How can I "block comment" SQL statements in Notepad++?
For example:
CREATE TABLE gmr_virtuemart_calc_categories (
id int(1) UNSIGNED NOT NULL,
virtuemart_calc_id int(1) UNSIGNED NOT NULL DEFAULT '0',
virtuemart_category_id int(1) UNSIGNED NOT NULL DEFAULT '0'
) ENGINE=MyISAM DEFAULT CHARSET=utf8;
It should be wrapped with /* at the start and */ at the end using regex in Notepad++ to produce:
/*CREATE TABLE ... (...) ENGINE=MyISAM DEFAULT CHARSET=utf8;*/
You only offer one sample input, so I am forced to build the pattern literally. If this pattern isn't suitable because there are alternative queries and/or other interfering text, then please update your question.
Tick the "Match case" box.
Find what: (CREATE[^;]+;) Replace with: /*$1*/
Otherwise, you can use this for sql query blocks that start with a capital and end in semicolon:
Find what: ([A-Z][^;]+;) Replace with: /*$1*/
To improve accuracy, you might include ^ start of line anchors or add \r\n after the semi-colon or match the CHARSET portion before the semi-colon. There are several adjustments that can be made. I cannot be confident of accuracy without knowing more about the larger body of text.
You could use a recursive regex.
I think NP uses boost or PCRE.
This works with both.
https://regex101.com/r/P75bXC/1
Find (?s)(CREATE\s+TABLE[^(]*(\((?:[^()']++|'.*?'|(?2))*\))(?:[^;']|'.*?')*;)
Replace /*$1*/
Explained
(?s) # Dot-all modifier
( # (1 start) The whole match
CREATE \s+ TABLE [^(]* # Create statement
( # (2 start), Recursion code group
\(
(?: # Cluster group
[^()']++ # Possesive, not parenth's or quotes
| # or,
' .*? ' # Quotes (can wrap in atomic group if need be)
| # or,
(?2) # Recurse to group 2
)* # End cluster, do 0 to many times
\)
) # (2 end)
# Trailer before colon statement end
(?: # Cluster group, can be atomic (?> ) if need be
[^;'] # Not quote or colon
| # or,
' .*? ' # Quotes
)* # End cluster, do 0 to many times
; # Colon at the end
) # (1 end)
Related
I have a Python script in which I'm trying to parse a string of the form:
one[two=three].four
Each word should be in its own capture group. The punctuation should not be captured.
Additionally, each part of the string is optional, and the part delimited by brackets can be repeated. So the above is the most complete example, but all of the following should also be valid matches:
one
.four
one[two=three][five=six]
[two=three]
[two].four
[two][five]
[]
In the case that one of the words is not present, instead of failing to capture, I'd like to capture a string of length 0.
The regex that I'm using is as follows:
pattern = re.compile(
r"""
^ # Assert start of string
(?P<cap1> # Start a new group for "one"
[a-z]* #
) #
(?: # Start a group for "two" and "three"
\[ # Match the "["
(?P<cap_2> # Start a group for "two"
[a-z]* #
) #
=? # Delimit two/three with "="
(?P<cap_3> # Start a group for "three"
[a-z]* #
) #
\] # Match the "]"
)* # End the two-three group, allowing repeats
\.? # Delimit three/four with "."
(?P<cap_4> # Begin a group for "four"
[a-z]* #
) #
$ # Assert end of string
""", re.IGNORECASE|re.VERBOSE)
What I've tried to do during that regex is, instead of allowing 0 or 1 of a group by appending ? to the entire group, I allowed any number of characters to be in the actual match itself by appending * to the character selection. Therefore, the match is forced to exist, but the string itself can have a length of 0.
The problem comes with the bracketed block. The package I'm using allows me to access all captures of a named group using match.captures(groupname). This way, I can access all matches for cap_2 using match.captures("cap_2"):
>>> pattern.match("one[two=three][five=six].four").captures("cap_2")
["two", "five"]
This works fine when the brackets are present. However, when they're not:
>>> pattern.match("one.four").captures("cap_2")
[]
Expected: [""]
I expect there to be at least an empty string present for cap_2 and cap_3. However, there's nothing.
