltone","Wiltone","04/10/2018","Marriage + - DOM"]]"
Why does string interpolation remove the quotes from JSON string and how can I resolve this?
I did find one solution to the problem, which was manually escaping the string:
json_str = _cases.to_json.gsub('"','\"')
path += " #{USER} #{PASS} \"#{json_str}\""
So basically I escape the double quotes generated in the to_json call. Then I manually add two escaped quotes around the interpolated variable. This will produce a desired result:
node superuser 123456 "[[\"FMCE\",\"Wiltone\",\"Wiltone\",\"04/10/2018\",\"Marriage + - DOM\"]]"
Notice how the outer quotes around the collection are not escaped, but the strings inside the collection are escaped. That will enable JavaScript to parse it with JSON.parse.
It is important to note that in this part:
json_str = _cases.to_json.gsub('"','\"')
it is adding a LITERAL backslash. Not an escape sequence.
But in this part:
path += " #{USER} #{PASS} \"#{json_str}\""
The \" wrapping the interpolated variable is an escape sequence and NOT a literal backslash.
Why do you think the first and last quote marks are part of the string? They do not belong to the JSON format. Your program’s behavior looks correct to me.
(Or more precisely, your program seems to be doing exactly what you told it to. Whether your instructions are any good is a question I can’t answer without more context.)
It's hard to tell with the small sample, but it looks like you might be getting quotes from your debugger output. assuming the output of .to_json is a string (usually is), then "#{json_str}" should be exactly equal to json_str. If it isn't, that's a bug in ruby somehow (doubtful).
If you need the quotes, you need to either add them manually or escape the string using whatever escape function is appropriate for your use case. You could use .to_json as your escape function even ("#{json_str.to_json}", for example).
For example, this regex
(.*)<FooBar>
will match:
abcde<FooBar>
But how do I get it to match across multiple lines?
abcde
fghij<FooBar>
Try this:
((.|\n)*)<FooBar>
It basically says "any character or a newline" repeated zero or more times.
It depends on the language, but there should be a modifier that you can add to the regex pattern. In PHP it is:
/(.*)<FooBar>/s
The s at the end causes the dot to match all characters including newlines.
The question is, can the . pattern match any character? The answer varies from engine to engine. The main difference is whether the pattern is used by a POSIX or non-POSIX regex library.
A special note about lua-patterns: they are not considered regular expressions, but . matches any character there, the same as POSIX-based engines.
Another note on matlab and octave: the . matches any character by default (demo): str = "abcde\n fghij<Foobar>"; expression = '(.*)<Foobar>*'; [tokens,matches] = regexp(str,expression,'tokens','match'); (tokens contain a abcde\n fghij item).
Also, in all of boost's regex grammars the dot matches line breaks by default. Boost's ECMAScript grammar allows you to turn this off with regex_constants::no_mod_m (source).
As for oracle (it is POSIX based), use the n option (demo): select regexp_substr('abcde' || chr(10) ||' fghij<Foobar>', '(.*)<Foobar>', 1, 1, 'n', 1) as results from dual
POSIX-based engines:
A mere . already matches line breaks, so there isn't a need to use any modifiers, see bash (demo).
The tcl (demo), postgresql (demo), r (TRE, base R default engine with no perl=TRUE, for base R with perl=TRUE or for stringr/stringi patterns, use the (?s) inline modifier) (demo) also treat . the same way.
