I'm trying to write a regex which gets everything but a specified pattern. I've been trying to use negative lookahead but whenever testing my expression, it never works.
I have files that are of this form:
(garbage info) filename (other garbage).extension
or
[garbage info] filename [other garbage].extension
For example, one of the files is [O2CXDR] report january [77012].pdf or
(XEW7CK) sales commissions (99723).xls
I'm using the regex.h library in C so I believe that it is a POSIX library.
I'm hoping on extracting "filename" and ".extension" so that I can write a script which will the files filename.extension
So far, I have a an expression to select the garbage info with the brackets and the spaces around it but I'm unable to select the rest.
\s*(\[|\().*?(\]|\))+\s*
and the negative lookahead I tried was:
.*(?!(\s*(\[|\().*?(\]|\))+\s*)).*
but it's just selecting everything in a single match.
I'm sure that I'm not understanding the lookaheads and lookbehind correctly. What do I have to do to fix my expression? Could somebody explain how they work since I'm a bit lost. Thanks!
Since you haven't specified a regex engine, I'll target a subset that can use the tags \K, \G, and \A (like PCRE).
The following uses a combination of match resets (\K), tempered greedy token, and start of match (without start of string) \G(?!\A), further explained below:
See regex in use here
Note: remove empty matches
\s*[[(].*?[])]\s*\K|\G(?!\A)(?:(?!\s*[[(].*?[])]\s*).)+
Match one of the following:
Option 1:
\s* Match any whitespace any number of times
[[(] Match either [ or (
.*? Match any character any number of times, but as few as possible (lazy matching)
[])] Match either ] or )
\s* Match any whitespace any number of times
\K Reset match - sets the given position in the regex as the new start of the match. This means that nothing preceding this tag will be captured in the overall match.
Option 2:
\G(?!\A) Match only at the starting point of the search or position of the previous successful match end, but not at the start of the string.
(?:(?!\s*[[(].*?[])]\s*).)+ Tempered greedy token matching anything more than once except the negative lookahead pattern (which is the same as the first option).
$ cat input_file
(garbage info) filename (other garbage).extension
(garbage info)filename(other garbage).extension
(garbage info)file name(other garbage).extension
[garbage info] filename [other garbage].extension
[garbage info]filename[other garbage].extension
[garbage info]file name[other garbage].extension
$ sed -re 's/^\s*(\([^\)]*\)|\[[^]]*\])\s*(.*\S)\s*(\([^\)]*\)|\[[^]]*\])(\..*)$/\2\4/' input_file
filename.extension
filename.extension
file name.extension
filename.extension
filename.extension
file name.extension
Maybe, as simple as
^(?:\(([^)]*)\)\s*([^(\r\n]*?)\s*\(([^)]*)\)|\[([^\]]*)\]\s*([^(\r\n]*?)\s*\[([^\]]*)\])\.(.*)$
we could extract those values.
Demo 1
RegEx Circuit
jex.im visualizes regular expressions:
If you don't need all of those capturing groups, we'd then simply remove those that we wouldn't want:
^(?:\([^)]*\)\s*([^(\r\n]*?)\s*\([^)]*\)|\[[^\]]*\]\s*([^(\r\n]*?)\s*\[[^\]]*\])\.(.*)$
Demo 2
Related
I'm working in notepad++, and using its find-replace dialog box.
NP++ documentation states: Notepad++ regular expressions use the Boost regular expression library v1.70, which is based on PCRE (Perl Compatible Regular Expression) syntax. ref: https://npp-user-manual.org/docs/searching
What I'm trying to do should be simple, but I'm a regex novice, and after 2-3 hrs of web searches and playing with online regex testers, I give up.
I want to replace all single quotes ' with double quote " , but if and only if the ' is to the RIGHT of one or more #, ie inside a python comment.
For example,
list1 = ['apple','banana','pear'] # All 'single quotes' to LEFT of # remained unchanged.
list2 = ['tomato','carrot'] # All 'single quotes' to RIGHT of one or more # are replaced
# # with "double quotes", like this.
The np++ file is over 800 lines, manual replacement would be tedious & error prone. Advice appreciated.
