I don't write many regular expressions so I'm going to need some help on the one.
I need a regular expression that can validate that a string is an alphanumeric comma delimited string.
Examples:
123, 4A67, GGG, 767 would be valid.
12333, 78787&*, GH778 would be invalid
fghkjhfdg8797< would be invalid
This is what I have so far, but isn't quite right: ^(?=.*[a-zA-Z0-9][,]).*$
Any suggestions?
Sounds like you need an expression like this:
^[0-9a-zA-Z]+(,[0-9a-zA-Z]+)*$
Posix allows for the more self-descriptive version:
^[[:alnum:]]+(,[[:alnum:]]+)*$
^[[:alnum:]]+([[:space:]]*,[[:space:]]*[[:alnum:]]+)*$ // allow whitespace
If you're willing to admit underscores, too, search for entire words (\w+):
^\w+(,\w+)*$
^\w+(\s*,\s*\w+)*$ // allow whitespaces around the comma
Try this pattern: ^([a-zA-Z0-9]+,?\s*)+$
I tested it with your cases, as well as just a single number "123". I don't know if you will always have a comma or not.
The [a-zA-Z0-9]+ means match 1 or more of these symbols
The ,? means match 0 or 1 commas (basically, the comma is optional)
The \s* handles 1 or more spaces after the comma
and finally the outer + says match 1 or more of the pattern.
This will also match
123 123 abc (no commas) which might be a problem
This will also match 123, (ends with a comma) which might be a problem.
Try the following expression:
/^([a-z0-9\s]+,)*([a-z0-9\s]+){1}$/i
This will work for:
test
test, test
test123,Test 123,test
I would strongly suggest trimming the whitespaces at the beginning and end of each item in the comma-separated list.
You seem to be lacking repetition. How about:
^(?:[a-zA-Z0-9 ]+,)*[a-zA-Z0-9 ]+$
I'm not sure how you'd express that in VB.Net, but in Python:
>>> import re
>>> x [ "123, $a67, GGG, 767", "12333, 78787&*, GH778" ]
>>> r = '^(?:[a-zA-Z0-9 ]+,)*[a-zA-Z0-9 ]+$'
>>> for s in x:
... print re.match( r, s )
...
<_sre.SRE_Match object at 0xb75c8218>
None
>>>>
You can use shortcuts instead of listing the [a-zA-Z0-9 ] part, but this is probably easier to understand.
Analyzing the highlights:
[a-zA-Z0-9 ]+ : capture one or more (but not zero) of the listed ranges, and space.
(?:[...]+,)* : In non-capturing parenthesis, match one or more of the characters, plus a comma at the end. Match such sequences zero or more times. Capturing zero times allows for no comma.
[...]+ : capture at least one of these. This does not include a comma. This is to ensure that it does not accept a trailing comma. If a trailing comma is acceptable, then the expression is easier: ^[a-zA-Z0-9 ,]+
Yes, when you want to catch comma separated things where a comma at the end is not legal, and the things match to $LONGSTUFF, you have to repeat $LONGSTUFF:
$LONGSTUFF(,$LONGSTUFF)*
If $LONGSTUFF is really long and contains comma repeated items itself etc., it might be a good idea to not build the regexp by hand and instead rely on a computer for doing that for you, even if it's just through string concatenation. For example, I just wanted to build a regular expression to validate the CPUID parameter of a XEN configuration file, of the ['1:a=b,c=d','2:e=f,g=h'] type. I... believe this mostly fits the bill: (whitespace notwithstanding!)
xend_fudge_item_re = r"""
e[a-d]x= #register of the call return value to fudge
(
0x[0-9A-F]+ | #either hardcode the reply
[10xks]{32} #or edit the bitfield directly
)
"""
xend_string_item_re = r"""
(0x)?[0-9A-F]+: #leafnum (the contents of EAX before the call)
%s #one fudge
(,%s)* #repeated multiple times
""" % (xend_fudge_item_re, xend_fudge_item_re)
xend_syntax = re.compile(r"""
\[ #a list of
'%s' #string elements
(,'%s')* #repeated multiple times
\]
$ #and nothing else
""" % (xend_string_item_re, xend_string_item_re), re.VERBOSE | re.MULTILINE)
Try ^(?!,)((, *)?([a-zA-Z0-9])\b)*$
Step by step description:
Don't match a beginning comma (good for the upcoming "loop").
