I am trying to work on regular expressions. I have a mainframe file which has several fields. I have a flat file parser which distinguishes several types of records based on the first three letters of every line. How do I write a regular expression where the first three letters are 'CTR'.
Beginning of line or beginning of string?
Start and end of string
/^CTR.*$/
/ = delimiter
^ = start of string
CTR = literal CTR
$ = end of string
.* = zero or more of any character except newline
Start and end of line
/^CTR.*$/m
/ = delimiter
^ = start of line
CTR = literal CTR
$ = end of line
.* = zero or more of any character except newline
m = enables multi-line mode, this sets regex to treat every line as a string, so ^ and $ will match start and end of line
While in multi-line mode you can still match the start and end of the string with \A\Z permanent anchors
/\ACTR.*\Z/m
\A = means start of string
CTR = literal CTR
.* = zero or more of any character except newline
\Z = end of string
m = enables multi-line mode
As such, another way to match the start of the line would be like this:
/(\A|\r|\n|\r\n)CTR.*/
or
/(^|\r|\n|\r\n)CTR.*/
\r = carriage return / old Mac OS newline
\n = line-feed / Unix/Mac OS X newline
\r\n = windows newline
Note, if you are going to use the backslash \ in some program string that supports escaping, like the php double quotation marks "" then you need to escape them first
so to run \r\nCTR.* you would use it as "\\r\\nCTR.*"
^CTR
or
^CTR.*
edit:
To be more clear: ^CTR will match start of line and those chars. If all you want to do is match for a line itself (and already have the line to use), then that is all you really need. But if this is the case, you may be better off using a prefab substr() type function. I don't know, what language are you are using. But if you are trying to match and grab the line, you will need something like .* or .*$ or whatever, depending on what language/regex function you are using.
Regex symbol to match at beginning of a line:
^
Add the string you're searching for (CTR) to the regex like this:
^CTR
Example: regex
That should be enough!
However, if you need to get the text from the whole line in your language of choice, add a "match anything" pattern .*:
^CTR.*
Example: more regex
If you want to get crazy, use the end of line matcher
$
Add that to the growing regex pattern:
^CTR.*$
Example: lets get crazy
Note: Depending on how and where you're using regex, you might have to use a multi-line modifier to get it to match multiple lines. There could be a whole discussion on the best strategy for picking lines out of a file to process them, and some of the strategies would require this:
Multi-line flag m (this is specified in various ways in various languages/contexts)
/^CTR.*/gm
Example: we had to use m on regex101
Try ^CTR.\*, which literally means start of line, CTR, anything.
This will be case-sensitive, and setting non-case-sensitivity will depend on your programming language, or use ^[Cc][Tt][Rr].\* if cross-environment case-insensitivity matters.
^CTR.*$
matches a line starting with CTR.
Not sure how to apply that to your file on your server, but typically, the regex to match the beginning of a string would be :
^CTR
The ^ means beginning of string / line
There's are ambiguities in the question.
What is your input string? Is it the entire file? Or is it 1 line at a time? Some of the answers are assuming the latter. I want to answer the former.
What would you like to return from your regular expression? The fact that you want a true / false on whether a match was made? Or do you want to extract the entire line whose start begins with CTR? I'll answer you only want a true / false match.
To do this, we just need to determine if the CTR occurs at either the start of a file, or immediately following a new line.
/(?:^|\n)CTR/
(?i)^[ \r\n]*CTR
(?i) -- case insensitive -- Remove if case sensitive.
[ \r\n] -- ignore space and new lines
* -- 0 or more times the same
CTR - your starts with string.
Related
There are tons of examples to do the conversion from C-style line comment to 1-line block comment. But I need to do the opposite: find a regex to replace multi-line block comment with line comments.
From:
This text must not be touched
/*
This
is
random
text
*/
This text must not be touched
To
This text must not be touched
// This
// is
// random
// text
This text must not be touched
I was thinking if there's a way to represent "each line" concept in regex, then just add // in front of each line. Something like
\/\*\n(?:(.+)\n)+\*\/ -> // $1
But the greediness nature of the regex engine makes $1 just match the last line before */. I know Perl and other languages have some advanced regex features like recursion, but I need to do this in a standard engine. Is there any trick to accomplish this?
EDIT: To clarify, I'm looking for pure regex solution, not involving any programming language. Should be testable on sites like https://regex101.com/.
