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python regex find not match returned unexpected result [duplicate]

I need a regular expression able to match everything but a string starting with a specific pattern (specifically index.php and what follows, like index.php?id=2342343).

Regex: match everything but:
a string starting with a specific pattern (e.g. any - empty, too - string not starting with foo):
Lookahead-based solution for NFAs:
^(?!foo).*$
^(?!foo)
Negated character class based solution for regex engines not supporting lookarounds:
^(([^f].{2}|.[^o].|.{2}[^o]).*|.{0,2})$
^([^f].{2}|.[^o].|.{2}[^o])|^.{0,2}$
a string ending with a specific pattern (say, no world. at the end):
Lookbehind-based solution:
(?<!world\.)$
^.*(?<!world\.)$
Lookahead solution:
^(?!.*world\.$).*
^(?!.*world\.$)
POSIX workaround:
^(.*([^w].{5}|.[^o].{4}|.{2}[^r].{3}|.{3}[^l].{2}|.{4}[^d].|.{5}[^.])|.{0,5})$
([^w].{5}|.[^o].{4}|.{2}[^r].{3}|.{3}[^l].{2}|.{4}[^d].|.{5}[^.]$|^.{0,5})$
a string containing specific text (say, not match a string having foo):
Lookaround-based solution:
^(?!.*foo)
^(?!.*foo).*$
POSIX workaround:
Use the online regex generator at www.formauri.es/personal/pgimeno/misc/non-match-regex
a string containing specific character (say, avoid matching a string having a | symbol):
^[^|]*$
a string equal to some string (say, not equal to foo):
Lookaround-based:
^(?!foo$)
^(?!foo$).*$
POSIX:
^(.{0,2}|.{4,}|[^f]..|.[^o].|..[^o])$
a sequence of characters:
PCRE (match any text but cat): /cat(*SKIP)(*FAIL)|[^c]*(?:c(?!at)[^c]*)*/i or /cat(*SKIP)(*FAIL)|(?:(?!cat).)+/is
Other engines allowing lookarounds: (cat)|[^c]*(?:c(?!at)[^c]*)* (or (?s)(cat)|(?:(?!cat).)*, or (cat)|[^c]+(?:c(?!at)[^c]*)*|(?:c(?!at)[^c]*)+[^c]*) and then check with language means: if Group 1 matched, it is not what we need, else, grab the match value if not empty
a certain single character or a set of characters:
Use a negated character class: [^a-z]+ (any char other than a lowercase ASCII letter)
Matching any char(s) but |: [^|]+
Demo note: the newline \n is used inside negated character classes in demos to avoid match overflow to the neighboring line(s). They are not necessary when testing individual strings.
Anchor note: In many languages, use \A to define the unambiguous start of string, and \z (in Python, it is \Z, in JavaScript, $ is OK) to define the very end of the string.
Dot note: In many flavors (but not POSIX, TRE, TCL), . matches any char but a newline char. Make sure you use a corresponding DOTALL modifier (/s in PCRE/Boost/.NET/Python/Java and /m in Ruby) for the . to match any char including a newline.
Backslash note: In languages where you have to declare patterns with C strings allowing escape sequences (like \n for a newline), you need to double the backslashes escaping special characters so that the engine could treat them as literal characters (e.g. in Java, world\. will be declared as "world\\.", or use a character class: "world[.]"). Use raw string literals (Python r'\bworld\b'), C# verbatim string literals #"world\.", or slashy strings/regex literal notations like /world\./.

You could use a negative lookahead from the start, e.g., ^(?!foo).*$ shouldn't match anything starting with foo.

You can put a ^ in the beginning of a character set to match anything but those characters.
[^=]*
will match everything but =

Just match /^index\.php/, and then reject whatever matches it.

In Python:
>>> import re
>>> p='^(?!index\.php\?[0-9]+).*$'
>>> s1='index.php?12345'
>>> re.match(p,s1)
>>> s2='index.html?12345'
>>> re.match(p,s2)
<_sre.SRE_Match object at 0xb7d65fa8>

Came across this thread after a long search. I had this problem for multiple searches and replace of some occurrences. But the pattern I used was matching till the end. Example below
import re
text = "start![image]xxx(xx.png) yyy xx![image]xxx(xxx.png) end"
replaced_text = re.sub(r'!\[image\](.*)\(.*\.png\)', '*', text)
print(replaced_text)
gave
start* end
Basically, the regex was matching from the first ![image] to the last .png, swallowing the middle yyy
Used the method posted above https://stackoverflow.com/a/17761124/429476 by Firish to break the match between the occurrence. Here the space is not matched; as the words are separated by space.
replaced_text = re.sub(r'!\[image\]([^ ]*)\([^ ]*\.png\)', '*', text)
and got what I wanted
start* yyy xx* end

Regexp - match pattern except if it's part of another pattern [duplicate]

I need a regular expression able to match everything but a string starting with a specific pattern (specifically index.php and what follows, like index.php?id=2342343).

