I need a regex that will only find matches where the entire string matches my query.
For instance if I do a search for movies with the name "Red October" I only want to match on that exact title (case insensitive) but not match titles like "The Hunt For Red October". Not quite sure I know how to do this. Anyone know?
Thanks!
Try the following regular expression:
^Red October$
By default, regular expressions are case sensitive. The ^ marks the start of the matching text and $ the end.
Generally, and with default settings, ^ and $ anchors are a good way of ensuring that a regex matches an entire string.
A few caveats, though:
If you have alternation in your regex, be sure to enclose your regex in a non-capturing group before surrounding it with ^ and $:
^foo|bar$
is of course different from
^(?:foo|bar)$
Also, ^ and $ can take on a different meaning (start/end of line instead of start/end of string) if certain options are set. In text editors that support regular expressions, this is usually the default behaviour. In some languages, especially Ruby, this behaviour cannot even be switched off.
Therefore there is another set of anchors that are guaranteed to only match at the start/end of the entire string:
\A matches at the start of the string.
\Z matches at the end of the string or before a final line break.
\z matches at the very end of the string.
But not all languages support these anchors, most notably JavaScript.
I know that this may be a little late to answer this, but maybe it will come handy for someone else.
Simplest way:
var someString = "...";
var someRegex = "...";
var match = Regex.Match(someString , someRegex );
if(match.Success && match.Value.Length == someString.Length){
//pass
} else {
//fail
}
Use the ^ and $ modifiers to denote where the regex pattern sits relative to the start and end of the string:
Regex.Match("Red October", "^Red October$"); // pass
Regex.Match("The Hunt for Red October", "^Red October$"); // fail
You need to enclose your regex in ^ (start of string) and $ (end of string):
^Red October$
If the string may contain regex metasymbols (. { } ( ) $ etc), I propose to use
^\QYourString\E$
\Q starts quoting all the characters until \E.
Otherwise the regex can be unappropriate or even invalid.
If the language uses regex as string parameter (as I see in the example), double slash should be used:
^\\QYourString\\E$
Hope this tip helps somebody.
Sorry, but that's a little unclear.
From what i read, you want to do simple string compare. You don't need regex for that.
string myTest = "Red October";
bool isMatch = (myTest.ToLower() == "Red October".ToLower());
Console.WriteLine(isMatch);
isMatch = (myTest.ToLower() == "The Hunt for Red October".ToLower());
You can do it like this Exemple if i only want to catch one time the letter minus a in a string and it can be check with myRegex.IsMatch()
^[^e][e]{1}[^e]$
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
I have a working regex under vim: /^ \{-}\a.*$\n
I implement a global search and replace as :%s/^ \{-}\a.*$\n//
This works great -- removes all lines that start with any number of spaces (matched non-greedily), followed by a letter and anything else to the end of the line including the newline.
I cannot (to save my soul) figure out the analogous regex in Python. Here's what make sense to me:
x = re.sub("^ *?\a.$\n","",y)
But this doesn't do anything.
Many thanks for your sagacious replies.
\a means the bell character (0x07) in Python, and $\n is a redundant bad idea, so:
x = re.sub(r"^ *[A-Za-z].*\n","",y)
Also, there's no reason to write ' *?' instead of ' *' here, as it's always going to be followed by a non-space if it's matching.
If you want to match any number of whitespace, you can also use the \s sequence.
Any letter will be matched by the [a-zA-Z] character class. You also don't need to use the $ and the \n, either will do.
Suggest the following:
x = re.sub(r"^\s*[a-zA-Z].*(\r|\n)","",y)
If you want at least one whitespace, use \s+ instead of \s*
Im trying to find c style comments in a c file but im having trouble if there happens to be // inside of quotations. This is the file:
/*My function
is great.*/
int j = 0//hello world
void foo(){
//tricky example
cout << "This // is // not a comment\n";
}
it will match with that cout. This is what i have so far (i can match the /**/ comments already)
fp = open(s)
p = re.compile(r'//(.+)')
txt = p.findall(fp.read())
print (txt)
The first step is to identify cases where // or /* must not be interpreted as the begining of a comment substring. For example when they are inside a string (between quotes). To avoid content between quotes (or other things), the trick is to put them in a capture group and to insert a backreference in the replacement pattern:
pattern:
(
"(?:[^"\\]|\\[\s\S])*"
|
'(?:[^'\\]|\\[\s\S])*'
)
|
//.*
|
/\*(?:[^*]|\*(?!/))*\*/
replacement:
\1
online demo
Since quoted parts are searching first, each time you find // or /*...*/, you can be sure that your are not inside a string.