Match optional comma and spaces.
Match characters you like.
The match of a word boundary make sure that a comma is necessary if more arguments are stacked in string.
Please use - ^((([a-zA-Z0-9\s]){1,45},)+([a-zA-Z0-9\s]){1,45})$
Here, I have set max word size to 45, as longest word in english is 45 characters, can be changed as per requirement
Related
I'd like to match three-character sequences of letters (only letters 'a', 'b', 'c' are allowed) separated by comma (last group is not ended with comma).
Examples:
abc,bca,cbb
ccc,abc,aab,baa
bcb
I have written following regular expression:
re.match('([abc][abc][abc],)+', "abc,defx,df")
However it doesn't work correctly, because for above example:
>>> print bool(re.match('([abc][abc][abc],)+', "abc,defx,df")) # defx in second group
True
>>> print bool(re.match('([abc][abc][abc],)+', "axc,defx,df")) # 'x' in first group
False
It seems only to check first group of three letters but it ignores the rest. How to write this regular expression correctly?
Try following regex:
^[abc]{3}(,[abc]{3})*$
^...$ from the start till the end of the string
[...] one of the given character
...{3} three time of the phrase before
(...)* 0 till n times of the characters in the brackets
What you're asking it to find with your regex is "at least one triple of letters a, b, c" - that's what "+" gives you. Whatever follows after that doesn't really matter to the regex. You might want to include "$", which means "end of the line", to be sure that the line must all consist of allowed triples. However in the current form your regex would also demand that the last triple ends in a comma, so you should explicitly code that it's not so.
Try this:
re.match('([abc][abc][abc],)*([abc][abc][abc])$'
This finds any number of allowed triples followed by a comma (maybe zero), then a triple without a comma, then the end of the line.
Edit: including the "^" (start of string) symbol is not necessary, because the match method already checks for a match only at the beginning of the string.
The obligatory "you don't need a regex" solution:
all(letter in 'abc,' for letter in data) and all(len(item) == 3 for item in data.split(','))
You need to iterate over sequence of found values.
data_string = "abc,bca,df"
imatch = re.finditer(r'(?P<value>[abc]{3})(,|$)', data_string)
for match in imatch:
print match.group('value')
So the regex to check if the string matches pattern will be
data_string = "abc,bca,df"
match = re.match(r'^([abc]{3}(,|$))+', data_string)
if match:
print "data string is correct"
Your result is not surprising since the regular expression
([abc][abc][abc],)+
tries to match a string containing three characters of [abc] followed by a comma one ore more times anywhere in the string. So the most important part is to make sure that there is nothing more in the string - as scessor suggests with adding ^ (start of string) and $ (end of string) to the regular expression.
An alternative without using regex (albeit a brute force way):
>>> def matcher(x):
total = ["".join(p) for p in itertools.product(('a','b','c'),repeat=3)]
for i in x.split(','):
if i not in total:
return False
return True
>>> matcher("abc,bca,aaa")
True
>>> matcher("abc,bca,xyz")
False
>>> matcher("abc,aaa,bb")
False
If your aim is to validate a string as being composed of triplet of letters a,b,and c:
for ss in ("abc,bbc,abb,baa,bbb",
"acc",
"abc,bbc,abb,bXa,bbb",
"abc,bbc,ab,baa,bbb"):
print ss,' ',bool(re.match('([abc]{3},?)+\Z',ss))
result
abc,bbc,abb,baa,bbb True
acc True
abc,bbc,abb,bXa,bbb False
abc,bbc,ab,baa,bbb False
\Z means: the end of the string. Its presence obliges the match to be until the very end of the string
By the way, I like the form of Sonya too, in a way it is clearer:
bool(re.match('([abc]{3},)*[abc]{3}\Z',ss))
To just repeat a sequence of patterns, you need to use a non-capturing group, a (?:...) like contruct, and apply a quantifier right after the closing parenthesis. The question mark and the colon after the opening parenthesis are the syntax that creates a non-capturing group (SO post).
For example:
(?:abc)+ matches strings like abc, abcabc, abcabcabc, etc.
(?:\d+\.){3} matches strings like 1.12.2., 000.00000.0., etc.
