I am trying to combine 2 regular expressions into 1 with the OR operator: |
I have one that checks for match of a letter followed by 8 digits:
Regex.IsMatch(s, "^[A-Z]\d{8}$")
I have another that checks for simply 9 digits:
Regex.IsMatch(s, "^\d{9}$")
Now, Instead of doing:
If Not Regex.IsMatch(s, "^[A-Z]\d{8}$") AndAlso
Not Regex.IsMatch(s, "^\d{9}$") Then
...
End If
I thought I could simply do:
If Not Regex.IsMatch(s, "^[A-Z]\d{8}|\d{9}$") Then
...
End If
Apparently I am not combining the two correctly and apparently I am horrible at regular expressions. Any help would be much appreciated.
And for those wondering, I did take a glance at How to combine 2 conditions and more in regex and I am still scratching my head.
The | operator has a high precedence and in your original regex will get applied first. You should be combining the two regex's w/ grouping parentheses to make the precedence clear. As in:
"^(([A-Z]\d{8})|(\d{9}))$"
How about using ^[A-Z0-9]\d{8}$ ?
I think you want to group the conditions:
Regex.IsMatch(s, "^(([A-Z]\d{8})|(\d{9}))$")
The ^ and $ represent the beginning and end of the line, so you don't want them considered in the or condition. The parens allow you to be explicit about "everything in this paren" or "anything in this other paren"
#MikeC's offering seems the best:
^[A-Z0-9]\d{8}$
...but as to why your expression is not working the way you might expect, you have to understand that the | "or" or "alternation" operator has a very high precedence - the only higher one is the grouping construct, I believe. If you use your example:
^[A-Z]\d{8}|\d{9}$
...you're basically saying "match beginning of string, capital letter, then 8 digits OR match 9 digits then end of string" -- if, instead you mean "match beginning of string, then a capital letter followed by 8 digits then the end of string OR the beginning of the string followed by 9 digits, then the end of string", then you want one of these:
^([A-Z]\d{8}|\d{9})$
^[A-Z]\d{8}$|^\d{9}$
Hope this is helpful for your understanding
I find the OR operator a bit weird sometimes as well, what I do I use groups to denote which sections I want to match, so your regex would become something like so: ^(([A-Z]\d{8})|(\d{9}))$
Related
My regex pattern looks something like
<xxxx location="file path/level1/level2" xxxx some="xxx">
I am only interested in the part in quotes assigned to location. Shouldn't it be as easy as below without the greedy switch?
/.*location="(.*)".*/
Does not seem to work.
You need to make your regular expression lazy/non-greedy, because by default, "(.*)" will match all of "file path/level1/level2" xxx some="xxx".
Instead you can make your dot-star non-greedy, which will make it match as few characters as possible:
/location="(.*?)"/
Adding a ? on a quantifier (?, * or +) makes it non-greedy.
Note: this is only available in regex engines which implement the Perl 5 extensions (Java, Ruby, Python, etc) but not in "traditional" regex engines (including Awk, sed, grep without -P, etc.).
location="(.*)" will match from the " after location= until the " after some="xxx unless you make it non-greedy.
So you either need .*? (i.e. make it non-greedy by adding ?) or better replace .* with [^"]*.
[^"] Matches any character except for a " <quotation-mark>
More generic: [^abc] - Matches any character except for an a, b or c
How about
.*location="([^"]*)".*
This avoids the unlimited search with .* and will match exactly to the first quote.
Use non-greedy matching, if your engine supports it. Add the ? inside the capture.
/location="(.*?)"/
Use of Lazy quantifiers ? with no global flag is the answer.
Eg,
If you had global flag /g then, it would have matched all the lowest length matches as below.
Here's another way.
Here's the one you want. This is lazy [\s\S]*?
The first item:
[\s\S]*?(?:location="[^"]*")[\s\S]* Replace with: $1
Explaination: https://regex101.com/r/ZcqcUm/2
For completeness, this gets the last one. This is greedy [\s\S]*
The last item:[\s\S]*(?:location="([^"]*)")[\s\S]*
Replace with: $1
Explaination: https://regex101.com/r/LXSPDp/3
There's only 1 difference between these two regular expressions and that is the ?