However, most POSIX-based tools process input line by line. Hence, . does not match the line breaks just because they are not in scope. Here are some examples how to override this:
sed - There are multiple workarounds. The most precise, but not very safe, is sed 'H;1h;$!d;x; s/\(.*\)><Foobar>/\1/' (H;1h;$!d;x; slurps the file into memory). If whole lines must be included, sed '/start_pattern/,/end_pattern/d' file (removing from start will end with matched lines included) or sed '/start_pattern/,/end_pattern/{{//!d;};}' file (with matching lines excluded) can be considered.
perl - perl -0pe 's/(.*)<FooBar>/$1/gs' <<< "$str" (-0 slurps the whole file into memory, -p prints the file after applying the script given by -e). Note that using -000pe will slurp the file and activate 'paragraph mode' where Perl uses consecutive newlines (\n\n) as the record separator.
gnu-grep - grep -Poz '(?si)abc\K.*?(?=<Foobar>)' file. Here, z enables file slurping, (?s) enables the DOTALL mode for the . pattern, (?i) enables case insensitive mode, \K omits the text matched so far, *? is a lazy quantifier, (?=<Foobar>) matches the location before <Foobar>.
pcregrep - pcregrep -Mi "(?si)abc\K.*?(?=<Foobar>)" file (M enables file slurping here). Note pcregrep is a good solution for macOS grep users.
See demos.
Non-POSIX-based engines:
php - Use the s modifier PCRE_DOTALL modifier: preg_match('~(.*)<Foobar>~s', $s, $m) (demo)
c# - Use RegexOptions.Singleline flag (demo): - var result = Regex.Match(s, #"(.*)<Foobar>", RegexOptions.Singleline).Groups[1].Value;- var result = Regex.Match(s, #"(?s)(.*)<Foobar>").Groups[1].Value;
powershell - Use the (?s) inline option: $s = "abcde`nfghij<FooBar>"; $s -match "(?s)(.*)<Foobar>"; $matches[1]
perl - Use the s modifier (or (?s) inline version at the start) (demo): /(.*)<FooBar>/s
python - Use the re.DOTALL (or re.S) flags or (?s) inline modifier (demo): m = re.search(r"(.*)<FooBar>", s, flags=re.S) (and then if m:, print(m.group(1)))
java - Use Pattern.DOTALL modifier (or inline (?s) flag) (demo): Pattern.compile("(.*)<FooBar>", Pattern.DOTALL)
kotlin - Use RegexOption.DOT_MATCHES_ALL : "(.*)<FooBar>".toRegex(RegexOption.DOT_MATCHES_ALL)
groovy - Use (?s) in-pattern modifier (demo): regex = /(?s)(.*)<FooBar>/
scala - Use (?s) modifier (demo): "(?s)(.*)<Foobar>".r.findAllIn("abcde\n fghij<Foobar>").matchData foreach { m => println(m.group(1)) }
javascript - Use [^] or workarounds [\d\D] / [\w\W] / [\s\S] (demo): s.match(/([\s\S]*)<FooBar>/)[1]
c++ (std::regex) Use [\s\S] or the JavaScript workarounds (demo): regex rex(R"(([\s\S]*)<FooBar>)");
vba vbscript - Use the same approach as in JavaScript, ([\s\S]*)<Foobar>. (NOTE: The MultiLine property of the RegExp object is sometimes erroneously thought to be the option to allow . match across line breaks, while, in fact, it only changes the ^ and $ behavior to match start/end of lines rather than strings, the same as in JavaScript regex)
behavior.)
ruby - Use the /m MULTILINE modifier (demo): s[/(.*)<Foobar>/m, 1]
rtrebase-r - Base R PCRE regexps - use (?s): regmatches(x, regexec("(?s)(.*)<FooBar>",x, perl=TRUE))[[1]][2] (demo)
ricustringrstringi - in stringr/stringi regex funtions that are powered with the ICU regex engine. Also use (?s): stringr::str_match(x, "(?s)(.*)<FooBar>")[,2] (demo)
go - Use the inline modifier (?s) at the start (demo): re: = regexp.MustCompile(`(?s)(.*)<FooBar>`)
swift - Use dotMatchesLineSeparators or (easier) pass the (?s) inline modifier to the pattern: let rx = "(?s)(.*)<Foobar>"
objective-c - The same as Swift. (?s) works the easiest, but here is how the option can be used: NSRegularExpression* regex = [NSRegularExpression regularExpressionWithPattern:pattern options:NSRegularExpressionDotMatchesLineSeparators error:®exError];
re2, google-apps-script - Use the (?s) modifier (demo): "(?s)(.*)<Foobar>" (in Google Spreadsheets, =REGEXEXTRACT(A2,"(?s)(.*)<Foobar>"))
NOTES ON (?s):
In most non-POSIX engines, the (?s) inline modifier (or embedded flag option) can be used to enforce . to match line breaks.