Note that the pattern is voluntary inefficient (due to (A|B)* subpatterns) to make it easier to understand. To make it more efficient you can rewrite it like this:
("(?=((?:[^"\\]+|\\[\s\S])*))\2"|'(?=((?:[^'\\]+|\\[\s\S])*))\3')|//.*|/\*(?=((?:[^*]+|\*(?!/))*))\4\*/
(?=(something+))\1 is only a way to emulate an atomic group (?>something+)
online demo
So, If you only want to find comments (and not to remove them), the most handy is to put the comments part of the pattern in capture group and to test if it isn't empty. The following pattern has been udapted (after Jonathan Leffler comment) to handle the trigraph ??/ that is interpreted as a backslash character by the preprocessor (I assume that the code isn't written for the -trigraphs option) and to handle the backslash followed by a newline character that allows to format a single line on several lines:
fp = open(s)
p = re.compile(r'''(?x)
(?=["'/]) # trick to make it faster, a kind of anchor
(?:
"(?=((?:[^"\\?]+|\?(?!\?/)|(?:\?\?/|\\)[\s\S])*))\1" # double quotes string
|
'(?=((?:[^'\\?]+|\?(?!\?/)|(?:\?\?/|\\)[\s\S])*))\2' # single quotes string
|
(
/(?:(?:\?\?/|\\)\n)*/(?:.*(?:\?\?|\\)/\n)*.* # single line comment
|
/(?:(?:\?\?/|\\)\n)*\* # multiline comment
(?=((?:[^*]+|\*+(?!(?:(?:\?\?/|\\)\n)*/))*))\4
\*(?:(?:\?\?/|\\)\n)*/
)
)
''')
for m in p.findall(fp.read()):
if (m[2]):
print m[2]
These changes would not affect the pattern efficiency since the main work for the regex engine is to find positions that begin with a quote or a slash. This task is simplify by the presence of a lookahead at the begining of the pattern (?=["'/]) that allows internals optimizations to quickly find the first character.
An other optimization is the use of emulated atomic groups, that reduces the backtracking to the minimum and allows to use greedy quantifiers inside repeated groups.
NB: a chance there is no heredoc syntax in C!
Python's re.findall method basically works the same way as most lexers do: it successively returns the longest match starting where the previous match finished. All that is required is to produce a disjunction of all the lexical patterns:
(<pattern 1>)|(<pattern 2>)|...|(<pattern n>)
Unlike most lexers, it doesn't require the matches to be contiguous, but that's not a significant difference since you can always just add (.) as the last pattern, in order to match all otherwise unmatched characters individually.
An important feature of re.findall is that if the regex has any groups, then only the groups will be returned. Consequently, you can exclude alternatives by simply leaving out the parentheses, or changing them to non-capturing parentheses:
(<pattern 1>)|(?:<unimportant pattern 2>)|(<pattern 3)
With that in mind, let's take a look at how to tokenize C just enough to recognize comments. We need to deal with:
Single-line comments: // Comment
Multi-line comments: /* Comment */
Double-quoted string: "Might include escapes like \n"
Single-quoted character: '\t'
(See below for a few more irritating cases)
With that in mind, let's create regexen for each of the above.
Two slashes followed by anything other than a newline: //[^\n]*
This regex is tedious to explain: /*[^*]*[*]+(?:[^/*][^*]*[*]+)*/
Note that it uses (?:...) to avoid capturing the repeated group.