Regex: match everything but:
a string starting with a specific pattern (e.g. any - empty, too - string not starting with foo):
Lookahead-based solution for NFAs:
^(?!foo).*$
^(?!foo)
Negated character class based solution for regex engines not supporting lookarounds:
^(([^f].{2}|.[^o].|.{2}[^o]).*|.{0,2})$
^([^f].{2}|.[^o].|.{2}[^o])|^.{0,2}$
a string ending with a specific pattern (say, no world. at the end):
Lookbehind-based solution:
(?<!world\.)$
^.*(?<!world\.)$
Lookahead solution:
^(?!.*world\.$).*
^(?!.*world\.$)
POSIX workaround:
^(.*([^w].{5}|.[^o].{4}|.{2}[^r].{3}|.{3}[^l].{2}|.{4}[^d].|.{5}[^.])|.{0,5})$
([^w].{5}|.[^o].{4}|.{2}[^r].{3}|.{3}[^l].{2}|.{4}[^d].|.{5}[^.]$|^.{0,5})$
a string containing specific text (say, not match a string having foo):
Lookaround-based solution:
^(?!.*foo)
^(?!.*foo).*$
POSIX workaround:
Use the online regex generator at www.formauri.es/personal/pgimeno/misc/non-match-regex
a string containing specific character (say, avoid matching a string having a | symbol):
^[^|]*$
a string equal to some string (say, not equal to foo):
Lookaround-based:
^(?!foo$)
^(?!foo$).*$
POSIX:
^(.{0,2}|.{4,}|[^f]..|.[^o].|..[^o])$
a sequence of characters:
PCRE (match any text but cat): /cat(*SKIP)(*FAIL)|[^c]*(?:c(?!at)[^c]*)*/i or /cat(*SKIP)(*FAIL)|(?:(?!cat).)+/is
Other engines allowing lookarounds: (cat)|[^c]*(?:c(?!at)[^c]*)* (or (?s)(cat)|(?:(?!cat).)*, or (cat)|[^c]+(?:c(?!at)[^c]*)*|(?:c(?!at)[^c]*)+[^c]*) and then check with language means: if Group 1 matched, it is not what we need, else, grab the match value if not empty
a certain single character or a set of characters:
Use a negated character class: [^a-z]+ (any char other than a lowercase ASCII letter)
Matching any char(s) but |: [^|]+
Demo note: the newline \n is used inside negated character classes in demos to avoid match overflow to the neighboring line(s). They are not necessary when testing individual strings.
Anchor note: In many languages, use \A to define the unambiguous start of string, and \z (in Python, it is \Z, in JavaScript, $ is OK) to define the very end of the string.
Dot note: In many flavors (but not POSIX, TRE, TCL), . matches any char but a newline char. Make sure you use a corresponding DOTALL modifier (/s in PCRE/Boost/.NET/Python/Java and /m in Ruby) for the . to match any char including a newline.
Backslash note: In languages where you have to declare patterns with C strings allowing escape sequences (like \n for a newline), you need to double the backslashes escaping special characters so that the engine could treat them as literal characters (e.g. in Java, world\. will be declared as "world\\.", or use a character class: "world[.]"). Use raw string literals (Python r'\bworld\b'), C# verbatim string literals #"world\.", or slashy strings/regex literal notations like /world\./.

You could use a negative lookahead from the start, e.g., ^(?!foo).*$ shouldn't match anything starting with foo.

You can put a ^ in the beginning of a character set to match anything but those characters.
[^=]*
will match everything but =

Just match /^index\.php/, and then reject whatever matches it.

In Python:
>>> import re
>>> p='^(?!index\.php\?[0-9]+).*$'
>>> s1='index.php?12345'
>>> re.match(p,s1)
>>> s2='index.html?12345'
>>> re.match(p,s2)
<_sre.SRE_Match object at 0xb7d65fa8>

Came across this thread after a long search. I had this problem for multiple searches and replace of some occurrences. But the pattern I used was matching till the end. Example below
import re
text = "start![image]xxx(xx.png) yyy xx![image]xxx(xxx.png) end"
replaced_text = re.sub(r'!\[image\](.*)\(.*\.png\)', '*', text)
print(replaced_text)
gave
start* end
Basically, the regex was matching from the first ![image] to the last .png, swallowing the middle yyy
Used the method posted above https://stackoverflow.com/a/17761124/429476 by Firish to break the match between the occurrence. Here the space is not matched; as the words are separated by space.
replaced_text = re.sub(r'!\[image\]([^ ]*)\([^ ]*\.png\)', '*', text)
and got what I wanted
start* yyy xx* end

