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, '.*?')
Related
I am trying to match a string in the format of domain\username using Lua and then mask the pattern with #.
So if the input is sample.com\admin; the output should be ######.###\#####;. The string can end with either a ;, ,, . or whitespace.
More examples:
sample.net\user1,hello -> ######.###\#####,hello
test.org\testuser. Next -> ####.###\########. Next
I tried ([a-zA-Z][a-zA-Z0-9.-]+)\.?([a-zA-Z0-9]+)\\([a-zA-Z0-9 ]+)\b which works perfectly with http://regexr.com/. But with Lua demo it doesn't. What is wrong with the pattern?
Below is the code I used to check in Lua:
test_text="I have the 123 name as domain.com\admin as 172.19.202.52 the credentials"
pattern="([a-zA-Z][a-zA-Z0-9.-]+).?([a-zA-Z0-9]+)\\([a-zA-Z0-9 ]+)\b"
res=string.match(test_text,pattern)
print (res)
It is printing nil.
Lua pattern isn't regular expression, that's why your regex doesn't work.
\b isn't supported, you can use the more powerful %f frontier pattern if needed.
In the string test_text, \ isn't escaped, so it's interpreted as \a.
. is a magic character in patterns, it needs to be escaped.
This code isn't exactly equivalent to your pattern, you can tweek it if needed:
test_text = "I have the 123 name as domain.com\\admin as 172.19.202.52 the credentials"
pattern = "(%a%w+)%.?(%w+)\\([%w]+)"
print(string.match(test_text,pattern))
Output: domain com admin
After fixing the pattern, the task of replacing them with # is easy, you might need string.sub or string.gsub.
Like already mentioned pure Lua does not have regex, only patterns.
Your regex however can be matched with the following code and pattern:
--[[
sample.net\user1,hello -> ######.###\#####,hello
test.org\testuser. Next -> ####.###\########. Next
]]
s1 = [[sample.net\user1,hello]]
s2 = [[test.org\testuser. Next]]
s3 = [[abc.domain.org\user1]]
function mask_domain(s)
s = s:gsub('(%a[%a%d%.%-]-)%.?([%a%d]+)\\([%a%d]+)([%;%,%.%s]?)',
function(a,b,c,d)
return ('#'):rep(#a)..'.'..('#'):rep(#b)..'\\'..('#'):rep(#c)..d
end)
return s
end
print(s1,'=>',mask_domain(s1))
print(s2,'=>',mask_domain(s2))
print(s3,'=>',mask_domain(s3))
The last example does not end with ; , . or whitespace. If it must follow this, then simply remove the final ? from pattern.
UPDATE: If in the domain (e.g. abc.domain.org) you need to also reveal any dots before that last one you can replace the above function with this one:
function mask_domain(s)
s = s:gsub('(%a[%a%d%.%-]-)%.?([%a%d]+)\\([%a%d]+)([%;%,%.%s]?)',
function(a,b,c,d)
a = a:gsub('[^%.]','#')
return a..'.'..('#'):rep(#b)..'\\'..('#'):rep(#c)..d
end)
return s
end
I've got two files:
1st: Entries.txt
confirmation.resend
send
confirmation.showResendForm
login.header
login.loginBtn
2nd: Used_Entries.txt
confirmation.showResendForm = some value
login.header = some other value
I want to find all entries from the first file (Entries.txt) that have not been asigned a value in the 2nd file (Used_Entries.txt)
In this example I'd like the following result:
confirmation.resend
send
login.loginBtn
In the result confirmation.showResendForm and login.header do not show up because these exist in the Used_Entries.txt
How do I do this? I've been playing around with regular expressions but haven't been able to solve it. A bash script or sth would be much appreciated!
You can do this with regex. But get your code mood ready, because you can't match both files with regex at once, and we do want to match both contents with regex at once. Well, that means you must have at least some understanding of your language, I would like you to concatenate the contents from the two files with at least a new line in between.
This regex solution expects your string to be matched to be in this format:
text (no equals sign)
text
text
...
key (no equals sign) ␣ (optional whitespace) = (literal equal) whatever (our regex will skip this part.)
key=whatever
key=whatever
Do I have your attention? Yes? Please see the following regex (using techniques accessible to most regex engines):
/(^[^=\n]+$)(?!(?s).*^\1\s*=)/m
Inspired from a recent answer I saw from zx81, you can switch to (?s) flag in the middle to switch to DOTALL mode suddenly, allowing you to start multiline matching with . in the middle of a RegExp. Using this technique and the set syntax above, here's what the regex does, as an explanation:
(^[^=\n]+$) Goes through all the text (no equals sign) elements. Enforces no equals signs or newlines in the capture. This means our regex hits every text element as a line, and tries to match it appropriately.
(?! Opens a negative lookahead group. Asserts that this match will not locate the following:
(?s).* Any number of characters or new lines - As this is a greedy match, will throw our matcher pointer to the very end of the string, skipping to the last parts of the document to backtrack and scoop up quickly.
