I need the same Regex for below two lines.
PieceID = NEW00000009
piece_id="NEW00000009"
find my pattern:
[pP]iece[_]*[iI][dD][" "]*=[" "]*["]*(?<PieceID>[A-Z0-9]["]*{11})"
You are not far from the good pattern, but you need several clarifications:
To make something optional use ? (that means {0,1}) instead of * (that means {0,}).
you don't need to put quotes around a space, the space character has nothing special. Even a character were special, you never need to protect it with quotes.
Your pattern can be written like this:
[pP]iece_?[iI][dD] *= *"?(?<PieceID>[A-Z0-9]{11})"?
You can be a little more rigorous, avoiding to match something like piece_id=NEW00000009":
[pP]iece_?[iI][dD] *= *("?)(?<PieceID>[A-Z0-9]{11})\\1
The optional quote is captured in group 1 and \\1 is a back-reference for the group 1
To finish you can make it more flexible if you replace spaces with \\s* (\\s is a character class for any white-spaces) or [^\\S\\r\\n]* that allows only horizontal white-spaces.
You'll need to worry about case-sensitivity with this one, but I think it will do what you're trying to do.
var s1 = "PieceID = NEW00000009";
var s2 = #"piece_id=""NEW00000009""";
var re = new Regex(#"^(PieceID|piece_id)\s*=\s*\""?(?<PieceID>[A-Z0-9]{11})\""?$");
Console.WriteLine(re.IsMatch(s1));
Console.WriteLine(re.IsMatch(s2));
Output:
True
True
Related
I'm trying to allow users to filter strings of text using a glob pattern whose only control character is *. Under the hood, I figured the easiest thing to filter the list strings would be to use Js.Re.test[https://rescript-lang.org/docs/manual/latest/api/js/re#test_], and it is (easy).
Ignoring the * on the user filter string for now, what I'm having difficulty with is escaping all the RegEx control characters. Specifically, I don't know how to replace the capture groups within the input text to create a new string.
So far, I've got this, but it's not quite right:
let input = "test^ing?123[foo";
let escapeRegExCtrl = searchStr => {
let re = [%re("/([\\^\\[\\]\\.\\|\\\\\\?\\{\\}\\+][^\\^\\[\\]\\.\\|\\\\\\?\\{\\}\\+]*)/g")];
let break = ref(false);
while (!break.contents) {
switch (Js.Re.exec_ (re, searchStr)) {
| Some(result) => {
let match = Js.Re.captures(result)[0];
Js.log2("Matching: ", match)
}
| None => {
break := true;
}
}
}
};
search -> escapeRegExCtrl
If I disregard the "test" portion of the string being skipped, the above output will produce:
Matching: ^ing
Matching: ?123
Matching: [foo
With the above example, at the end of the day, what I'm trying to produce is this (with leading and following .*:
.*test\^ing\?123\[foo.*
But I'm unsure how to achieve creating a contiguous string from the matched capture groups.
(echo "test^ing?123[foo" | sed -r 's_([\^\?\[])_\\\1_g' would get the work done on the command line)
EDIT
Based on Chris Maurer's answer, there is a method in the JS library that does what I was looking for. A little digging exposed the ReasonML proxy for that method:
https://rescript-lang.org/docs/manual/latest/api/js/string#replacebyre
Let me see if I have this right; you want to implement a character matcher where everything is literal except *. Presumably the * is supposed to work like that in Windows dir commands, matching zero or more characters.
Furthermore, you want to implement it by passing a user-entered character string directly to a Regexp match function after suitably sanitizing it to only deal with the *.
If I have this right, then it sounds like you need to do two things to get the string ready for js.re.test:
Quote all the special regex characters, and
Turn all instances of * into .* or maybe .*?
Let's keep this simple and process the string in two steps, each one using Js.re.replace. So the list of special characters in regex are [^$.|?*+(). Suitably quoting these for replace:
str.replace(/[\[\\\^\$\.\|\?\+\(\)]/g, '\$&')
This is just all those special characters quoted. The $& in the replacement specifications says to insert whatever matched.
Then pass that result to a second replace for the * to .*? transformation.
str.replace(/*+/g, '.*?')
i have following statement {$("#aprilfoolc").val("HoliWed27"); $("#UgadHieXampp").val("ugadicome");}.and i want to get the string with combination.i have written following regex but it is not working.
please help!
(?=[\$("#]?)[\w]*(?<=[")]?)
Your lookaround assertions are using character classes by mistake, and you've confused lookbehind and lookahead. Try the following:
(?<=\$\(")\w*(?="\))
You could use this simpler one :
'{$("#aprilfoolc").val("HoliWed27");}'.match(/\$\(\"#(\w+)\"[^"]*"(\w+)"/)
This returns
["$("#aprilfoolc").val("HoliWed27"", "aprilfoolc", "HoliWed27"]
where the strings you want are at indexes 1 and 2.