Here, you can use
^[abc]{3}(?:,[abc]{3})*$
^^
Note that using a capturing group is fraught with unwelcome effects in a lot of Python regex methods. See a classical issue described at re.findall behaves weird post, for example, where re.findall and all other regex methods using this function behind the scenes only return captured substrings if there is a capturing group in the pattern.
In Pandas, it is also important to use non-capturing groups when you just need to group a pattern sequence: Series.str.contains will complain that this pattern has match groups. To actually get the groups, use str.extract. and
the Series.str.extract, Series.str.extractall and Series.str.findall will behave as re.findall.
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:
I want to validate a decimal number (decimal[19,3]). I used this
#"[\d]{1,16}|[\d]{1,16}[\.]\d{1,3}"
but it didn't work.
Below are valid values:
1234567890123456.123
1234567890123456.12
1234567890123456.1
1234567890123456
1234567
0.0
.1
Simplification:
The \d doesn't have to be in []. Use [] only when you want to check whether a character is one of multiple characters or character classes.
. doesn't need to be escaped inside [] - [\.] appears to just allow ., but allowing \ to appear in the string in the place of the . may be a language dependent possibility (?). Or you can just take it out of the [] and keep it escaped.
So we get to:
\d{1,16}|\d{1,16}\.\d{1,3}
(which can be shortened using the optional / "once or not at all" quantifier (?)
to \d{1,16}(\.\d{1,3})?)
Corrections:
You probably want to make the second \d{1,16} optional, or equivalently simply make it \d{0,16}, so something like .1 is allowed:
\d{1,16}|\d{0,16}\.\d{1,3}
If something like 1. should also be allowed, you'll need to add an optional . to the first part:
\d{1,16}\.?|\d{0,16}\.\d{1,3}
Edit: I was under the impression [\d] matches \ or d, but it actually matches the character class \d (corrected above).
This would match your 3 scenarios
^(\d{1,16}|(\d{0,16}\.)?\d{1,3})$
first part: a 0 to 16 digit number
second: a 0 to 16 digit number with 1 to 3 decimals
third: nothing before a dot and then 1 to 3 decimals
the ^ and $ are anchorpoints that match start of line and end of line, so if you need to search for numbers inside lines of text, your should remove those.
Testdata:
Usage in C#
string resultString = null;
try {
resultString = Regex.Match(subjectString, #"\d{1,16}\.?|\d{0,16}\.\d{1,3}").Value;
} catch (ArgumentException ex) {
// Syntax error in the regular expression
}
Slight optimization
A bit more complicated regex, but a bit more correct would be to have the ?: notation in the "inner" group, if you are not using it, to make that a non-capture group, like this:
^(\d{1,16}|(?:\d{0,16}\.)?\d{1,3})$
Following Regex will help you out -
#"^(\d{1,16}(\.\d{1,3})?|\.\d{1,3})$"
Try something like that
(\d{0,16}\.\d{0,3})|(\d{0,16})
It work with all your examples.
edit. new version ;)
You can try:
^\d{0,16}(?:\.|$)(?:\d{0,3}|)$
match 0 to 16 digits
then match a dot or end of string
and then match 3 more digits
How can I match any characters between a period and a semi-colon in Regex, including whitespace, carriage returns, and other periods and special characters including brackets?
For example, given:
SomeMethod().AnotherMethod("argument")
.AnotherMethod(1, "arg");
I want to be able to match:
.AnotherMethod("argument")
.AnotherMethod(1, "arg");
\.([^;]*); seems to fit the bill.
Note that the question you asked is probably simpler than what you really need: presumably you want to get a similar result from SomeMethod().AnotherMethod("foo; bar; and baz").AnotherMethod(1);, but the regex you've asked for will stop at the first ;.
Either [.](.*?); or [.]([^;]*);.
[.] is just another way to escape the . char, traditionally escaped using \.
\.([^;]*); works per your original spec, but you'll quickly run into problems...
In practice you'll need named groups and a recursive regex to parse this properly.