If placed at the start of the pattern, (?s) changes the bahavior of all . in the pattern. If the (?s) is placed somewhere after the beginning, only those .s will be affected that are located to the right of it unless this is a pattern passed to Python's re. In Python re, regardless of the (?s) location, the whole pattern . is affected. The (?s) effect is stopped using (?-s). A modified group can be used to only affect a specified range of a regex pattern (e.g., Delim1(?s:.*?)\nDelim2.* will make the first .*? match across newlines and the second .* will only match the rest of the line).
POSIX note:
In non-POSIX regex engines, to match any character, [\s\S] / [\d\D] / [\w\W] constructs can be used.
In POSIX, [\s\S] is not matching any character (as in JavaScript or any non-POSIX engine), because regex escape sequences are not supported inside bracket expressions. [\s\S] is parsed as bracket expressions that match a single character, \ or s or S.
If you're using Eclipse search, you can enable the "DOTALL" option to make '.' match any character including line delimiters: just add "(?s)" at the beginning of your search string. Example:
(?s).*<FooBar>
In many regex dialects, /[\S\s]*<Foobar>/ will do just what you want. Source
([\s\S]*)<FooBar>
The dot matches all except newlines (\r\n). So use \s\S, which will match ALL characters.
We can also use
(.*?\n)*?
to match everything including newline without being greedy.
This will make the new line optional
(.*?|\n)*?
In Ruby you can use the 'm' option (multiline):
/YOUR_REGEXP/m
See the Regexp documentation on ruby-doc.org for more information.
"." normally doesn't match line-breaks. Most regex engines allows you to add the S-flag (also called DOTALL and SINGLELINE) to make "." also match newlines.
If that fails, you could do something like [\S\s].
For Eclipse, the following expression worked:
Foo
jadajada Bar"
Regular expression:
Foo[\S\s]{1,10}.*Bar*
Note that (.|\n)* can be less efficient than (for example) [\s\S]* (if your language's regexes support such escapes) and than finding how to specify the modifier that makes . also match newlines. Or you can go with POSIXy alternatives like [[:space:][:^space:]]*.
Use:
/(.*)<FooBar>/s
The s causes dot (.) to match carriage returns.
Use RegexOptions.Singleline. It changes the meaning of . to include newlines.
Regex.Replace(content, searchText, replaceText, RegexOptions.Singleline);
In notepad++ you can use this
<table (.|\r\n)*</table>
It will match the entire table starting from
rows and columns
You can make it greedy, using the following, that way it will match the first, second and so forth tables and not all at once
<table (.|\r\n)*?</table>
In a Java-based regular expression, you can use [\s\S].
This works for me and is the simplest one:
(\X*)<FooBar>
Generally, . doesn't match newlines, so try ((.|\n)*)<foobar>.
In JavaScript you can use [^]* to search for zero to infinite characters, including line breaks.
$("#find_and_replace").click(function() {
var text = $("#textarea").val();
search_term = new RegExp("[^]*<Foobar>", "gi");;
replace_term = "Replacement term";
var new_text = text.replace(search_term, replace_term);
$("#textarea").val(new_text);
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<button id="find_and_replace">Find and replace</button>
<br>
<textarea ID="textarea">abcde
fghij<Foobar></textarea>
Solution:
Use pattern modifier sU will get the desired matching in PHP.
