This question already has answers here:
Regex Last occurrence?
(7 answers)
Closed 3 years ago.
I have the following RegEx syntax that will match the first date found.
([0-9]+)/([0-9]+)/([0-9]+)
However, I would like to start from the end of the content and search backwards. In other words, in the below example, my syntax will always match the first date, but I want it to match the last instead.
Some Text here
01/02/15
Some additional
text here.
10/04/14
Ending text
here
I believe this is possible by using a negative lookahead, but all my attempts failed at this because I don't understand RegEx enough. Help would be appreciated.
Note: my application uses RegEx PCRP.
You could make the dot match a newline using for example an inline modifier (?s) and match until the end of the string.
Then make use of backtracking until the last occurrence of the date like pattern and precede the first digit with a word boundary.
Use \K to forget what was matched and match the date like pattern.
^(?s).*\b\K[0-9]+/[0-9]+/[0-9]+
Regex demo
Note that the pattern is a very broad match and does not validate a date itself.
Related
This question already has answers here:
Regex.Match whole words
(4 answers)
Closed 3 years ago.
I have the following regex:
^USD|AUD|BRL|GBP|CAD|CNY|DKK|AED|EUR|HKD|INR|MYR|MXN|NZD|PHP|SGD|THB|ARS|COP|CLP|PEN|VEF$
When using this example string: 16ccf52b144~~refCode-3-d5779a89-d437-448a-bf53-efad2cdd66f6~20191020T16:00~20191026T16:00~USD~305.81~~~~**8294A2B49CD60ABE4FC7081F05CD06AA17E837CCADEB0ABC57B6AC94B09882FB
I am expecting the regex to return USD, instead it is returning CAD. How can I edit the regex so that it returns USD ...Ideally regex should look at ~currencyCode~ ...instead right now it is looking at currencyCode without tilde.
The ^ and $ assertions are unnecessary in your regex since the substring you are trying to match is at neither the beginning nor the end of the string, and the fact that ^ is preceding USD means that the pattern can only match USD if it is at the beginning of the string.
Instead, group the alternations and surround them with word boundary assertions:
\b(?:USD|AUD|BRL|GBP|CAD|CNY|DKK|AED|EUR|HKD|INR|MYR|MXN|NZD|PHP|SGD|THB|ARS|COP|CLP|PEN|VEF)\b
You haven't said which language or framework you're using, so I'll assume you want a generally applicable regex.
If you know that ~ will precede and follow your currency, then you can use a zero-width assertion to find text between ~ characters like so:
(?<=~)(USD|AUD|BRL|GBP|CAD|CNY|DKK|AED|EUR|HKD|INR|MYR|MXN|NZD|PHP|SGD|THB|ARS|COP|CLP|PEN|VEF)(?=~)
This will match the USD in 6:00~USD~305 because it's surrounded by ~, but not the CAD in 7CCADEB0 because it's not surrounded by them.
This question already has answers here:
How to match only strings that do not contain a dot (using regular expressions)
(3 answers)
Closed 3 years ago.
I have numerous SELECT statements conjoined by UNION keyword in a single file. What I want to do is to extract all the db.table strings only? How can I delete all words not containing period (.) using regex in notepad++ editor? Database and table are the only ones with a period.
It's okay with me even if new lines are not removed. Though, as a learning bonus for everyone seeing this post, you can also show the regex that trims the new lines, that will show this output:
db.table1
db.table2
...
db.tablen
You may try the following find and replace, in regex mode:
Find: (?<=^|\s)[^.]+(?=$|\s)
Replace: <empty string>
Demo
Note that my replacement only removes the undesired terms in the query; it does not make an effort to remove stray or leftover whitespace. To do that, you can easily do a quick second replacement to remove whitespace you don't want.
Edit:
It appears that Notepad++ doesn't like the variable width lookbehinds I used in the pattern. Here is a refactored, and more verbose version, which uses strictly fixed width lookbehinds:
(^[^.]+$)|(^[^.]+(?=\s))|((?<=\s)[^.]+$)|((?<=\s)[^.]+(?=\s))
Demo
The logic in both of the above patterns is to match a word consisting entirely of non dot characters, which are surrounded on either side by one or more of the following:
start of the string (^)
end of the string ($)
any type of whitespace (\s)
My guess is that maybe this expression:
([\s\S]*?)(\S*(\.)\S*)
being replaced with $2\n or:
(\S*(\.)\S*)|(.+?)
with $1 might work.