Here's a starting point (in ruby):
re = %r{
(?<sstr> '(?:[^']|(?<=\\)')*?') {0} # single quoted
(?<dstr> "(?:[^"]|(?<=\\)")*?") {0} # double quoted
(?<nstr> [^;]*?) {0} # non-string
(?<arg> (?:\g<sstr>|\g<dstr>|\g<nstr>|)) {0} # optional arg
(?<func> \s*[a-zA-Z0-9_]+\s*\(\s*\g<arg>\s*\)\s*) {0} # foo(arg)
(?<chain> \g<func>(?:\.\g<func>)*) {0} # chained foo(arg)
^\g<chain>; # match until ;
}x
pp re.match 'SomeMethod("With (a; Funky\', arg\"ument");'
Note that the above is only a partial solution. For instance, it wouldn't cover things like:
foo(cond()? 'even' : "more", "; funky")
But it'll hopefully get you started in the right direction, if you fail to find a built-in function/argument parser in the language that you're coding into.
Sorry in advance that this might be a little challenging to read...
I'm trying to parse a line (actually a subject line from an IMAP server) that looks like this:
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?=
It's a little hard to see, but there are two =?/?= pairs in the above line. (There will always be one pair; there can theoretically be many.) In each of those =?/?= pairs, I want the third argument (as defined by a ? delimiter) extracted. (In the first pair, it's "Here is som", and in the second it's "e text.")
Here's the regex I'm using:
=\?(.+)\?.\?(.*?)\?=
I want it to return two matches, one for each =?/?= pair. Instead, it's returning the entire line as a single match. I would have thought that the ? in the (.*?), to make the * operator lazy, would have kept this from happening, but obviously it doesn't.
Any suggestions?
EDIT: Per suggestions below to replace ".?" with "[^(\?=)]?" I'm now trying to do:
=\?(.+)\?.\?([^(\?=)]*?)\?=
...but it's not working, either. (I'm unsure whether [^(\?=)]*? is the proper way to test for exclusion of a two-character sequence like "?=". Is it correct?)
Try this:
\=\?([^?]+)\?.\?(.*?)\?\=
I changed the .+ to [^?]+, which means "everything except ?"
A good practice in my experience is not to use .*? but instead do use the * without the ?, but refine the character class. In this case [^?]* to match a sequence of non-question mark characters.
You can also match more complex endmarkers this way, for instance, in this case your end-limiter is ?=, so you want to match nonquestionmarks, and questionmarks followed by non-equals:
([^?]*\?[^=])*[^?]*
At this point it becomes harder to choose though. I like that this solution is stricter, but readability decreases in this case.
One solution:
=\?(.*?)\?=\s*=\?(.*?)\?=
Explanation:
=\? # Literal characters '=?'
(.*?) # Match each character until find next one in the regular expression. A '?' in this case.
\?= # Literal characters '?='
\s* # Match spaces.
=\? # Literal characters '=?'
(.*?) # Match each character until find next one in the regular expression. A '?' in this case.
\?= # Literal characters '?='
Test in a 'perl' program:
use warnings;
use strict;
while ( <DATA> ) {
printf qq[Group 1 -> %s\nGroup 2 -> %s\n], $1, $2 if m/=\?(.*?)\?=\s*=\?(.*?)\?=/;
}
__DATA__
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?=
Running:
perl script.pl
Results:
Group 1 -> utf-8?Q?Here is som
Group 2 -> utf-8?Q?e text.
EDIT to comment:
I would use the global modifier /.../g. Regular expression would be:
/=\?(?:[^?]*\?){2}([^?]*)/g
Explanation:
=\? # Literal characters '=?'
(?:[^?]*\?){2} # Any number of characters except '?' with a '?' after them. This process twice to omit the string 'utf-8?Q?'
([^?]*) # Save in a group next characters until found a '?'
/g # Repeat this process multiple times until end of string.
Tested in a Perl script:
use warnings;
use strict;
while ( <DATA> ) {
printf qq[Group -> %s\n], $1 while m/=\?(?:[^?]*\?){2}([^?]*)/g;
}
__DATA__
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?= =?utf-8?Q?more text?=
Running and results:
Group -> Here is som
Group -> e text.
Group -> more text
Thanks for everyone's answers! The simplest expression that solved my issue was this:
=\?(.*?)\?.\?(.*?)\?=
The only difference between this and my originally-posted expression was the addition of a ? (non-greedy) operator on the first ".*". Critical, and I'd forgotten it.