This construction
(?=[\$*"#]?)
will match a lookahead, but only optional -- the character set is followed by a ?. This kind of defeats the next part,
[\w]
which matches word characters only. So the lookahead will never match. Similar, this part
(?<=[")])
will also never match, because logically there can never be one of the characters " or ) at the end of a string that matches \w only. Again, since this portion is optional (that ? at the end again) it will simply never match.
It's a bit unclear what you are after. Strings inside double quotes, yes, but in the first one you want to skip the hash -- why? Given your input and desired output, this ought to work:
\w+(?=")
Also possible:
/\("[#]?(.*?)"\)/
import re
s='{$("#aprilfoolc").val("HoliWed27");}'
f = re.findall(r'\("[#]?(.*?)"\)',s)
for m in f:
print m
I don't know why, but if you want capturing of two groups simultaneously, so:
/\("#(.*?)"\).*?\("(.*?)"\)/
import re
s='{$("#aprilfoolc").val("HoliWed27");}'
f = re.findall(r'\("#(.*?)"\).*?\("(.*?)"\)',s)
for m in f:
print m[0],m[1]
In JavaScript:
var s='{$("#aprilfoolc").val("HoliWed27")';
var re=/\("#(.*?)"\).*?\("(.*?)"\)/;
alert(s.match(re));
Example data:
029Extract this specific string. Do not capture anything else.
In the example above, I would like to capture the first n characters immediately after the 3 digit entry which defines the value of n. I.E. the 29 characters "Extract this specific string."
I can do this within a loop, but it is slow. I would like (if it is possible) to achieve this with a single regex statement instead, using some kind of backreference. Something like:
(\d{3})(.{\1})
With perl, you can do:
my $str = '029Extract this specific string. Do not capture anything else.';
$str =~ s/^(\d+)(.*)$/substr($2,0,$1)/e;
say $str;
output:
Extract this specific string.
You can not do it with single regex, while you can use knowledge where regex stop processing to use substr. For example in JavaScript you can do something like this http://jsfiddle.net/75Tm5/
var input = "blahblah 011I want this, and 029Extract this specific string. Do not capture anything else.";
var regex = /(\d{3})/g;
var matches;
while ((matches = regex.exec(input)) != null) {
alert(input.substr(regex.lastIndex, matches[0]));
}
This will returns both lines:
I want this
Extract this specific string.
Depending on what you really want, you can modify Regex to match only numbers starting from line beginning, match only first match etc
Are you sure you need a regex?
From https://stackoverflow.com/tags/regex/info:
Fools Rush in Where Angels Fear to Tread
The tremendous power and expressivity of modern regular expressions
can seduce the gullible — or the foolhardy — into trying to use
regular expressions on every string-related task they come across.
This is a bad idea in general, ...
Here's a Python three-liner:
foo = "029Extract this specific string. Do not capture anything else."
substr_len = int(foo[:3])
print foo[3:substr_len+3]
And here's a PHP three-liner:
$foo = "029Extract this specific string. Do not capture anything else.";
$substr_len = (int) substr($foo,0,3);
echo substr($foo,3,substr_len+3);
How do I use regex to convert
11111aA$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$%%';
var patt = /^1+/g;
str = str.replace(patt,"");
I have string like this "first#second", and I wonder how to get "second" part without "#" symbol as result of RegEx, not as match capture using brackets
upd: I forgot to add one more special char at the end of string, real string is "first#second*"
Simple regex:
/#(.*)$/
If you really don't want it to be a match capture, and you know there's a # in the string but none in the part you want, you can do
/[^#]*$/
and the whole regex is what you want.
If you must use regex, and you insist on not using capturing groups, you can use lookbehind in flavors that support them like this:
(?<=#).*
Or you can also capture just anything but #, to the end of the string, so something like this:
[^#]*$
The capturing group option, of course, is:
#(.*)
\__/
1
This matches the # too, but group 1 captures the part that you want.
Lastly, a non-regex alternative may look something like this:
secondPart = wholeString.substring( wholeString.indexOf("#") + 1 )
There may be issues with some of these solutions if # can also appear (perhaps escaped) anywhere else in the string.
References
regular-expressions.info
Lookarounds, Brackets for Capturing, Anchors
/[a-z]+#([a-z]+)/
You can use lookaround to exclude parts of an expression.
http://www.regular-expressions.info/lookaround.html
if your using java then
you can consider using Pattern & Matcher class. Pattern gives you a compiled, optimizer version of Regular expression. Matcher gives a complete internals of RE Matches.
Both Pattern.match & String.spilt gives same result where in first is compartively faster.
for e.g)
String s = "first#second#third";
String re = "#";
Pattern p = Pattern.compile(re);
Matcher m = p.matcher();
int ms = 0;
int me = 0;
while( m.find() ) {
System.out.println("start "+m.start()+" end "+ m.end()+" group "+m.group());
me = m.start();
System.out.println(s.substring(ms,me));
ms = m.end();
}
if other language u can consider using back-reference & groups also. if you find any repetitions.