A quote, any repetition of a character other than quote and backslash, or a backslash followed by any character whatsoever. That's not a precise definition of an escape sequence, but it's good enough to detect when a " terminates the string, which is all we care about: "(?:[^"\\]|\\.*)"
The same as (3) but with single quotes: '(?:[^'\\]|\\.)*'
Finally, the goal was to find the text of C-style comments. So we just need to avoid captures in any of the other groups. Hence:
p = re.compile('|'.join((r"(//[^\n])*"
,r"/*[^*]*[*]+(?:[^/*][^*]*[*]+)*/"
,'"'+r"""(?:[^"\\]|\\.)*"""+'"'
,r"'(?:[^'\\]|\\.)*'")))
return [c[2:] for c in p.findall(text) if c]
Above, I left out some obscure cases which are unlikely to arise:
In an #include <...> directive, the <...> is essentially a string. In theory, it could contain quotes or sequences which look like comments, but in practice you will never see:
#include </*This looks like a comment but it is a filename*/>
A line which ends with \ is continued on the next line; the \ and following newline character are simply removed from the input. This happens before any lexical scanning is performed, so the following is a perfectly legal comment (actually two comments):
/\
**************** Surprise! **************\
//////////////////////////////////////////
To make the above worse, the trigraph ??/ is the same as a \, and that replacement happens before the continuation handling.
/************************************//??/
**************** Surprise! ************??/
//////////////////////////////////////////
Outside of obfuscation contests, no-one actually uses trigraphs. But they're still in the standard. The easiest way to deal with both of these issues would be to prescan the string:
return [c[2:]
for c in p.findall(text.replace('//?','\\').replace('\\\n',''))
if c]
The only way to deal with the #include <...> issue, if you really cared about it, would be to add one more pattern, something like #define\s*<[^>\n]*>.
I'm trying to add some light markdown support for a javascript preprocessor which I'm writing in Python.
For the most part it's working, but sometimes the regex I'm using is acting a little odd, and I think it's got something to do with raw-strings and escape sequences.
The regex is: (?<!\\)\"[^\"]+\"
Yes, I am aware that it only matches strings beginning with a " character. However, this project is born out of curiosity more than anything, so I can live with it for now.
To break it down:
(?<\\)\" # The group should begin with a quotation mark that is not escaped
[^\"]+ # and match any number of at least one character that is not a quotation mark (this is the biggest problem, I know)
\" # and end at the first quotation mark it finds
That being said, I (obviously) start hitting problems with things like this:
"This is a string with an \"escaped quote\" inside it"
I'm not really sure how to say "Everything but a quotation mark, unless that mark is escaped". I tried:
([^\"]|\\\")+ # a group of anything but a quote or an escaped quote
, but that lead to very strange results.
I'm fully prepared to hear that I'm going about this all wrong. For the sake of simplicity, let's say that this regex will always start and end with double quotes (") to avoid adding another element in the mix. I really want to understand what I have so far.
Thanks for any assistance.
EDIT
As a test for the regex, I'm trying to find all string literals in the minified jQuery script with the following code (using the unutbu's pattern below):
STRLIT = r'''(?x) # verbose mode
(?<!\\) # not preceded by a backslash
" # a literal double-quote
.*? # non-greedy 1-or-more characters
(?<!\\) # not preceded by a backslash
" # a literal double-quote
'''
f = open("jquery.min.js","r")
jq = f.read()
f.close()
literals = re.findall(STRLIT,jq)
The answer below fixes almost all issues. The ones that do arise are within jquery's own regular expressions, which is a very edge case. The solution no longer misidentifies valid javascript as markdown links, which was really the goal.
I think I first saw this idea in... Jinja2's source code? Later transplanted it to Mako.
r'''(\"\"\"|\'\'\'|\"|\')((?<!\\)\\\1|.)*?\1'''
Which does the following:
(\"\"\"|\'\'\'|\"|\') matches a Python opening quote, because this happens to be taken from code for parsing Python. You probably don't need all those quote types.
((?<!\\)\\\1|.) matches: EITHER a matching quote that was escaped ONLY ONCE, OR any other character. So \\" will still be recognized as the end of the string.
*? non-greedily matches as many of those as possible.
And \1 is just the closing quote.
Alas, \\\" will still incorrectly be detected as the end of the string. (The template engines only use this to check if there is a string, not to extract it.) This is a problem very poorly suited for regular expressions; short of doing insane things in Perl, where you can embed real code inside a regex, I'm not sure it's possible even with PCRE. Though I'd love to be proven wrong. :) The killer is that (?<!...) has to be constant-length, but you want to check that there's any even number of backslashes before the closing quote.
