I couldnt find any previously asked question similar to this. I need a regex to get all the data before second last special character.
For example:
suite 1, street 1, zip city, country
I need only suite 1, street 1.
I know how to get the data before just the last special character using [^,]*$ but not the second last one.
You can use the following regex and the first capturing group will have your desired substring:
(.*)(?:,[^,]*){2}$
Demo: https://regex101.com/r/AWpsL3/1
Or if the tool you're using does support capturing groups, you can use the following regex with lookahead instead:
.*(?=(?:,[^,]*){2}$)
Demo: https://regex101.com/r/AWpsL3/4
you can use look ahead
.+(?=.*,.*,)
explanation
.+ matches everything until the position look head starts , if the look ahead does not fail
Positive Look ahead (?=.*,.*,)
asserts two commas exist at the end
check demo
Depending on the implementation of regex, it may not support lookaround (which is what the above solution uses). A work around for this would be to perform a string split on your delimiter character (in this case, comma). Then perform a string join of the first two elements.
mystr = 'suite 1, street 1, zip city, country';
parts = mystr.split(',');
return parts[0]+','+parts[1];
Try (([ \w]+,?){2})(?=,) and dont use global flag (doesn't return after first match)
regex
Related
I'm having an issue with Regex.
I'm trying to match T0000001 (2, 3 and so on).
However, some of the lines it searches has what I can describe as positioners. These are shown as a question mark, followed by 2 digits, such as ?21.
These positioners describe a new position if the document were to be printed off the website.
Example:
T123?214567
T?211234567
I need to disregard ?21 and match T1234567.
From what I can see, this is not possible.
I have looked everywhere and tried numerous attempts.
All we have to work off is the linked image. The creators cant even confirm the flavour of Regex it is - they believe its Python but I'm unsure.
Regex Image
Update
Unfortunately none of the codes below have worked so far. I thought to test each code in live (Rather than via regex thinking may work different but unfortunately still didn't work)
There is no replace feature, and as mentioned before I'm not sure if it is Python. Appreciate your help.
Do two regex operations
First do the regex replace to replace the positioners with an empty string.
(\?[0-9]{2})
Then do the regex match
T[0-9]{7}
If there's only one occurrence of the 'positioners' in each match, something like this should work: (T.*?)\?\d{2}(.*)
This can be tested here: https://regex101.com/r/XhQXkh/2
Basically, match two capture groups before and after the '?21' sequence. You'll need to concatenate these two matches.
At first, match the ?21 and repace it with a distinctive character, #, etc
\?21
Demo
and you may try this regex to find what you want
(T(?:\d{7}|[\#\d]{8}))\s
Demo,,, in which target string is captured to group 1 (or \1).
Finally, replace # with ?21 or something you like.
Python script may be like this
ss="""T123?214567
T?211234567
T1234567
T1234434?21
T5435433"""
rexpre= re.compile(r'\?21')
regx= re.compile(r'(T(?:\d{7}|[\#\d]{8}))\s')
for m in regx.findall(rexpre.sub('#',ss)):
print(m)
print()
for m in regx.findall(rexpre.sub('#',ss)):
print(re.sub('#',r'?21', m))
Output is
T123#4567
T#1234567
T1234567
T1234434#
T123?214567
T?211234567
T1234567
T1234434?21
If using a replace functionality is an option for you then this might be an approach to match T0000001 or T123?214567:
Capture a T followed by zero or more digits before the optional part in group 1 (T\d*)
Make the question mark followed by 2 digits part optional (?:\?\d{2})?
Capture one or more digits after in group 2 (\d+).
Then in the replacement you could use group1group2 \1\2.
Using word boundaries \b (Or use assertions for the start and the end of the line ^ $) this could look like:
\b(T\d*)(?:\?\d{2})?(\d+)\b
Example Python
Is the below what you want?
Use RegExReplace with multiline tag (m) and enable replace all occurrences!
Pattern = (T\d*)\?\d{2}(\d*)
replace = $1$2
Usage Example:
I'm trying to looking for Street|St|Drive|Dr and then get all the contents of the line to extract the address:
(?:(?!\s{2,}|\$).)*(Street|St|Drive|Dr).*?(?=\s{2,})
.. but it also matches:
Full match 420-442 ` Tax Invoice/Statement`
Group 1. 433-435 `St`
Full match 4858-4867 `163.66 DR`
Group 1. 4865-4867 `DR`
Full match 11053-11089 ` Permanent Water Saving Plan, please`
Group 1. 11077-11079 `Pl`
How do i match only whole words and not substrings so it ignores words that contain those words (the first match for example).
One option is to use the the word-boundary anchor, \b, to accomplish this:
(?:(?!\s{2,}|\$).)*\b(Street|St|Drive|Dr)\b.*?(?=\s{2,})
If you provide an example of the raw text you're parsing, I'll be able to give additional help if this doesn't work.
Edit:
From the link you posted in a comment, it seems that the \b solution solves your question:
How do i match only whole words and not substrings so it ignores words that contain those words (the first match for example).
However, it seems like there are additional issues with your regex.
Lets say I have a string something-123.
I need to get last 5 (or less) characters of it but only up to - if there is one in the string, so the result would be like thing, but if string has no - in it, like something123 then the result would be ng123, and if string is like 123 then the result would be 123.
