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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

Can I mix character classes in Python RegEx?

Special sequences (character classes) in Python RegEx are escapes like \w or \d that matches a set of characters.
In my case, I need to be able to match all alpha-numerical characters except numbers.
That is, \w minus \d.
I need to use the special sequence \w because I'm dealing with non-ASCII characters and need to match symbols like "Æ" and "Ø".
One would think I could use this expression: [\w^\d] but it doesn't seem to match anything and I'm not sure why.
So in short, how can I mix (add/subtract) special sequences in Python Regular Expressions?
EDIT: I accidentally used [\W^\d] instead of [\w^\d]. The latter does indeed match something, including parentheses and commas which are not alpha-numerical characters as far as I'm concerned.

You can use r"[^\W\d]", ie. invert the union of non-alphanumerics and numbers.

You cannot subtract character classes, no.
Your best bet is to use the regex project, which offers additional functionality while remaining backwards compatible with the re module in in the standard library. It supports character classes based on Unicode properties:
\p{IsAlphabetic}
This will match any character that the Unicode specification states is an alphabetic character.
Even better, regex does support character class subtraction; it views such classes as sets and allows you to create a difference with the -- operator:
[\w--\d]
matches everything in \w except anything that also matches \d.

You can exclude classes using a negative lookahead assertion, such as r'(?!\d)[\w]' to match a word character, excluding digits. For example:
>>> re.search(r'(?!\d)[\w]', '12bac')
<_sre.SRE_Match object at 0xb7779218>
>>> _.group(0)
'b'
To exclude more than one group, you can use the usual [...] syntax in the lookahead assertion, for example r'(?![0-5])[\w]' would match any alphanumeric character except for digits 0-5.
As with [...], the above construct matches a single character. To match multiple characters, add a repetition operator:
>>> re.search(r'((?!\d)[\w])+', '12bac15')
<_sre.SRE_Match object at 0x7f44cd2588a0>
>>> _.group(0)
'bac'

I don't think you can directly combine (boolean and) character sets in a single regex, whether one is negated or not. Otherwise you could simply have combined [^\d] and \w.
Note: the ^ has to be at the start of the set, and applies to the whole set. From the docs: "If the first character of the set is '^', all the characters that are not in the set will be matched.".
Your set [\w^\d] tries to match an alpha-numerical character, followed by a caret, followed by a digit. I can imagine that doesn't match anything either.
I would do it in two steps, effectly combining the regular expressions. First match by non-digits (inner regex), then match by alpha-numerical characters:
re.search('\w+', re.search('([^\d]+)', s).group(0)).group(0)
or variations to this theme.
Note that would need to surround this with a try: except: block, as it will throw an AttributeError: 'NoneType' object has no attribute 'group' in case one of the two regexes fails. But you can, of course, split this single line up in a few more lines.

Regular expression for letters, dash, underscore, numbers, and space

This is my attempt
def matcher(ex):
if re.match(r'^[\w|\d][A-Za-z0-9_-]+$', ex):
print 'yes'
My goal is to match only submission that satisfy all the followings
begins with only a letter or a numeric digit, and
only letter, space, dash, underscore and numeric digit are allowed
all ending spaces are stripped
In my regex, matcher('__') is considered valid. How can I modify to achieve what I want really want? I believe \w also includes underscore. But matcher('_') is not matched...

def matcher(ex):
ex = ex.rstrip()
if re.match(r'^[a-zA-Z0-9][ A-Za-z0-9_-]*$', ex):
print 'yes'
Problems in your original regex:
| doesn't mean alternation in a character class, it means a pipe character literally.
You used + for your following characters, meaning one or more, so a one-character string like '_' wouldn't match.
You used \w in your first character, which accepted underscores.

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..

Looking for a regular expression including alphanumeric + "&" and ";"

Here's the problem:
split=re.compile('\\W*')
This regular expression works fine when dealing with regular words, but there are occasions where I need the expression to include words like käyttäj&aml;auml;.
What should I add to the regex to include the & and ; characters?

I would treat the entities as a unit (since they also can contain numerical character codes), resulting in the following regular expression:
(\w|&(#(x[0-9a-fA-F]+|[0-9]+)|[a-z]+);)+
This matches
either a word character (including “_”), or
an HTML entity consisting of
the character “&”,
the character “#”,
the character “x” followed by at least one hexadecimal digit, or
at least one decimal digit, or
at least one letter (= named entity),
a semicolon
at least once.
/EDIT: Thanks to ΤΖΩΤΖΙΟΥ for pointing out an error.

You probably want to take the problem reverse, i.e. finding all the character without the spaces:
[^ \t\n]*
Or you want to add the extra characters:
[a-zA-Z0-9&;]*
In case you want to match HTML entities, you should try something like:
(\w+|&\w+;)*

you should make a character class that would include the extra characters. For example:
split=re.compile('[\w&;]+')
This should do the trick. For your information
\w (lower case 'w') matches word characters (alphanumeric)
\W (capital W) is a negated character class (meaning it matches any non-alphanumeric character)
* matches 0 or more times and + matches one or more times, so * will match anything (even if there are no characters there).

Looks like this RegEx did the trick:
split=re.compile('(\\\W+&\\\W+;)*')
Thanks for the suggestions. Most of them worked fine on Reggy, but I don't quite understand why they failed with re.compile.