Compiling user defined regex object and searching/matching in Python [duplicate]

This question already has answers here:
How to create raw string from string variable in python?
(3 answers)
Closed 4 years ago.
I am writing an online regex checker, which takes input from the user in the form of a pattern, and flags, and uses that to compile a regex object. The regex object is then used to check if the test string matches within the format provided by the regex pattern or not. As of this moment, the compile function looks like this:
class RegexObject:
...
def compile(self):
flags = ''
if self.multiline_flag:
flags = re.M
if self.dotall_flag:
flags |= re.S
if self.verbose_flag:
flags |= re.X
if self.ignorecase_flag:
flags |= re.I
if self.unicode_flag:
flags |= re.U
regex = re.compile(self.pattern, flags)
return regex
Please note, the self.pattern and all the flags are class attributes defined by the user using a simple form. However, one thing I noticed in the docs is that there is usually an r before the pattern in the compile functions, like this:
re.compile(r'(?<=abc)def')
How do I place that r in my code before my variable name? Also, if I want to tell the user if the test input is valid or not, should I be using the match method, or the search method?
Thanks for any help.
Edit: This question is not a duplicate of this one, because that question has nothing to do with regular expressions.

Don't worry about the r, you don't need it here.
The r stands for "raw", not "regex". In an r string, you can put backslashes without escaping them. R strings are often used in regexes because there are often many backslashes in regexes. Escaping the backslashes can be annoying. See this shell output:
>>> s = r"\a"
>>> s2 = "\a"
>>> s2
'\x07'
>>> s
'\\a'
And you should use search, as match only looks at the start of the string. Look at the docs.
re.search(pattern, string, flags=0)
Scan through string looking for
the first location where the regular expression pattern produces a
match, and return a corresponding match object. Return None if no
position in the string matches the pattern; note that this is
different from finding a zero-length match at some point in the
string.
re.match(pattern, string, flags=0)
If zero or more characters at the
beginning of string match the regular expression pattern, return a
corresponding match object. Return None if the string does not match
the pattern; note that this is different from a zero-length match.
Note that even in MULTILINE mode, re.match() will only match at the
beginning of the string and not at the beginning of each line.
If you want to locate a match anywhere in string, use search() instead
(see also search() vs. match()).

You need not use r.Instead you should use re.escape.match or search again should be user input.

How to ignore \n in regular expressions in python?

So i have a regex telling if a number is integer.
regex = '^(0|[1-9][0-9]*)$'
import re
bool(re.search(regex, '42\n'))
returns True, and it is not supposed to?
Where does the problem come from ?

From the documentation:
'$'
Matches the end of the string or just before the newline at the end of the string
Try \Z instead.
Also, any time you find yourself writing a regular expression that starts with ^ or \A and ends with $ or \Z, if your intent is to only match the entire string, you should probably use re.fullmatch() instead of re.search() (and omit the boundary markers from the regex). Or if you're using a version of Python that's too old to have re.fullmatch(), (you really need to upgrade but) you can use re.match() and omit the beginning-of-string boundary marker.

regex ahould be regex = '\b^(0|[1-9][0-9]*)$\b'

The regex in the question matches ->start of line, numbers and end of line. And the given string matches that, thats why it is returning true. If you want it to return False when there is a number present, you can use "!" to indicate NOT.
Refer https://docs.python.org/2/library/re.html
regex = '!(0|[1-9][0-9]*)$'
bool(re.search(regex, '42\n')) => (Returns false)

Yeah, that $ matching one \n before the end is kind of trap/inconsistency. Check out my list of regex traps for python: http://www.cofoh.com/advanced-regex-tutorial-python/traps

Using anchors in python regex to get exact match

I need to validate a version number consisting of 'v' plus positive int, and nothing else
eg "v4", "v1004"
I have
import re
pattern = "\Av(?=\d+)\W"
m = re.match(pattern, "v303")
if m is None:
print "noMatch"
else:
print "match"
But this doesn't work! Removing the \A and \W will match for v303 but will also match for v30G, for example
Thanks

Pretty straightforward. First, put anchors on your pattern:
"^patternhere$"
Now, let's put together the pattern:
"^v\d+$"
That should do it.

I think you may want \b (word boundary) rather than \A (start of string) and \W (non word character), also you don't need to use lookahead (the (?=...)).
Try: "\bv(\d+)" if you need to capture the int, "\bv\d+" if you don't.
Edit: You probably want to use raw string syntax for Python regexes, r"\bv\d+\b", since "\b" is a backspace character in a regular string.
Edit 2: Since + is "greedy", no trailing \b is necessary or desired.

Simply use
\bv\d+\b
Or enclosed it with ^\bv\d+\b$
to match it entirely..