Example:
preg_match('/(.*)/sU', $content, $match);
Sources:
Pattern Modifiers
In the context of use within languages, regular expressions act on strings, not lines. So you should be able to use the regex normally, assuming that the input string has multiple lines.
In this case, the given regex will match the entire string, since "<FooBar>" is present. Depending on the specifics of the regex implementation, the $1 value (obtained from the "(.*)") will either be "fghij" or "abcde\nfghij". As others have said, some implementations allow you to control whether the "." will match the newline, giving you the choice.
Line-based regular expression use is usually for command line things like egrep.
Try: .*\n*.*<FooBar> assuming you are also allowing blank newlines. As you are allowing any character including nothing before <FooBar>.
I had the same problem and solved it in probably not the best way but it works. I replaced all line breaks before I did my real match:
mystring = Regex.Replace(mystring, "\r\n", "")
I am manipulating HTML so line breaks don't really matter to me in this case.
I tried all of the suggestions above with no luck. I am using .NET 3.5 FYI.
I wanted to match a particular if block in Java:
...
...
if(isTrue){
doAction();
}
...
...
}
If I use the regExp
if \(isTrue(.|\n)*}
it included the closing brace for the method block, so I used
if \(!isTrue([^}.]|\n)*}
to exclude the closing brace from the wildcard match.
Often we have to modify a substring with a few keywords spread across lines preceding the substring. Consider an XML element:
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>81</PercentComplete>
</TASK>
Suppose we want to modify the 81, to some other value, say 40. First identify .UID.21..UID., then skip all characters including \n till .PercentCompleted.. The regular expression pattern and the replace specification are:
String hw = new String("<TASK>\n <UID>21</UID>\n <Name>Architectural design</Name>\n <PercentComplete>81</PercentComplete>\n</TASK>");
String pattern = new String ("(<UID>21</UID>)((.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
String replaceSpec = new String ("$1$2$440$6");
// Note that the group (<PercentComplete>) is $4 and the group ((.|\n)*?) is $2.
String iw = hw.replaceFirst(pattern, replaceSpec);
System.out.println(iw);
<TASK>
<UID>21</UID>
<Name>Architectural design</Name>
<PercentComplete>40</PercentComplete>
</TASK>
The subgroup (.|\n) is probably the missing group $3. If we make it non-capturing by (?:.|\n) then the $3 is (<PercentComplete>). So the pattern and replaceSpec can also be:
pattern = new String("(<UID>21</UID>)((?:.|\n)*?)(<PercentComplete>)(\\d+)(</PercentComplete>)");
replaceSpec = new String("$1$2$340$5")
and the replacement works correctly as before.
Typically searching for three consecutive lines in PowerShell, it would look like:
$file = Get-Content file.txt -raw
$pattern = 'lineone\r\nlinetwo\r\nlinethree\r\n' # "Windows" text
$pattern = 'lineone\nlinetwo\nlinethree\n' # "Unix" text
$pattern = 'lineone\r?\nlinetwo\r?\nlinethree\r?\n' # Both
$file -match $pattern
# output
True
Bizarrely, this would be Unix text at the prompt, but Windows text in a file:
$pattern = 'lineone
linetwo
linethree
'
Here's a way to print out the line endings:
'lineone
linetwo
linethree
' -replace "`r",'\r' -replace "`n",'\n'
# Output
lineone\nlinetwo\nlinethree\n
Option 1
One way would be to use the s flag (just like the accepted answer):
/(.*)<FooBar>/s
Demo 1
Option 2
A second way would be to use the m (multiline) flag and any of the following patterns:
/([\s\S]*)<FooBar>/m
or
/([\d\D]*)<FooBar>/m
or
/([\w\W]*)<FooBar>/m
Demo 2
RegEx Circuit
jex.im visualizes regular expressions:
I have tried using something like "struct a {" and "struct a {" to look for the declaration of "a". But it seems opengrok just ignores the curly brackets. Is there a way to search for the phrase "struct a {"?