This is because of the * I place after the two+three section of the regex, in order to allow multiple of those groups - this is allowing that part of the regex to be skipped altogether.
Changing that * to + breaks the regex, as now it won't match the above example at all because it's trying to match the brackets. Adding a ? after each bracket means that cap_1 and cap_2 are not delimited and includes what should be in cap_4 in cap_3.
What's the solution here? How can I allow a group containing two capturing groups to be executed multiple times, but match only empty strings when the brackets are not present?
You may solve the problem by replacing * after the (?:\[(?P<cap_2>[a-z]*)=?(?P<cap_3>[a-z]*)\])* repeated group with + and adding an alternative with a second occurrence of groups cap_2 and cap_3 (note that PyPi regex module supports multiple identically named groups in the same regex):
import regex as re
s = 'one.four'
pattern = re.compile(
r"""
^ # Assert start of string
(?P<cap1> # Start a new group for "one"
[a-z]* #
) #
(?:
(?: # Start a group for "two" and "three"
\[ # Match the "["
(?P<cap_2> # Start a group for "two"
[a-z]* #
) #
=? # Delimit two/three with "="
(?P<cap_3> # Start a group for "three"
[a-z]* #
) #
\] # Match the "]"
)+ # End the two-three group, allowing repeats
|
(?P<cap_2>)(?P<cap_3>)
)
\.? # Delimit three/four with "."
(?P<cap_4> # Begin a group for "four"
[a-z]* #
) #
$ # Assert end of string
""", re.IGNORECASE|re.VERBOSE)
print ( pattern.match("one.four").captures("cap_2") )
# => ['']
See the Python demo
The thing is, the (?:\[(?P<cap_2>[a-z]*)=?(?P<cap_3>[a-z]*)\])* part matches by all means since it can match an empty string, and if you just add the alternatives without changing the modifier, the expected results won't be achieved. So, if there is no [...]s, the second cap_2 and cap_3 groups with empty patterns willmatch by all means capturing an empty string.
if you want it to either match the empty string or something else, you need the OR operator: |
if you want your regexp to match an empty string, you need something that matches the empty string: e.g. () or (not empty|)
Combined and applied to your case, that would look like this (simplified):
((?:\[stuff inside the brackets\])+|)
The outermost group captures the whole bracket construct (e.g. [two][three]) if it's present or the empty string. Notice that the left part of the | operator now has to match at least once (+).
For migrating a system to PDO, we would like to replace some queries. Via a regular expression, we can improve the progress. We are looking for a method to match:
mysqli_query($db, [expression follows here])
So, we made use of the following regex:
mysqli_query\(\$db, (.*?)\)
The problem is that we've a problem when we have for example a join query with an opening and closing () parameter.
Example: mysqli_query($db, "SELECT users.id FROM users JOIN (... ) on .. WHERE users.id=1")
Is it possible to edit the regex so we allow a ) when it is opened? So, the number of ( and ) should be equal.
You need an editor with PCRE-compatible regexes (Dreamweaver probably only supports JavaScript-style regexes); then you can use a recursive regex like this:
mysqli_query\(\$db, ((?:[^()]++|\((?1)\))*)\)
Test it live on regex101.com.
Explanation:
( # Match and capture in group 1:
(?: # Start of non-capturing group that either matches
[^()]++ # a sequence of characters except parentheses
| # or
\( # an opening parenthesis
(?1) # followed by an expression that follows the same rules as group 1
\) # and a closing parenthesis.
)* # Do this any number of times (including 0)
) # End of group 1
Try the following: mysqli_query\(\$db, ('|")(.+?)\1\)[\s;]. This assumes that the expression will be between single or double quotes. This might fail depending on what content comes after it though, and for some notations within the SQL query itself (e.g. if it encounters \')).
https://regex101.com/r/1NRYUw/1
Following pattern: (v[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2})(-[0-9]{1,2})?((-schema)?(-dev)?)((-schema)?(-dev)?) from http://regexr.com/ is meant to be used in a shell script with grep and does match the following strings (working example):
Hello I am a text and this is my v1.12.33-32 version
Hello I am a text and this is my v1.12.33-dev version
Hello I am a text and this is my v1.12.33-dev-schema version
Hello I am a text and this is my v1.12.33-schema version
Hello I am a text and this is my v1.12.33-3-schema version
and so forth
So I made the words schema and dev optional. They can be ommitted or used in a arbitrary order. What I don't what is this:
Hello I am a text and this is my v1.12.33-foo version
or Hello I am a text and this is my v1.12.33-asfs version
to match.