This regex should do what you want:
(^[^#]*#|(?<!^)\G)[^'\n]*\K'
It looks for a ' which is preceded by either
^[^#]*# : start of line and some number of non-# characters followed by a #; or
(?<!^)\G : the start of line or the end of the previous match (\G), with a negative lookbehind for start of line (?<!^), meaning that it only matches at the end of the previous match
and then some number of non ' or newline (to prevent the match wrapping around the end of the previous line) characters [^'\n]*.
We then use \K to reset the match, so that everything before that is discarded from the match, and the regex only matches the '.
That can then be replaced with ".
Demo on regex101
Update
You can avoid matching apostrophes within words by only matching ones that are either preceded or followed by a non-word character:
(^[^#]*#|(?<!^)\G)[^'\n]*\K('(?=\W)|(?<=\W)')
Demo on regex101
Update 2
You can also deal with the case where there are # characters in strings by qualifying the first part of the regex with the requirement for there to be matched pairs of quotes beforehand:
(?:^[^'#]*(?:'[^']*'[^#']*)*[^'#]*#|(?<!^)\G)[^'\n]*\K(?:'(?=\W)|(?<=\W)')
Demo on regex101
I want to catch strings which respond to a pattern in a subject string.
Patterns examples: ##name##, ##address##, ##bankAccount##, ...
Subject example: This is the template with patterns : ##name##Your bank account is : ##bankAccount##Your address is : ##address##
With the following regex: .*(#{2}[a-zA-Z]*#{2}).*, only the last pattern is matched.
How to capture all the patterns, not just the last or first ?
Now that I've formatted your regex properly, the problem shows. A * in your regex was hidden since markdown took it to make the text italics.
Your opening .* matches greedily as much as it can, only backing up enough to let (#{2}[a-zA-Z]*#{2}) match. This matches the last pattern found in the line, everything before it having been matched by the .*.
You need to remove .* as I mentioned in my comment, and use preg_match_all:
$re = '~#{2}[a-zA-Z]*#{2}~';
preg_match_all($re, "##name##, ##address##, ##bankAccount##", $m);
print_r($m);
See the PHP demo
The .*#{2}[a-zA-Z]*#{2}.* matched 0 or more characters other than a newline at first, grabbing the whole line, and then backtracked until the last occurrence of #{2}[a-zA-Z]*#{2} pattern, and the last .* only grabbed the rest of the line. Removing the .* and using preg_match_all, all substrings matching the #{2}[a-zA-Z]*#{2} pattern can be extracted.
I am attempting to edit a csv file, below is a sample line from this file.
|MIGRATE|;|10000|;|2ACC0003|;|30/09/13|;|Positive Adjmt.|;||;|MIGRATE|;|95004U
The beginning of the line |MIGRATE| needs to be modified without changing the second MIGRATE so the line would read
|MIGRATE|;|MIG_IN|;|10000|;|2ACC0003|;|30/09/13|;|Positive Adjmt.|;||;|MIGRATE|;|95004U
There are 7700 or so lines so if I am forced to do this manually I will probably cry a little.
Thanks in advance!
Just replace all the ones you want not changed with another word temporarily, then replace the rest with what you want. I'm not sure what you're asking here, but from what I can guess this might help.
It seems like you could just search for Just search for:
^\|MIGRATE\|
And replace with:
|MIGRATE|;|MIG_IN|
Make sure you've checked 'Regular expression' in the 'Search Mode' options.
Explanation: The ^ is a begin anchor; it will match the beginning of the line, ensuring that it does not match the second |MIGRATE|. The \ characters are required to escape the | characters since they normally have special meaning in regular expressions, and you want to match a literal |.
You can use beginning of line anchors:
Find:
^(\|MIGRATE\|)
Replace with:
$1;|MIG_IN|
regex101 demo
Just make sure that you are using the regular expression mode of the Search&Replace.
If you want to be a bit fancier, you can use a positive lookbehind:
Find:
(?<=^\|MIGRATE\|)
Replace with:
;|MIG_IN|
^ Will match only at the beginning of a line.
( ... ) is called a capture group, and will save the contents of the match in variable you can use (in the first regex, I accessed the variable using $1 in the replace. The first capture gets stored to $1, the second to $2, etc.)
| is a special character meaning 'or' in regex (to match a character or group of characters or another, e.g. a|b matches a or b. As such, you need to escape it with a backslash to make a regex match a literal |.