Grok supports escaping special characters that are part of the query syntax.
The current list of special characters are
+ - && || ! ( ) { } [ ] ^ " ~ * ? : \
To escape these character use the \ before the character.
For example to search for (1+1):2 use the query: \(1\+1\)\:2
You should be able to search with "struct a {" (with quotes)
From OpenGrok documentation:
Escaping special characters:
Opengrok supports escaping special characters that are part of the query syntax. Current special characters are:
+ - && || ! ( ) { } [ ] ^ " ~ * ? : \ /
To escape these character use the \ before the character. For example to search for (1+1):2 use the query: (1+1):2
NOTE on analyzers: Indexed words are made up of Alpha-Numeric and Underscore characters. One letter words are usually not indexed as symbols!
Most other characters (including single and double quotes) are treated as "spaces/whitespace" (so even if you escape them, they will not be found, since most analyzers ignore them).
The exceptions are: # $ % ^ & = ? . : which are mostly indexed as separate words.
Because some of them are part of the query syntax, they must be escaped with a reverse slash as noted above.
So searching for +1 or + 1 will both find +1 and + 1.
Valid FIELDs are
full
Search through all text tokens (words,strings,identifiers,numbers) in
index.
defs
Only finds symbol definitions (where e.g. a variable (function, ...)
is defined).
refs
Only finds symbols (e.g. methods, classes, functions, variables).
path
path of the source file (no need to use dividers, or if, then use "/"
Windows users, "" is an escape key in Lucene query syntax! Please don't use "", or replace it with "/"). Also note that if you want
just exact path, enclose it in "", e.g. "src/mypath", otherwise
dividers will be removed and you get more hits.
hist
History log comments.
type
Type of analyzer used to scope down to certain file types (e.g. just C
sources). Current mappings: [ada=Ada, asm=Asm, bzip2=Bzip(2), c=C,
clojure=Clojure, csharp=C#, cxx=C++, eiffel=Eiffel, elf=ELF,
erlang=Erlang, file=Image file, fortran=Fortran, golang=Golang,
gzip=GZIP, haskell=Haskell, jar=Jar, java=Java, javaclass=Java class,
javascript=JavaScript, json=Json, kotlin=Kotlin, lisp=Lisp, lua=Lua,
mandoc=Mandoc, pascal=Pascal, perl=Perl, php=PHP, plain=Plain Text,
plsql=PL/SQL, powershell=PowerShell script, python=Python, ruby=Ruby,
rust=Rust, scala=Scala, sh=Shell script, sql=SQL, swift=Swift,
tar=Tar, tcl=Tcl, troff=Troff, typescript=TypeScript,
uuencode=UUEncoded, vb=Visual Basic, verilog=Verilog, xml=XML,
zip=Zip] The term (phrases) can be boosted (making it more relevant)
using a caret ^ , e.g. help^4 opengrok - will make term help boosted
Opengrok search is powered by Lucene, for more detail on query syntax refer to Lucene docs.
I've got a two column CSV with a name and a number. Some people's name use commas, for example Joe Blow, CFA. This comma breaks the CSV format, since it's interpreted as a new column.
I've read up and the most common prescription seems to be replacing that character, or replacing the delimiter, with a new value (e.g. this|that|the, other).
I'd really like to keep the comma separator (I know excel supports other delimiters but other interpreters may not). I'd also like to keep the comma in the name, as Joe Blow| CFA looks pretty silly.
Is there a way to include commas in CSV columns without breaking the formatting, for example by escaping them?
To encode a field containing comma (,) or double-quote (") characters, enclose the field in double-quotes:
field1,"field, 2",field3, ...
Literal double-quote characters are typically represented by a pair of double-quotes (""). For example, a field exclusively containing one double-quote character is encoded as """".
For example:
Sheet: |Hello, World!|You "matter" to us.|
CSV: "Hello, World!","You ""matter"" to us."