The other answers here fail to spell out a full solution for regex versions which don't support non-greedy matching. The greedy quantifiers (.*?, .+? etc) are a Perl 5 extension which isn't supported in traditional regular expressions.
If your stopping condition is a single character, the solution is easy; instead of
a(.*?)b
you can match
a[^ab]*b
i.e specify a character class which excludes the starting and ending delimiiters.
In the more general case, you can painstakingly construct an expression like
start(|[^e]|e(|[^n]|n(|[^d])))end
to capture a match between start and the first occurrence of end. Notice how the subexpression with nested parentheses spells out a number of alternatives which between them allow e only if it isn't followed by nd and so forth, and also take care to cover the empty string as one alternative which doesn't match whatever is disallowed at that particular point.
Of course, the correct approach in most cases is to use a proper parser for the format you are trying to parse, but sometimes, maybe one isn't available, or maybe the specialized tool you are using is insisting on a regular expression and nothing else.
Because you are using quantified subpattern and as descried in Perl Doc,
By default, a quantified subpattern is "greedy", that is, it will
match as many times as possible (given a particular starting location)
while still allowing the rest of the pattern to match. If you want it
to match the minimum number of times possible, follow the quantifier
with a "?" . Note that the meanings don't change, just the
"greediness":
*? //Match 0 or more times, not greedily (minimum matches)
+? //Match 1 or more times, not greedily
Thus, to allow your quantified pattern to make minimum match, follow it by ? :
/location="(.*?)"/
import regex
text = 'ask her to call Mary back when she comes back'
p = r'(?i)(?s)call(.*?)back'
for match in regex.finditer(p, str(text)):
print (match.group(1))
Output:
Mary
This may be a very simple question but I have not much experience with regex expressions. This page is a good source of regex expressions but could not figure out how to include them into my following code:
data %>% filter(grepl("^A01H1", icl))
Question
I would like to extract the values in one column of my data frame starting with this A01H1 up to 2 more digits, for example A01H100, A01H140, A01H110. I could not find a solution despite my few attempts:
Attempts
I looked at this question from which I used ^A01H1[0-9].{2} to select up tot two more digits.
I tried with adding any character ^A01H1[0-9][0-9][x-y] to stop after two digits.
Any help would be much appreciated :)
You can use "^A01H1\\d{1,2}$".
The first part ("^A01H1"), you figured out yourself, so what are we doing in the second part ("\\d{1,2}$")?
\d includes all digits and is equivalent to [0-9], since we are working in R you need to escape \ and thus we use \\d
{1,2} indicates we want to have 1 or 2 matches of \\d
$ specifies the end of the string, so nothing should come afterwards and this prevents to match more than 2 digits
It looks as if you want to match a part of a string that starts with A01H1, then contains 1 or 2 digits and then is not followed with any digit.
You may use
^A01H1\d{1,2}(?!\d)
See the regex demo. If there can be no text after two digits at all, replace (?!\d) with $.
Details
^ - start of strinmg
A01H1 - literal string
\d{1,2} - one to two digits
(?!\d) - no digit allowed immediately to the right
$ - end of string
In R, you could use it like
grepl("^A01H1\\d{1,2}(?!\\d)", icl, perl=TRUE)
Or, with the string end anchor,
grepl("^A01H1\\d{1,2}$", icl)
Note the perl=TRUE is only necessary when using PCRE specific syntax like (?!\d), a negative lookahead.
I try to find a regex that matches the string only if the string does not end with at least three '0' or more. Intuitively, I tried:
.*[^0]{3,}$
But this does not match when there one or two zeroes at the end of the string.
If you have to do it without lookbehind assertions (i. e. in JavaScript):
^(?:.{0,2}|.*(?!000).{3})$
Otherwise, use hsz's answer.
Explanation:
^ # Start of string
(?: # Either match...