In my second regex, I used (?<= ... ) which is called a positive lookbehind. It makes sure that the part to be matched has what's inside before it. For instance, (?<=a)b matches a b only if it has an a before it. So that the b in ab matches but not in bb.
The website I linked also explains the details of the regex and you can try out some regex yourself!
I'm parsing through code using a Perl-REGEX parsing engine in my IDE and I want to grab any variables that look like
$hash->{ hash_key04}
and nuke the rest of the code..
So far my very basic REGEX doesnt do what I expected
(.*)(\$hash\-\>\{[\w\s]+\})(.*)
(
\$
hash
\-\>
\{
[\w\s]+
\}
)
I know to use replace for this ($1,$2,etc), but match (.*) before and after the target string doesnt seem to capture all the rest of the code!
UPADTED:
tried matching null but of course thats too greedy.
([^\0]*)
What expression in regex should i use to look only for the string pattern and remove the rest?
The problem is I want to be left with the list of $hash->{} strings after the replace runs in the IDE.
This is better approached from the other direction. Instead of trying to delete everything you don't want, what about extracting everything you do want?
my #vars = $src_text =~ /(\$hash->\{[\w\s]+\})/g;
Breaking down the regex:
/( # start of capture group
\$hash-> # prefix string with $ escaped
\{ # opening escaped delimiter
[\w\s]+ # any word characters or space
\} # closing escaped delimiter
)/g; # match repeatedly returning a list of captures
Here is another way that might fit within your IDE better:
s/(\$hash->\{[\w\s]+\})|./$1/gs;
This regex tries to match one of your hash variables at each location, and if it fails, it deletes the next character and then tries again, which after running over the whole file will have deleted everything you don't want.
Depends on your coding language. What you want is group 2 (The second set of characters in parenthesis). In perl that would be $2, in VIM it would be \2, etc ...
It depends on the platform, but generally, replace the pattern with an empty string.
In javascript,
// prints "the la in ing"
console.log('the latest in testing'.replace(/test/g, ''));
In bash
$ echo 'the latest in testing' | sed 's/test//g'
the la in ing
In C#
Console.WriteLine(Regex.Replace("the latest in testing", "test", ""));
etc
By default the wildcard . won't match newlines. You can enable newlines in its matching set using a flag depending on what regex standard you're using and under what language/api. Or you can add them explicitly yourself by defining a character set:
[.\n\r]* <- Matches any character including newline, carriage return.
Combine this with capture groups to grab desired variables from your code and skip over lines which contain no capture group.
If you want help constructing the proper regex for your context you'll need to paste some input text and specify what the output should be.
I think you want to add a ^ to the beginning of the regex s/^.(PATTERN)(.)$/$1/ so that it starts at the beginning of the line and goes to the end, removing anything except that pattern.
I am doing this in groovy.
Input:
hip_abc_batch hip_ndnh_4_abc_copy_from_stgig abc_copy_from_stgig
hiv_daiv_batch hip_a_de_copy_from_staging abc_a_de_copy_from_staging
I want to get the last column. basically anything that starts with abc_.
I tried the following regex (works for second line but not second.
\abc_.*\
but that gives me everything after abc_batch
I am looking for a regex that will fetch me anything that starts with abc_
but I can not use \^abc_.*\ since the whole string does not start with abc_
It sounds like you're looking for "words" (i.e., sequences that don't include spaces) that begin with abc_. You might try:
/\babc_.*\b/
The \b means (in some regular expression flavors) "word boundary."
Try this:
/\s(abc_.*)$/m
Here is a commented version so you can understand how it works:
\s # match one whitepace character
(abc_.*) # capture a string that starts with "abc_" and is followed
# by any character zero or more times
$ # match the end of the string
Since the regular expression has the "m" switch it will be a multi-line expression. This allows the $ to match the end of each line rather than the end of the entire string itself.
You don't need to trim the whitespace as the second capture group contains just the text. After a cursory scan of this tutorial I believe this is the way to grab the value of a capture group using Groovy:
matcher = (yourString =~ /\s(abc_.*)$/m)
// this is how you would extract the value from
// the matcher object
matcher[0][1]
I think you are looking for this: \s(abc_[a-zA-Z_]*)$
If you are using perl and you read all lines into one string, don't forget to set the the m option on your regex (that stands for "Treat string as multiple lines").
Oh, and Regex Coach is your free friend.