If you are interested in a single regex pass in the modern JavaScript engine (and other regex engines supporting infinite length patterns in lookbehinds), you can use
/(?<=^(\/)\*(?:(?!^\/\*)[\s\S])*?\r?\n)(?=[\s\S]*?^\*\/)|(?:\r?\n)?(?:^\/\*|^\*\/)/gm
Replace with $1$1, see the regex demo.
Details
(?<=^(\/)\*(?:(?!^\/\*)[\s\S])*?\r?\n) - a positive lookbehind that matches a location that is immediately preceded with
^(\/)\* - /* substring at the start of a line (with / captured into Group 1)
(?:(?!^\/\*)[\s\S])*? - any char, zero or more occurrences, as few as possible, not starting a /* char sequence that appears at the start of a line
\r?\n - a CRLF or LF ending
(?=[\s\S]*?^\*\/) - a positive lookahead that requires any 0 or more chars as few as possible followed with */ at the start of a line, immediately to the right of the current location
| - or
(?:\r?\n)? - an optional CRLF or LF linebreak
(?:^\/\*|^\*\/) - and then either /* or */ at the start of a line.
As usual in such cases, two regular expressions—the second applied to the matches of the first—can do what one cannot achieve.
const txt = `This text must not be touched
/*
This
is
random
text
*/
This text must not be touched`;
const to1line = str => str.replace(
/\/\*\s*(.*?)\s*\*\//gs,
(_, comment) => comment.replace( /^/mg, '//')
);
console.log( to1line( txt ));
I am trying to parse a file that contains parameter attributes. The attributes are setup like this:
w=(nf*40e-9)*ng
but also like this:
par_nf=(1) * (ng)
The issue is, all of these parameter definitions are on a single line in the source file, and they are separated by spaces. So you might have a situation like this:
pd=2.0*(84e-9+(1.0*nf)*40e-9) nf=ng m=1 par=(1) par_nf=(1) * (ng) plorient=0
The current algorithm just splits the line on spaces and then for each token, the name is extracted from the LHS of the = and the value from the RHS. My thought is if I can create a Regex match based on spaces within parameter declarations, I can then remove just those spaces before feeding the line to the splitter/parser. I am having a tough time coming up with the appropriate Regex, however. Is it possible to create a regex that matches only spaces within parameter declarations, but ignores the spaces between parameter declarations?
Try this RegEx:
(?<=^|\s) # Start of each formula (start of line OR [space])
(?:.*?) # Attribute Name
= # =
(?: # Formula
(?!\s\w+=) # DO NOT Match [space] Word Characters = (Attr. Name)
[^=] # Any Character except =
)* # Formula Characters repeated any number of times
When checking formula characters, it uses a negative lookahead to check for a Space, followed by Word Characters (Attribute Name) and an =. If this is found, it will stop the match. The fact that the negative lookahead checks for a space means that it will stop without a trailing space at the end of the formula.
Live Demo on Regex101
Thanks to #Andy for the tip:
In this case I'll probably just match on the parameter name and equals, but replace the preceding whitespace with some other "parse-able" character to split on, like so:
(\s*)\w+[a-zA-Z_]=
Now my first capturing group can be used to insert something like a colon, semicolon, or line-break.
You need to add Perl tag. :-( Maybe this will help:
I ended up using this in C#. The idea was to break it into name value pairs, using a negative lookahead specified as the key to stop a match and start a new one. If this helps
var data = #"pd=2.0*(84e-9+(1.0*nf)*40e-9) nf=ng m=1 par=(1) par_nf=(1) * (ng) plorient=0";
var pattern = #"
(?<Key>[a-zA-Z_\s\d]+) # Key is any alpha, digit and _
= # = is a hard anchor
(?<Value>[.*+\-\\\/()\w\s]+) # Value is any combinations of text with space(s)
(\s|$) # Soft anchor of either a \s or EOB
((?!\s[a-zA-Z_\d\s]+\=)|$) # Negative lookahead to stop matching if a space then key then equal found or EOB
";
Regex.Matches(data, pattern, RegexOptions.IgnorePatternWhitespace | RegexOptions.ExplicitCapture)
.OfType<Match>()
.Select(mt => new
{
LHS = mt.Groups["Key"].Value,
RHS = mt.Groups["Value"].Value
});
Results:
This is not correct use of wildcards ? I'm attempting to match String that contains a date. I don't want to include the date in the returned String or the String value that prepends the matched String.
object FindText extends App{
val toFind = "find1"
val line = "this is find1 the line 1 \n 21/03/2015"
val find = (toFind+".*\\d{2}/\\d{2}/\\d{4}").r
println(find.findFirstIn(line))
}
Output should be : "find1 the line 1 \n "
but String is not found.