I want the option to be a bit more constrained. At the moment the Regex is still matching the stuff that...well actually matches.
This for example:
Hello I am a text and this is my v1.123.33
results in an empty string while this:
`Hello I am a text and this is my v1.12.33-bla"
still results in v.1.12.33
Is this because of the grouping I made? So at least the fully matching groups will be taken for the returned match-string?
To match only the version string, disallow extra trailing tags, yet allow trailing unmatched text, you need a regex language that supports lookahead. Standard grep / egrep regexes do not support lookahead.
You have two options:
Since you seem to be relying on GNU grep anyway, you could use a Perl regex, such as
v[0-9]{1,2}(\.[0-9]{1,2}){2}(-[0-9]{1,2})?((-schema(-dev)?)?|(-dev(-schema)?)?)?(?!\S)
The negative lookahead at the end allows the match to appear at the end of the line, but also requires that if it does not end the line then the next character following the match must be whitespace (which is not itself included in the match).
You could give up on completely isolating the target text via -o, and instead allow the pattern to match the trailing context, too:
v[0-9]{1,2}(\.[0-9]{1,2}){2}(-[0-9]{1,2})?((-schema(-dev)?)?|(-dev(-schema)?)?)?(\s.*)?$
In this case, you could isolate the target text in a second step, by stripping off any tail beginning with whitespace.
Note that neither of these pays attention to text preceeding the match. You have similar options for handling that portion as you do for handling the trailing portion.
The problem seems to be all the optional expressions lurking at the
edge (end).
You can solve that a few ways, but none are %100 because you'd need
more rules to control what matches.
It's not like you can say no - is allowed afterword, the engine will
backtrack to one of the range digits {1,2} to make a match.
What seems to work for now is passing on a whitespace end edge
or matching the dev/schema items.
(v[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2})(-[0-9]{1,2})?(?:(?!\S)|(-(schema|dev)(?:-(schema|dev))?))
Expanded
( # (1 start)
v [0-9]{1,2}
\. [0-9]{1,2}
\. [0-9]{1,2}
) # (1 end)
( - [0-9]{1,2} )? # (2)
(?:
(?! \S ) # Whitespace boundary
| # or,
( # (3 start)
-
( schema | dev ) # (4)
(?:
-
( schema | dev ) # (5)
)?
) # (3 end)
)
edit
If you want to avoid matching the same schema|dev word twice, just add
a negative assertion of group 4, before capture group 5 above.
(v[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2})(-[0-9]{1,2})?(?:(?!\S)|(-(schema|dev)(?:-(?!\4)(schema|dev))?))
Expanded
( # (1 start)
v [0-9]{1,2}
\. [0-9]{1,2}
\. [0-9]{1,2}
) # (1 end)
( - [0-9]{1,2} )? # (2)
(?:
(?! \S ) # Whitespace boundary
| # or,
( # (3 start)
-
( schema | dev ) # (4)
(?:
-
(?! \4 ) # Not same word twice
( schema | dev ) # (5)
)?
) # (3 end)
)
Since regular expressions are open-ended, you need to specify with $ where you want the match to end, so you don't let the regex engine silently ignore trailing junk.
With only two tags in the optional set, I would just enumerate the 4 possibilities:
(v[0-9]{1,2}\.[0-9]{1,2}\.[0-9]{1,2})(-[0-9]{1,2})?(-schema|-dev|-dev-schema|-schema-dev)?$
My version:
grep --perl-regexp \
'\bv(?:\d{1,2}\.){2}\d{1,2}(?:\-\d{1,2})?(?:\-(?:schema|dev))?(?:\s|$)' \
path/to/file
Where
the first \b is a word boundary(you might want to make it stricter);
(?: ... ) expressions are non-capturing groups;
\s|$ is either a space character, or the end of line
The rest is just refactored for simplicity.