.{0,2} # a string of up to two characters
| # or
.* # any string
(?!000) # (unless followed by three zeroes)
.{3} # followed by three characters
) # End of alternation
$ # End of string
You can try using a negative look-behind, i.e.:
(?<!000)$
Tests:
Test Target String Matches
1 654153640 Yes
2 5646549800 Yes
3 848461158000 No
4 84681840000 No
5 35450008748 Yes
Please keep in mind that negative look-behinds aren't supported in every language, however.
What wrong with the no-look-behind, more general-purpose ^(.(?!.*0{3,}$))*$?
The general pattern is ^(.(?!.* + not-ending-with-pattern + $))*$. You don't have to reverse engineer the state machine like Tim's answer does; you just insert the pattern you don't want to match at the end.
This is one of those things that RegExes aren't that great at, because the string isn't very regular (whatever that means). The only way I could come up with was to give it every possibility.
.*[^0]..$|.*.[^0].$|.*..[^0]$
which simplifies to
.*([^0]|[^0].|[^0]..)$
That's fine if you only want strings not ending in three 0s, but strings not ending in ten 0s would be long. But thankfully, this string is a bit more regular than some of these sorts of combinations, and you can simplify it further.
.*[^0].{0,2}$
I'm trying to build a somewhat REGEX expression of the of only numbers including decimal with a maximum of 3 numbers to the right of the decimal (thousandths) and 50 to the left. Valid entries would like something like these.
1
1.0
.1
1.011
.011
1202938.123
1237923782.0
So far I have ^([0-9]*|\d*\.\d{1}?\d*){1,999}$.. Any help appreciated. Thanks.
I believe this should suffice:
^(?=.)\d{0,50}(?:\.\d{0,3})?$
See the regex demo. Note this will also match 1., if this is undesired change \d{0,3} to \d{1,3}. Similarely, this regex will match .5 (with no integer part), if you dont want this then use \d{1,50} instead of \d{0,50}.
You could try:
^(?=.+)\d{0,50}(?:\.\d{1,3})?$
Demonstration here at regex101.com
Explanation -
^ tells the regex that the match will begin at the start of the string,
\d{0, 50} matches 0 - 50 digits,
(?=.+) is a positive look-ahead, that tells the regex that the matching should only start if the line contains some characters in it (as rightly pointed out in the comments!),
(?:\.\d{1,3})? matches an optional dot (.), followed by 1 - 3 digits,
$ tells the regex that whatever it has matched so far will be followed by the end of the string.
Other way: You can check if the string isn't empty and if the dot is always followed by digits, putting a word-boundary at a strategic place:
^\d{0,50}\.?\b\d{0,3}$
As you can see, all is optional in the pattern except the word-boundary that does the magic.
demo
Hi I'm trying to right a regular expression that will take a string and ensure it starts with an 'R' and is followed by 4 numeric digits then anything
eg. RXXXX.................
Can anybody help me with this? This is for ASP.NET
You want it to be at the beginning of the line, not anywhere. Also, for efficiency, you dont want the .+ or .* at the end because that will match unnecessary characters. So the following regex is what you really want:
^R\d{4}
This should do it...
^R\d{4}.*$
\d{4} matches 4 digits
.* is simply a way to match any character 0 or more times
the beginning ^ and end $ anchors ensure that nothing precedes or follows
As Vincent suggested, for your specific task it could even be simplified to this...
^R\d{4}
Because as you stated, it doesn't really matter what follows.
/^R\d{4}.*/ and set the case insensitive option unless you only want capital R's
^R\d{4}.*
The caret ^ matches the position before the first character in the string.
\d matches any numeric character (it's the same as [0-9])
{4} indicates that there must be exactly 4 numbers, and
.* matches 0 or more other characters
To use:
string input = "R0012 etc..";
Match match = Regex.Match(input, #"^R\d{4}.*", RexOptions.IgnoreCase);
if (match.Success)
{
// Success!
}
Note the use of RexOptions.IgnoreCase to ignore the case of the letter R (so it'll match strings which start with r. Leave this out if you don't want to undertake a case insensitive match.