Demo 1
Demo 2
This question already has answers here:
Regex match entire words only
(7 answers)
Closed 3 years ago.
I would like to conduct regex substitution. Here is the pattern I am using:
.*?fee.*?$|.*?charge.*?$
The matches the desired lines
"fees credit card"
"charges for interest"
However, it is also matching on coffee and feeder (I want to be specific that it does not match "coffee" or "feed" lines, how can I specifically prevent these matches but still handle cases like fee, fees)
"coffee shop"
feeder cattle
You could use an alternation with 2 word boundaries \b to prevent the words being part of a larger word.
For you example data, if you want to match the single or single or plural version you can make the s at the end optional by using a question mark.
^.*\b(?:fees?|charges?)\b.*$
^ Start of the string
.*\b Match any char except a newline followed by a word boundary
(?:fees?|charges?) Match any of the listed followed by an optional s
\b.* Word boundary, match any char except a newline 0+ times
$ Assert end of the string
Regex demo
If you are just trying to match those two lines, you can simply use an expression similar to this:
^(fees|charges).+$
If you wish to match certain words, you might add boundaries to group one similar to this expression:
^\b(fees|fee|charge|charges)\b(.+)$
If your pattern might be in the middle of string inputs, you can add another group in the left, similar to this expression:
(?:.+|)\b(fees|fee|charge|charges)\b(?:.+|)$
This graph shows how an expression like that would work:
Regular expression design can be achieved much easier, if/when there is real data.
This question already has answers here:
Regex: match everything but a specific pattern
(6 answers)
Closed 4 years ago.
After countless hours of trying to get this regex to work (including looking all over StackOverflow), I thought I'd reach out for help on here as I have not been successful).
I have tried creating a regex to match everything and to not match any parameters that look like this:
text=3242ffs3F34
The data after the = sign can be random (its a mixture of numeric and string characters) and is never the same. So far I have created the following regex below, which is almost doing what I am after but it does not work.
\b(?!text=.*)\b\S+
Assistance is much appreciated!
EDIT:
I will be using the regex to match everything in a file but to filter out all parameters that look like this:
text=3242ffs3F34
Below is an example of how the config file will look like:
This is a test
test=asda
test2=22rr2
text=3242ffs3F34
test5=hello
To match everything except strings containing LAST_DOMINO_TIME= as substring you can use the expression:
(?!.*\bLAST_DOMINO_TIME=.*$)^.*$
(?! Negative lookahead.
.* Match anything.
\b Word boundary.
LAST_DOMINO_TIME= Literal substring.
.*$ Anything up to end of string.
) Close lookahead.
^.*$ Assert position beginning of line, match anything up to end of line.
You can try it here.
This question already has answers here:
Regular expression for a string containing one word but not another
(5 answers)
Closed 3 years ago.
Have regex in our project that matches any url that contains the string
"/pdf/":
(.+)/pdf/.+
Need to modify it so that it won't match urls that also contain "help"
Example:
Shouldn't match: "/dealer/help/us/en/pdf/simple.pdf"
Should match: "/dealer/us/en/pdf/simple.pdf"
If lookarounds are supported, this is very easy to achieve:
(?=.*/pdf/)(?!.*help)(.+)
See a demo on regex101.com.
(?:^|\s)((?:[^h ]|h(?!elp))+\/pdf\/\S*)(?:$|\s)
First thing is match either a space or the start of a line
(?:^|\s)
Then we match anything that is not a or h OR any h that does not have elp behind it, one or more times +, until we find a /pdf/, then match non-space characters \S any number of times *.
((?:[^h ]|h(?!elp))+\/pdf\/\S*)
If we want to detect help after the /pdf/, we can duplicate matching from the start.
((?:[^h ]|h(?!elp))+\/pdf\/(?:[^h ]|h(?!elp))+)
Finally, we match a or end line/string ($)
(?:$|\s)
The full match will include leading/trailing spaces, and should be stripped. If you use capture group 1, you don't need to strip the ends.
Example on regex101