The expression allows only schema, or dev at the "end".
I am trying to validate the Name entered by a user. I need to make sure the Name entered is 'literally' valid. I tried many regular expressions within my limited knowledge, none of them seem to work fully. For example /^[^.'-]+([a-zA-Z]+|[ .'-]{1})$/
Since the PHP website I'm working on, is fully in English, only English names are allowed.
The rules applicable to filtering the name are:
A name may contain any of these characters: [a-zA-Z .'-]
The name may start only with an Alphabet or an Apostrophe
Any of the characters in [ .'-] may not occur more than once in a stretch, ie., no '---' or '--'
A space should not follow - or ' nor come immediately before . or -
Can anyone please provide the proper regular expression to implement these?
Here's a regex to solve your problem (demo at regex101):
/^(?!.*(''| |--|- |' | \.| -|\.\.|\n))['a-z][- '.a-z]*$/gi
Breakdown:
(?!.*(''| |--|- |' | \.| -|\.\.|\n)) negative lookahead to ensure that no doubled characters are found
['a-z] start with one of these characters
[- '.a-z]* the rest can also include spaces and dashes, and are not required (* instead of +)
This should work for you:
/^'?([a-z]+((\. )|( ')|['\- \.])?)+$/i
Demo at regex101.com
Explanation:
'? Optionally start with an apostrophe
([a-z]+((\. )|( ')|['\- \.])?)+ Afterwards allow 1-n groups of at least 1 alphabetic character, followed by certain (both) valid combinations of special characters. These are ". ", " '" (quotes to see the spaces) or any single special character.
/i Match case insensitive, otherwise you'd just specify [a-zA-Z] instead of [a-z].
I am not sure if John,.Doe or ' should be considered valid names. With my expression they would not.
You could probably get better performance by limiting all the backtracking a giant assertion would have to do.
Something like this..
# (?i)^(?=[a-z'])(?:[a-z]+|([ ](?!-)(?<!'[ ])|[.'-])(?!\1))*$
(?i) # Modifier, case insensitive
^ # BOS
(?= [a-z'] ) # Starts with an alpha or apos '
(?: # Cluster
[a-z]+ # 1 to many alpha's
| # or
( # (1 start), Capture group, one of [ .'-]
[ ] # Single space
(?! - ) # not a dash - ahead
(?<! ' [ ] ) # nor apos space '[ ] behind
| # or,
[.'-] # Single dot . or apos ' or dash -
) # (1 end)
(?! \1 ) # Backref capt grp 1,
# not adjacent duplicate of one of [ .'-]
)* # Cluster end, do 0 to many times
$ # EOS
I am attemping to use regex to parse chat logs (namely Skype messages). So the regex I am currently using matches Skype logs correctly....as long as they don't have new lines.
So I tried adding the s modifier to the end, but this now makes it match everything (because its now multiline). So I was wondering if there was a way to both allow multiline, but stop before the [ at the beginning of Skype messages.
My regex is here: https://regex101.com/r/nL0vO9/1
You can use a tempered greedy Token:
\[(?:(?!\n\[).)*
Note you also need to include g modifier so you don't stop on first match
See Demo
Like #sln pointed out, if you want to keep new lines use this instead:
\[(?:.(?<!\n\[))*
Set the flags to //mg multi-line and global. Don't use the s Dot all flag.
edit: I guess you could use something simpler if you don't care about validating/parsing out the time/name/message parts. Either #CrayonViolent or #RodrigoLópez should work for that.
# ^\[([^\r\n\]]*)\]([^:\r\n]*):((?:(?!^\[).*(?:\r?\n)*)*)
^ # BOL
\[
( [^\r\n\]]* ) # (1), Time
\]
( [^:\r\n]* ) # (2), Person
:
( # (3 start), Message
(?: # Cluster group
(?! ^ \[ ) # Assert, not BOL and [
.* # Get all to end of line
(?: \r? \n )* # Optional, Get 1 to many line-breaks
)* # End cluster, do 1 to many times
) # (3 end)