regex to match strings not ending with a pattern? - regex

I am trying to form a regular expression that will match strings that do NOT end a with a DOT FOLLOWED BY NUMBER.
eg.
abcd1
abcdf12
abcdf124
abcd1.0
abcd1.134
abcdf12.13
abcdf124.2
abcdf124.21
I want to match first three.
I tried modifying this post but it didn't work for me as the number may have variable length.
Can someone help?

You can use something like this:
^((?!\.[\d]+)[\w.])+$
It anchors at the start and end of a line. It basically says:
Anchor at the start of the line
DO NOT match the pattern .NUMBERS
Take every letter, digit, etc, unless we hit the pattern above
Anchor at the end of the line
So, this pattern matches this (no dot then number):
This.Is.Your.Pattern or This.Is.Your.Pattern2012
However it won't match this (dot before the number):
This.Is.Your.Pattern.2012
EDIT: In response to Wiseguy's comment, you can use this:
^((?!\.[\d]+$)[\w.])+$ - which provides an anchor after the number. Therefore, it must be a dot, then only a number at the end... not that you specified that in your question..

If you can relax your restrictions a bit, you may try using this (extended) regular expression:
^[^.]*.?[^0-9]*$
You may omit anchoring metasymbols ^ and $ if you're using function/tool that matches against whole string.
Explanation: This regex allows any symbols except dot until (optional) dot is found, after which all non-numerical symbols are allowed. It won't work for numbers in improper format, like in string: abcd1...3 or abcd1.fdfd2. It also won't work correctly for some string with multiple dots, like abcd.ab123cd.a (the problem description is a bit ambigous).
Philosophical explanation: When using regular expressions, often you don't need to do exactly what your task seems to be, etc. So even simple regex will do the job. An abstract example: you have a file with lines are either numbers, or some complicated names(without digits), and say, you want to filter out all numbers, then simple filtering by [^0-9] - grep '^[0-9]' will do the job.
But if your task is more complex and requires validation of format and doing other fancy stuff on data, why not use a simple script(say, in awk, python, perl or other language)? Or a short "hand-written" function, if you're implementing stand-alone application. Regexes are cool, but they are often not the right tool to use.

I would just use a simple negative look-behind anchored at the end:
.*(?<!\\.\\d+)$

Related

Regex everything after, but not including

I am trying to regex the following string:
https://www.amazon.com/Tapps-Top-Apps-and-Games/dp/B00VU2BZRO/ref=sr_1_3?ie=UTF8&qid=1527813329&sr=8-3&keywords=poop
I want only B00VU2BZRO.
This substring is always going to be a 10 characters, alphanumeric, preceded by dp/.
So far I have the following regex:
[d][p][\/][0-9B][0-9A-Z]{9}
This matches dp/B00VU2BZRO
I want to match only B00VU2BZRO with no dp/
How do I regex this?
Here is one regex option which would produce an exact match of what you want:
(?<=dp\/)(.*)(?=\/)
Demo
Note that this solution makes no assumptions about the length of the path fragment occurring after dp/. If you want to match a certain number of characters, replace (.*) with (.{10}), for example.
Depending on your language/method of application, you have a couple of options.
Positive look behind. This will make your regex more complicated, but will make it match what you want exactly:
(<=dp/)[0-9A-Z]{10}
The construct (<=...) is called a positive look behind. It will not consume any of the string, but will only allow the match to happen if the pattern between the parens is matched.
Capture group. This will make the regex itself slightly simpler, but will add a step to the extraction process:
dp/([0-9A-Z]{10})
Anything between plain parens is a capture group. The entire pattern will be matched, including dp/, but most languages will give you a way of extracting the portion you are interested in.
Depending on your language, you may need to escape the forward slash (/).
As an aside, you never need to create a character class for single characters: [d][p][\/] can equally well be written as just dp\/.

positive look ahead and replace

Recently I'm writing/testing regexps on https://regex101.com/.
My question is: Is it possible to do a positive look-ahead AND a replacement in the same "replacement"? Or just limited kind of replacement is possible.
Input is several lines with phone numbers. Let's say the correct phone number where the number of "numbers" are 11. No matter how the numbers are divided/group together with - / characters, no matter if starts with + 00 or it is omitted.
Some example lines:
+48301234567
+48/30/1234567
+48-30-12-345-67
+483011223344556677
0048301234567
+(48)30/1234567
Positive look-ahead able to check if from the beginning until the end of line there are only 11 digits, regardless how many other, above specified character separating them. This works perfectly.
Where the positive look-ahead check is fine, I would like to delete every character but numbers. The replacement works fine until I'm not involving look-ahead.
Checking the regexp itself working perfectly ("gm" modes):
^(?:\+|00)?(?:[\-\/\(\)]?\d){11}$
Checking the replace part works perfectly (replace to nothing):
[^\d\n]
Put this into look-ahead, after the deletion of non new-line and non-digit characters from the matching lines:
(?=^(?:\+|00)?(?:[\-\/\(\)]?\d){11}$)[^\d\n]
Even I put the ^ $ into look-ahead, seems the replacement working only from beginning of the lines until the very first digit.
I know in real life the replacement and the check should/would go separate ways, however I'm curious if I could mix look-ahead/look-behind with string operations like replace, delete, take the string apart and put together as I like.
UPDATE: This is what would do the trick, however I feel this one "ugly" a bit. Is there any prettier solution?
https://regex101.com/r/yT5dA4/2
Or the version which I asked originally, where only digits remains: regex101.com/r/yT5dA4/3
You cannot replace/delete text with regex. Regex is just a tool for matching certain strings and then taking certain action depending on the matching text, eg. perform a substitution, retrieve the second capture group.
However it is possible to perform certain decisions within a regex engine, by using conditionals. The common syntax for this, with a lookahead assertion, is (?(?=regex)then|else).
With conditionals you can change the behaviour depending on how the text matches the regex. For your example you could do something like:
^(\+)?(?(1)\(|\d)
If the phone number starts with a plus it must be followed by a bracket, else it should start with a digit. Although in your situation, this is not very useful.
If you want to read up more on conditionals in regex you can do so here.