I know how to mach last 5 characters:
/.{5}$/
I know how to mach everything up to first -:
/[^-]*/
But I can not figure out how to combine them, and to make things worse I need to get the match without extracting it from specific groups and similar advanced regex stuff because I want to use it in SQL Anywhere, please help.
Tank you all for the help, but looks like a complete regex solution is going to be too complicated for my problem, so I did it very simple: SELECT right(regexp_substr('something-123', '[^-]*'), 4).
One option is to group the result:
(.{4})-
Now you have captured the result but without the -.
Or using lookarounds you can:
.{4}(?=-)
which matches any 4 characters that appears before "-".
You can use:
.{5}(?=(?:-[^-]*)?$)
See the regex demo
We match 5 symbols other than a newline only before the last - in the string or at the very end of the string ((?=(?:-[^-]*)?$)). You only need to collect the matches, no need checking groups/submatches.
UPDATE
To match any 1 to 5 characters other than a hyphen before the first hyphen (if present in the string), you can use
([^-]{1,5})(?:(?:-[^-]*)*)?$
See demo. We rely on a lookahead here, that checks if there are -+non-hyphen sequences are after the expected substring.
An faster alternative:
^[^-]*?([^-]{1,5})(?:-|$)
This regex will search for any characters other than - up to 1 to 5 such characters.
Note that here, the value we need is in Group 1.
How about:
(.{5})(?:-[^-]+)?$
The result is in group 1
Try this regex:
(.{1,5})(?:-.*|$)
Group 1 has the result you need
demo
I have a string that looks something like this:
"Element 1 | Element 2| Element 3: element 4"
I want to substring the portion of the source string that follows the colon (to the end of the source string), but if there is no colon, then I want to grab the whole string.
What I've tried so far are variations around this:
:.*|.*
:?.*
etc.
However, while they'll match if either the colon is present or not, they don't prefer the substring when the colon is found.
I've been playing with this on http://regexpal.com.
Ultimately, this will be used in a CMDB tool for matching CIs - so a general solution would be ideal, rather than language- or engine-specific.
You can use the following:
(:.*|[^:]*)$
See DEMO
Explanation:
if there is no colon, then I want to grab the whole string
This if condition can be specified using a negitive character class of colon
You can use:
(?:^|:)[^:\n]*$
RegEx Demo
I am using this pattern to extract confirmation dates from a text file and converting them to a date object (see my post here Extract/convert date from string in MS Access).
The current pattern matches all strings that look like a date, but may not be the confirmation date (which is always preceded by Confirmed by), and moreover, may not have complete date information (e.g. no AM or PM).
Pattern: (\d+/\d+/\d+\s+\d+:\d+:\d+\s+\w+|\d+-\w+-\d+\s+\d+:\d+:\d+)
Sample text:
WHEN COMPARED WITH RESULT OF 7/13/12 09:06:42 NO SIGNIFICANT
CHANGE; Confirmed by SMITH, MD, JOHN (2242) on 7/14/2012 3:46:21 PM;
The above pattern matches the following:
WHEN COMPARED WITH RESULT OF 7/13/12 09:06:42 NO SIGNIFICANT
^^^^^^^^^^^^^^^^^^^^
CHANGE; Confirmed by SMITH, MD, JOHN (2242) on 7/14/2012 3:46:21 PM;
^^^^^^^^^^^^^^^^^^^^
I want the pattern to look for the date in the segment of the text file that begins with Confirmed by and ends with a semi-colon. Also, in order to properly convert the time, the pattern should match only AM or PM at the end. How can I restrict the pattern to this segment and add the additional AM or PM criteria?
Can anyone help?
In order to match the end of the string, use $ at the end of your regex. To match the entire phrase "Confirmed by <someone> on <date>", use plain text (remember that plain text can be used in a regex as well -- if you aren't using special characters, the matcher will match your query verbatim). You need to use a negative look-ahead to exclude entire words.So maybe something like this:
Confirmed by (?!\ on\ )(\d+/\d+/\d+\s+\d+:\d+:\d+\s+\w+|\d+-\w+-\d+\s+\d+:\d+:\d+)$
Which will allow you to match a string that starts with "Confirmed by", followed by anything except for " on ", followed by the date that you capture, and the end of the string.
Edit: the negative look-ahead part is tricky, look at the answer below for more reference:
A regular expression to exclude a word/string
I don't see any need for a lookahead here, positive or negative. This works correctly on your sample string:
Confirmed by [^;]*(\d+/\d+/\d+\s+\d+:\d+:\d+(?:\s+(?:AM|PM))?|\d+-\w+-\d+\s+\d+:\d+:\d+);
The [^;]* effectively corrals the match between a Confirmed by sequence and its closing semicolon. (I'm assuming the semicolon will always be present.)
+(?:\s+(?:AM|PM))? makes the AM/PM optional, along with its leading whitespace.
The actual date will be stored in capturing group #1.
Try this:
(\d+/\d+/\d+\s+\d+:\d+:\d+\s+(?:AM|PM));
The simplest answer is more than often a good enough solution. By turning of the default greedy behavior (using the question mark: .*?) the regular expression will instead try to find the shortest match that matches the pattern. A pattern never matches the same string more than once, this means that each Confirmed by can only be coupled with one date which in this case is the next to follow.
Confirmed by.*?(\d+/\d+/\d+\s+\d+:\d+:\d+\s+(?:AM|PM));