If you want to get this correct, and not just mostly-correct, you might have to use a real parser. Have a look at parsley, pyparsing, or any of these tools.
edit: By the way, there's no need to check that the opening quote doesn't have a backslash before it. That's not valid syntax outside a string in JS (or Python).
Perhaps use two negative look behinds:
import re
text = r'''"This is a string with an \"escaped quote\" inside it". While ""===r?+r:wt.test(r)?st.parseJSON(r) :r}catch(o){}st.data(e,n,r)}else r=t}return r}function s(e){var t;for(t in e)if(("data" '''
for match in (re.findall(r'''(?x) # verbose mode
(?<!\\) # not preceded by a backslash
" # a literal double-quote
.*? # 1-or-more characters
(?<!\\) # not preceded by a backslash
" # a literal double-quote
''', text)):
print(match)
yields
"This is a string with an \"escaped quote\" inside it"
""
"data"
The question mark in .+? makes the pattern non-greedy. The non-greediness causes the pattern to match when it encounters the first unescaped double quotation mark.
Using python, the correct regex matching double quoted string is:
pattern = r'"(\.|[^"])*"'
It describes strings starts and ends with ". For each character inside the two double quotes, it's either an escaped character OR any character expect ".
unutbu's ansever is wrong because for valid string "\\\\", cannot matched by that pattern.
I have this weirdly formatted URL. I have to extract the contents in '()'.
Sample URL : http://sampleurl.com/(K(ThinkCode))/profile/view.aspx
If I can extract ThinkCode out of it, I will be a happy man! I am having a tough time with regexing special chars like '(' and '/'.
>>> foo = re.compile( r"(?<=\(K\()[^\)]*" )
>>> foo.findall( r"http://sampleurl.com/(K(ThinkCode))/profile/view.aspx" )
['ThinkCode']
Explanation
In regex-world, a lookbehind is a way of saying "I want to match ham, but only if it's preceded by spam. We write this as (?<=spam)ham. So in this case, we want to match [^\)]*, but only if it's preceded by \(K\(.
Now \(K\( is a nice, easy regex, because it's plain text! It means, match exactly the string (K(. Notice that we have to escape the brackets (by putting \ in front of them), since otherwise the regex parser would think they were part of the regex instead of a character to match!
Finally, when you put something in square brackets in regex-world, it means "any of the characters in here is OK". If you put something inside square brackets where the first character is ^, it means "any character not in here is OK". So [^\)] means "any character that isn't a right-bracket", and [^\)]* means "as many characters as possible that aren't right-brackets".
Putting it all together, (?<=\(K\()[^\)]* means "match as many characters as you can that aren't right-brackets, preceded by the string (K(.
Oh, one last thing. Because \ means something inside strings in Python as well as inside regexes, we use raw strings -- r"spam" instead of just "spam". That tells Python to ignore the \'s.
Another way
If lookbehind is a bit complicated for you, you can also use capturing groups. The idea behind those is that the regex matches patterns, but can also remember subpatterns. That means that you don't have to worry about lookaround, because you can match the entire pattern and then just extract the subpattern inside it!
To capture a group, simply put it inside brackets: (foo) will capture foo as the first group. Then, use .groups() to spit out all the groups that you matched! This is the way the other answer works.
It's not too hard, especially since / isn't actually a special character in Python regular expressions. You just backslash the literal parens you want. How about this:
s = "http://sampleurl.com/(K(ThinkCode))/profile/view.aspx"
mo = re.match(r"http://sampleurl\.com/\(K\(([^)]+)\)\)/profile.view\.aspx", s);
print mo.group(1)
Note the use of r"" raw strings to preserve the backslashes in the regular expression pattern string.
If you want to have special characters in a regex, you need to escape them, such as \(, \/, \\.
Matching things inside of nested parenthesis is quite a bit of a pain in regex. if that format is always the same, you could use this:
\(.*?\((.*?)\).*?\)
Basically: find a open paren, match characters until you find another open paren, group characters until I see a close paren, then make sure there are two more close paren somewhere in there.
mystr = "http://sampleurl.com/(K(ThinkCode))/profile/view.aspx"
import re
re.sub(r'^.*\((\w+)\).*',r'\1',mystr)