Regular expression using negative lookbehind not working in Notepad++

I have a source file with literally hundreds of occurrences of strings flecha.jpg and flecha1.jpg, but I need to find occurrences of any other .jpg image (i.e. casa.jpg, moto.jpg, whatever)
I have tried using a regular expression with negative lookbehind, like this:
(?<!flecha|flecha1).jpg
but it doesn't work! Notepad++ simply says that it is an invalid regular expression.
I have tried the regex elsewhere and it works, here is an example so I guess it is a problem with NPP's handling of regexes or with the syntax of lookbehinds/lookaheads.
So how could I achieve the same regex result in NPP?
If useful, I am using Notepad++ version 6.3 Unicode
As an extra, if you are so kind, what would be the syntax to achieve the same thing but with optional numbers (in this case only '1') as a suffix of my string? (even if it doesn't work in NPP, just to know)...
I tried (?<!flecha[1]?).jpg but it doesn't work. It should work the same as the other regex, see here (RegExr)
Notepad++ seems to not have implemented variable-length look-behinds (this happens with some tools). A workaround is to use more than one fixed-length look-behind:
(?<!flecha)(?<!flecha1)\.jpg
As you can check, the matches are the same. But this works with npp.
Notice I escaped the ., since you are trying to match extensions, what you want is the literal .. The way you had, it was a wildcard - could be any character.
About the extra question, unfortunately, as we can't have variable-length look-behinds, it is not possible to have optional suffixes (numbers) without having multiple look-behinds.
Solving the problem of the variable-length-negative-lookbehind limitation in Notepad++
Given here are several strategies for working around this limitation in Notepad++ (or any regex engine with the same limitation)
Defining the problem
Notepad++ does not support the use of variable-length negative lookbehind assertions, and it would be nice to have some workarounds. Let's consider the example in the original question, but assume we want to avoid occurrences of files named flecha with any number of digits after flecha, and with any characters before flecha. In that case, a regex utilizing a variable-length negative lookbehind would look like (?<!flecha[0-9]*)\.jpg.
Strings we don't want to match in this example
flecha.jpg
flecha1.jpg
flecha00501275696.jpg
aflecha.jpg
img_flecha9.jpg
abcflecha556677.jpg
The Strategies
Inserting Temporary Markers
Begin by performing a find-and-replace on the instances that you want to avoid working with - in our case, instances of flecha[0-9]*\.jpg. Insert a special marker to form a pattern that doesn't appear anywhere else. For this example, we will insert an extra . before .jpg, assuming that ..jpg doesn't appear elsewhere. So we do:
Find: (flecha[0-9]*)(\.jpg)
Replace with: $1.$2
Now you can search your document for all the other .jpg filenames with a simple regex like \w+\.jpg or (?<!\.)\.jpg and do what you want with them. When you're done, do a final find-and-replace operation where you replace all instances of ..jpg with .jpg, to remove the temporary marker.
Using a negative lookahead assertion
A negative lookahead assertion can be used to make sure that you're not matching the undesired file names:
(?<!\S)(?!\S*flecha\d*\.jpg)\S+\.jpg
Breaking it down:
(?<!\S) ensures that your match begins at the start of a file name, and not in the middle, by asserting that your match is not preceded by a non-whitespace character.
(?!\S*flecha\d*\.jpg) ensures that whatever is matched does not contain the pattern we want to avoid
\S+\.jpg is what actually gets matched -- a string of non-whitespace characters followed by .jpg.
Using multiple fixed-length negative lookbehinds
This is a quick (but not-so-elegant) solution for situations where the pattern you don't want to match has a small number of possible lengths.
For example, if we know that flecha is only followed by up to three digits, our regex could be:
(?<!flecha)(?<!flecha[0-9])(?<!flecha[0-9][0-9])(?<!flecha[0-9][0-9][0-9])\.jpg
Are you aware that you're only matching (in the sense of consuming) the extension (.jpg)? I would think you wanted to match the whole filename, no? And that's much easier to do with a lookahead:
\b(?!flecha1?\b)\w+\.jpg
The first \b anchors the match to the beginning of the name (assuming it's really a filename we're looking at). Then (?!flecha1?\b) asserts that the name is not flecha or flecha1. Once that's done, the \w+ goes ahead and consumes the name. Then \.jpg grabs the extension to finish off the match.