Dot does not match newline characters by default. You can set a DOTALL flag to make it happen (I have also added a "positive look-ahead - the (?=...) thingy - since you did not want the date to be included in the match": val find = (toFind+"""(?s).*(?=\d{2}/\d{2}/\d{4})""").r
(Note also, that in scala you do not need to escape special characters in strings, enclosed in a triple-quote pairs ... pretty neat).
The problem lies with the newline in the test string. A .* does not match newlines apparently. Replacing this with .*\\n?.* should fix it. One could also use a multiline flag in the regex such as:
val find = ("(?s)"+toFind+".*\\d{2}/\\d{2}/\\d{4}").r
I'm very new to regular expression. I want to extract the following string
"109_Admin_RegistrationResponse_20130103.txt"
from this file content, the contents is selected per line:
01-10-13 10:44AM 47 107_Admin_RegistrationDetail_20130111.txt
01-10-13 10:40AM 11 107_Admin_RegistrationResponse_20130111.txt
The regular expression should not pick the second line, only the first line should return a true.
Your Regex has a lot of different mistakes...
Your line does not start with your required filename but you put an ^ there
missing + in your character group [a-zA-Z], hence only able to match a single character
does not include _ in your character group, hence it won't match Admin_RegistrationResponse
missing \ and d{2} would match dd only.
As per M42's answer (which I left out), you also need to escape your dot . too, or it would match 123_abc_12345678atxt too (notice the a before txt)
Your regex should be
\d+_[a-zA-Z_]+_\d{4}\d{2}\d{2}\.txt$
which can be simplified as
\d+_[a-zA-Z_]+_\d{8}\.txt$
as \d{2}\d{2} really look redundant -- unless you want to do with capturing groups, then you would do:
\d+_[a-zA-Z_]+_(\d{4})(\d{2})(\d{2})\.txt$
Remove the anchors and escape the dot:
\d+[a-zA-Z_]+\d{8}\.txt
I'm a newbie in php but i think you can use explode() function in php or any equivalent in your language.
$string = "01-09-13 10:17AM 11 109_Admin_RegistrationResponse_20130103.txt";
$pieces = explode("_", $string);
$stringout = "";
foreach($i = 0;$i<count($pieces);i++){
$stringout = $stringout.$pieces[$i];
}
How do I use regex to convert
11111aA$xx1111xxdj$%%`
to
aA$xx1111xxdj$%%
So, in other words, I want to remove (or match) the FIRST grouping of 1's.
Depending on the language, you should have a way to replace a string by regex. In Java, you can do it like this:
String s = "11111aA$xx1111xxdj$%%";
String res = s.replaceAll("^1+", "");
The ^ "anchor" indicates that the beginning of the input must be matched. The 1+ means a sequence of one or more 1 characters.
Here is a link to ideone with this running program.
The same program in C#:
var rx = new Regex("^1+");
var s = "11111aA$xx1111xxdj$%%";
var res = rx.Replace(s, "");
Console.WriteLine(res);
(link to ideone)
In general, if you would like to make a match of anything only at the beginning of a string, add a ^ prefix to your expression; similarly, adding a $ at the end makes the match accept only strings at the end of your input.
If this is the beginning, you can use this:
^[1]*
As far as replacing, it depends on the language. In powershell, I would do this:
[regex]::Replace("11111aA$xx1111xxdj$%%","^[1]*","")
This will return:
aA$xx1111xxdj$%%
If you only want to replace consecutive "1"s at the beginning of the string, replace the following with an empty string:
^1+
If the consecutive "1"s won't necessarily be the first characters in the string (but you still only want to replace one group), replace the following with the contents of the first capture group (usually \1 or $1):
1+(.*)
Note that this is only necessary if you only have a "replace all" capability available to you, but most regex implementations also provide a way to replace only one instance of a match, in which case you could just replace 1+ with an empty string.
I'm not sure but you can try this
[^1](\w*\d*\W)* - match all as a single group except starting "1"(n) symbols
In Javascript
var str = '11111aA$xx1111xxdj$%%';
var patt = /^1+/g;
str = str.replace(patt,"");