More examples (sheet → csv):
regular_value → regular_value
Fresh, brown "eggs" → "Fresh, brown ""eggs"""
" → """"
"," → ""","""
,,," → ",,,"""
,"", → ","""","
""" → """"""""
See wikipedia.
I found that some applications like Numbers in Mac ignore the double quote if there is space before it.
a, "b,c" doesn't work while a,"b,c" works.
The problem with the CSV format, is there's not one spec, there are several accepted methods, with no way of distinguishing which should be used (for generate/interpret). I discussed all the methods to escape characters (newlines in that case, but same basic premise) in another post. Basically it comes down to using a CSV generation/escaping process for the intended users, and hoping the rest don't mind.
Reference spec document.
If you want to make that you said, you can use quotes. Something like this
$name = "Joe Blow, CFA.";
$arr[] = "\"".$name."\"";
so now, you can use comma in your name variable.
You need to quote that values.
Here is a more detailed spec.
In addition to the points in other answers: one thing to note if you are using quotes in Excel is the placement of your spaces. If you have a line of code like this:
print '%s, "%s", "%s", "%s"' % (value_1, value_2, value_3, value_4)
Excel will treat the initial quote as a literal quote instead of using it to escape commas. Your code will need to change to
print '%s,"%s","%s","%s"' % (value_1, value_2, value_3, value_4)
It was this subtlety that brought me here.
You can use Template literals (Template strings)
e.g -
`"${item}"`
CSV files can actually be formatted using different delimiters, comma is just the default.
You can use the sep flag to specify the delimiter you want for your CSV file.
Just add the line sep=; as the very first line in your CSV file, that is if you want your delimiter to be semi-colon. You can change it to any other character.
This isn't a perfect solution, but you can just replace all uses of commas with ‚ or a lower quote. It looks very very similar to a comma and will visually serve the same purpose. No quotes are required
in JS this would be
stringVal.replaceAll(',', '‚')
You will need to be super careful of cases where you need to directly compare that data though
Depending on your language, there may be a to_json method available. That will escape many things that break CSVs.
I faced the same problem and quoting the , did not help. Eventually, I replaced the , with +, finished the processing, saved the output into an outfile and replaced the + with ,. This may seem ugly but it worked for me.
May not be what is needed here but it's a very old question and the answer may help others. A tip I find useful with importing into Excel with a different separator is to open the file in a text editor and add a first line like:
sep=|
where | is the separator you wish Excel to use.
Alternatively you can change the default separator in Windows but a bit long-winded:
Control Panel>Clock & region>Region>Formats>Additional>Numbers>List separator [change from comma to your preferred alternative]. That means Excel will also default to exporting CSVs using the chosen separator.
You could encode your values, for example in PHP base64_encode($str) / base64_decode($str)
IMO this is simpler than doubling up quotes, etc.
https://www.php.net/manual/en/function.base64-encode.php
The encoded values will never contain a comma so every comma in your CSV will be a separator.
You can use the Text_Qualifier field in your Flat file connection manager to as ". This should wrap your data in quotes and only separate by commas which are outside the quotes.
First, if item value has double quote character ("), replace with 2 double quote character ("")
item = item.ToString().Replace("""", """""")
Finally, wrap item value:
ON LEFT: With double quote character (")
ON RIGHT: With double quote character (") and comma character (,)
csv += """" & item.ToString() & ""","
Double quotes not worked for me, it worked for me \". If you want to place a double quotes as example you can set \"\".
You can build formulas, as example:
fprintf(strout, "\"=if(C3=1,\"\"\"\",B3)\"\n");
will write in csv:
=IF(C3=1,"",B3)
A C# method for escaping delimiter characters and quotes in column text. It should be all you need to ensure your csv is not mangled.
private string EscapeDelimiter(string field)
{
if (field.Contains(yourEscapeCharacter))
{
field = field.Replace("\"", "\"\"");
field = $"\"{field}\"";
}
return field;
}