Regex to pull set of numbers from a string

This is my string:
a:1:{i:0;a:4:{s:1:"i";s:5:"19807";s:1:"c";s:19:"7025462932180014917";s:1:"a";
d:88.719999999999998863131622783839702606201171875;s:2:"ba";
d:88.719999999999998863131622783839702606201171875;}}
Im just looking to pull out the portion that starts with 702546 and ends before that double quote after the 7. The number may be a varied length but will always start with 702546 and always end at the quote mark.
So my final selection would be 7025462932180014917.
use this regex (?<=")(702546\d+)(?=")
if youe regex engine doesn't support lookbehind use this regex "(702546\d+)" match will be in group 1
This could vary a bit depending on the technology/language you are using, but basically you can use something like this
(?<=")702546\d+(?=")
The first (?<=") may be overkill if there is no risk of other numbers in your string starting with 702546.
Breaking this down:
(?<=") // a zero-width assertion (lookbehind), so we will only match if preceded by double-quotes
702546 // literal match
\d+ // one or more additional digits
(?=") // another zero-width assertion (lookahead), so we will only match if followed by double-quotes
Basically, a zero-width assertion just means that the value must be there in order for a match to succeed, but the value does not become part of the match. We use them here because we want to be sure that the value is enclosed in quotes, without making the quotes part of our match.
The basic regex is 702546[^"]* to match any string like that, but you probably want to use a programming language of some sort to actually pull those out.
vnix$ grep -o '702546[^"]*' file
Or in a scripting language, something like this;
perl -nle 'print $1 while m/(702546[^"]*)/g' file
Your problem description leads me to believe that this is not actually what you should be doing, though. What problem are you really trying to solve?

How can I "inverse match" with regex?

I'm processing a file, line-by-line, and I'd like to do an inverse match. For instance, I want to match lines where there is a string of six letters, but only if these six letters are not 'Andrea'. How should I do that?
I'm using RegexBuddy, but still having trouble.
(?!Andrea).{6}
Assuming your regexp engine supports negative lookaheads...
...or maybe you'd prefer to use [A-Za-z]{6} in place of .{6}
Note that lookaheads and lookbehinds are generally not the right way to "inverse" a regular expression match. Regexps aren't really set up for doing negative matching; they leave that to whatever language you are using them with.
For Python/Java,
^(.(?!(some text)))*$
http://www.lisnichenko.com/articles/javapython-inverse-regex.html
In PCRE and similar variants, you can actually create a regex that matches any line not containing a value:
^(?:(?!Andrea).)*$
This is called a tempered greedy token. The downside is that it doesn't perform well.
The capabilities and syntax of the regex implementation matter.
You could use look-ahead. Using Python as an example,
import re
not_andrea = re.compile('(?!Andrea)\w{6}', re.IGNORECASE)
To break that down:
(?!Andrea) means 'match if the next 6 characters are not "Andrea"'; if so then
\w means a "word character" - alphanumeric characters. This is equivalent to the class [a-zA-Z0-9_]
\w{6} means exactly six word characters.
re.IGNORECASE means that you will exclude "Andrea", "andrea", "ANDREA" ...
Another way is to use your program logic - use all lines not matching Andrea and put them through a second regex to check for six characters. Or first check for at least six word characters, and then check that it does not match Andrea.
Negative lookahead assertion
(?!Andrea)
This is not exactly an inverted match, but it's the best you can directly do with regex. Not all platforms support them though.
If you want to do this in RegexBuddy, there are two ways to get a list of all lines not matching a regex.
On the toolbar on the Test panel, set the test scope to "Line by line". When you do that, an item List All Lines without Matches will appear under the List All button on the same toolbar. (If you don't see the List All button, click the Match button in the main toolbar.)
On the GREP panel, you can turn on the "line-based" and the "invert results" checkboxes to get a list of non-matching lines in the files you're grepping through.
I just came up with this method which may be hardware intensive but it is working:
You can replace all characters which match the regex by an empty string.
This is a oneliner:
notMatched = re.sub(regex, "", string)
I used this because I was forced to use a very complex regex and couldn't figure out how to invert every part of it within a reasonable amount of time.
This will only return you the string result, not any match objects!
(?! is useful in practice. Although strictly speaking, looking ahead is not a regular expression as defined mathematically.
You can write an inverted regular expression manually.
Here is a program to calculate the result automatically.
Its result is machine generated, which is usually much more complex than hand writing one. But the result works.
If you have the possibility to do two regex matches for the inverse and join them together you can use two capturing groups to first capture everything before your regex
^((?!yourRegex).)*
and then capture everything behind your regex
(?<=yourRegex).*
This works for most regexes. One problem I discovered was when I had a quantifier like {2,4} at the end. Then you gotta get creative.
In Perl you can do:
process($line) if ($line =~ !/Andrea/);