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.
Related
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.
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 to allow alphanumeric and dot
(3 answers)
Closed 4 years ago.
I am trying to match the string when there's 0 or multiple dots. The regex that I can only match multiple dots but not 0 dot.
(\w*)((\w*\.)+\w*)
These are the test string I am using
dial.check.Catch.Url
dial.check.Catch.Url.Dial.check.Catch.Url
32443.324342.23423424.23.423.423.42.34.234.32.4..2..2.342.4
234dfasfd2aa4234234.234aa341.4.123daaadf.df.af....
12fd.dafd
.
abc
The Regex will match these
dial.check.Catch.Url
dial.check.Catch.Url.Dial.check.Catch.Url
32443.324342.23423424.23.423.423.42.34.234.32.4..2..2.342.4
234dfasfd2aa4234234.234aa341.4.123daaadf.df.af....
12fd.dafd
.
But not this one:
abc
https://regexr.com/?38ed7
If you really must use a regex, here is one (but it is inefficient):
/^(?![^.]*\.[^.]*$).*$/
It says:
Match a string so that the beginning of the string is not followed by a whole string with a single dot.
It does some backtracking when parsing the negative lookahead.
As mentioned in the comments to the question, I do think, unless you must have a regex, that a simple function might be better. But if you like the conciseness of a regex and performance is not a huge concern, you can go with the one I gave above. Regexes with "nots" in them are generally a tad messy, but once you understand lookarounds they do become doable. Cheers.
/\..*\.|^[^.]*$/
Or, in plain English:
Match EITHER a dot, then any number of characters, then another dot; OR the beginning of the string, then any number of non-dots, then the end of the string.
This question already has answers here:
Java RegEx that matches exactly 8 digits
(3 answers)
Closed 5 years ago.
Basically I'm looking for a regex that matches some simple phone numbers.
I want to match numbers in a longer string of text like 123 4567, 891-0111, or 21314151, something that is (hopefully) identified by (\d{3,4}[- ]\d{3,4}|\d{4,8}), but I don't want to match them if they're part of a longer number like 3919503570275.
If I require the next character to be a non-digit or the end of a line, then that next character is also included in the match, which I don't want.
Surround your regex with a lookahead and a lookbehind to reject \d on both sides:
(?<!\d)(\d{3,4}[- ]\d{3,4}|\d{4,8})(?!\d)
Demo.
Note that this would accept a string that looks like a phone number preceded or followed by letters.
Depending on what programming language you use, I suggest to either use negative look-ahead or to use groups to extract the number.
See https://www.regular-expressions.info/lookaround.html for information about lookaround pattern.
This question already has answers here:
Regular expression for a string containing one word but not another
(5 answers)
Closed 3 years ago.
Have regex in our project that matches any url that contains the string
"/pdf/":
(.+)/pdf/.+
Need to modify it so that it won't match urls that also contain "help"
Example:
Shouldn't match: "/dealer/help/us/en/pdf/simple.pdf"
Should match: "/dealer/us/en/pdf/simple.pdf"
If lookarounds are supported, this is very easy to achieve:
(?=.*/pdf/)(?!.*help)(.+)
See a demo on regex101.com.
(?:^|\s)((?:[^h ]|h(?!elp))+\/pdf\/\S*)(?:$|\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