^\1\s*= The captured key, followed by an equals sign after some optional whitespaces, in its own line.
) Ends our group.
View a Regex Demo!
A regex demo with more test cases
I'm stupid. I could had just put this:
/(^[^=\n]+$)(?!.*^\1\s*=)/sm
I've been going at this a little bit to complex and just solved it with a small script in scala:
import scala.io.Source
object HelloWorld {
def main(args: Array[String]) {
val entries = (for(line <- Source.fromFile("Entries.txt").getLines()) yield {
line
}).toList
val usedEntries = (for(line <- Source.fromFile("Used_Entries.txt").getLines()) yield {
line.dropRight(line.length - line.indexOf(' '))
}).toList
println(entries)
println(usedEntries)
val missingEntries = (for {
entry <- entries
if !usedEntries.exists(_ == entry)
} yield {
entry
}).toList
println(missingEntries)
println("Missing Entries: ")
println()
for {
missingEntry <- missingEntries
} yield {
println(missingEntry)
}
}
}
import re
e=open("Entries.txt",'r')
m=e.readlines()
u=open("Used_Entries.txt",'r')
s=u.read()
y=re.sub(r"= .*","",s)
for i in m:
if i.strip() in [k.strip() for k in y.split("\n")] :
pass
else:
print i.strip()
I need to come up with a regular expression to parse my input string. My input string is of the format:
[alphanumeric].[alpha][numeric].[alpha][alpha][alpha].[julian date: yyyyddd]
eg:
A.A2.ABC.2014071
3.M1.MMB.2014071
I need to substring it from the 3rd position and was wondering what would be the easiest way to do it.
Desired result:
A2.ABC.2014071
M1.MMB.2014071
(?i) will be considered as case insensitive.
(?i)^[a-z\d]\.[a-z]\d\.[a-z]{3}\.\d{7}$
Here a-z means any alphabet from a to z, and \d means any digit from 0 to 9.
Now, if you want to remove the first section before dot, then use this regex and replace it with $1 (or may be \1)
(?i)^[a-z\d]\.([a-z]\d\.[a-z]{3}\.\d{7})$
Another option is replace below with empty:
(?i)^[a-z\d]\.
If the input string is just the long form, then you want everything except the first two characters. You could arrange to substitute them with nothing:
s/^..//
Or you could arrange to capture everything except the first two characters:
/^..(.*)/
If the expression is part of a larger string, then the breakdown of the alphanumeric components becomes more important.
The details vary depending on the language that is hosting the regex. The notations written above could be Perl or PCRE (Perl Compatible Regular Expressions). Many other languages would accept these regexes too, but other languages would require tweaks.
Use this regex:
\w.[A-Z]\d.[A-Z]{3}.\d{7}
Use the above regex like this:
String[] in = {
"A.A2.ABC.2014071", "3.M1.MMB.2014071"
};
Pattern p = Pattern.compile("\\w.[A-Z]\\d.[A-Z]{3}.\\d{7}");
for (String s: in ) {
Matcher m = p.matcher(s);
while (m.find()) {
System.out.println("Result: " + m.group().substring(2));
}
}
Live demo: http://ideone.com/tns9iY
I have a list of email addresses which take various forms:
john#smith.com
Angie <angie#aol.com>
"Mark Jones" <mark#jones.com>
I'm trying to cut only the email portion from each. Ex: I only want the angie#aol.com from the second item in the list. In other words, I want to match everything between < and > or match everything if it doesn't exist.
I know this can be done in 2 steps:
Capture on (?<=\<)(.*)(?=\>).
If there is no match, use the entire text.
But now I'm wondering: Can both steps be reduced into one simple regular expression?
What about:
(?<=\<).*(?=\>)|^[^<]*$
^[^>]*$ will match the entire string, but only if it doesn't contain a <. And that's OR'ed (|) with what you had.
Explanation:
^ - start of string
[^<] - not-< character
[^<]* - zero or more not-< characters
$ - end of string
You're after an exclusive or operator. Have a look here.
(\<.+\#.+\..+\>) matches those email addresses in side <> only...
(\<.+\#.+\..+\>)|(.+) matches everything instead of matching the first condition in the OR then skipping the second.
Depending on what language you are using to implement this regex, you might be able to use an inbuilt exclusive or operator. Otherwise, you might need to put a bit of logic in there to use the string if no matches are found. E.g. (pseudo type code):
string = 'your data above';
if( regex_finds_match ( '(\<.+\#.+\..+\>)', string ) ) {
// found match, use the match
str_to_use = regex_match(es);
} else {
// didn't find a match:
str_to_use = string;
}
It is possible, but your current logic is probably simpler. Here is what I came up with, email address will always be in the first capturing group:
^(?:.*<|)(.*?)(?:>|$)
Example: http://rubular.com/r/8tKHaYYY4T
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);