I am trying to come up with a smallest regex possible for extracting parts of a string with the last section being an optional one. The string will look something like:
jack:Bill(23):Space Force (23, Apple;Orange)
or
jack:Bill(23):Space Force
I need to extract as follows:
Jack
Bill(23)
Space Force
23
Apple;Orange
The last 2 items may or may not appear based on the source string. I am trying with a regex like:
(.*?):(.*?):(.*?)(\\(([0-9]+),([^\\)]*)?\\))?
But this does not seem to work.
I got it working with (.*?):(.*?):([^\\(]*)(\\(([0-9]+), ([^\\)]*)?\\))?
Related
I am at the beginning of learning Regex, and I use every opportunity to understand how it's working. Currently I am trying to extract dates from a text file (which is in fact a vnt-file type from my mobile phone). It looks like following:
BEGIN:VNOTE
VERSION:1.1
BODY;ENCODING=QUOTED-PRINTABLE;CHARSET=UTF-8:18.07.=0A14.08.=0A15.09.=0A15.10.=
=0A13.11.=0A13.12.=0A12.01.=0A03.02. Grippe=0A06.03.=0A04.04.2015=0A0=
5.05.2015=0A03.06.2015=0A03.07.2015=0A02.08.2015=0A30.08.2015=0A28.09=
17.11.2017=0A
DCREATED:20171118T095601
X-IRMC-LUID:150
END:VNOTE
I want to extract all dates, so that the final list is like that:
18.07.
14.08.
15.09.
15.10.
and so on. If the date has also a year, it should also be displayed.
I almost found out how to detect the dates by the following regex:
.+(\d\d\.\d\d\.(2015|2016|2017)?).+
But it only detect very few of the dates. The result is this:
BEGIN:VNOTE
VERSION:1.1
15.10.
04.04.2015
30.08.2015
24.01.2016
DCREATED:20171118T075601
X-IRMC-LUID:150
END:VNOTE
Then I tried to add a question mark which makes the .+ not greedy, as far as I read in tutorials. Then the regex looks like:
.+?(\d\d\.\d\d\.(2015|2016|2017)?).+?
But the result is still not what I am looking for:
BEGIN:VNOTE
VERSION:1.1
21.03.20.04.18.05.18.06.18.07.14.08.15.09.15.10.
13.11.13.12.12.01.03.02.06.03.04.04.20150A0=
03.06.201503.07.201502.08.201530.08.20150A28.09=
28.10.201525.11.201528.12.201524.01.20160A
DCREATED:20171118T075601
X-IRMC-LUID:150
END:VNOTE
For someone who is familiar with regex I am pretty sure this is very easy to solve, but I don't get it. It's very confusing when you are new to regex. I tried to find a hint in some tutorials or stackoverflow posts, but all I found is this: Notepad++ how to extract only the text field which is needed?
But it doesn't work for me. I assume it might have something to do with the fact that my text file is not one single line.
I have my example on regex101 too.
I would be very thankful if maybe someone can give me a hint what else I can try.
Edit: I would like to detect the dates with the regex and as a result have a list with only the dates (maybe it is called substitute?)
Edit 2: Sorry for not mentioning it earlier: I just want to use the regex in e.g. Notepad++ or an online regex test website. Just to get the result of the dates and save the result in a new txt-file. I don't want to use the regex in an programming language. My apologies for not being precisely before.
Edit 3: The result should be a list with the dates, and each date in a new line:
I want to extract all dates, so that the final list is like that:
18.07.
14.08.
15.09.
15.10.
I suggest this pattern:
(?:.*?|\G)(\d\d\.\d\d\.(?:\d{4})?)
This makes use of the \G flag that, in this case, allows for multiple matches from the very start of the match without letting any single unmatched character in the text, thus allowing the removal of all but what's wanted.
If you want to remove the extra matches as well, add |.* at the end:
(?:.*?|\G)(\d\d\.\d\d\.(?:\d{4})?)|.*
regex101 demo
In N++, make sure the options underlined are selected, and that the cursor is at the beginning. In the picture below, I replaced then undid the replacement, only to show that matches were identified (16 replacements).
You can try using the following pattern:
\d{2}\.\d{2}\.(?:\d{4})?
This will match day.month dates of the form 18.07., but it also allows such a date to be followed by a four digit year, e.g. 18.07.2017. While it would be nice to make the pattern more restrictive, to avoid false fire matches, I do not see anything obvious which can be added to the above pattern. Follow the demo link below to see the pattern in action.
Demo
I am trying to extract data from semi structured text, it is an email composed of tab delimited tables. Users have entered time stamp on top of each table and within the table they list security identifiers that I am looking for.
The goal is to extract correctsecurity and the time stamp on top of the table that correctsecurity is located at.
For example...
10:00 AM
not it
not it
9:00 AM
not it
correctsecurity
..is supposed to return 9:00 AM correctsecurity. However my current regex is returning 10:00 AM correctsecurity, meaning right item, but not the right time.
Here is my regex so far:
((1[0-2]|[0-9]):[0-5][0-9](\s?(AM|PM))?)(?:(.*\n)+)(correctsecurity)
Note that the last part correctsecurity is being created dynamically based on other criteria so even if I were to provide the actual item in this question it would do little help(because it is one of many), for simplicity sake please assume that correctsecurity is exactly the item I am looking for.
Lastly I am doing this in VBA so maybe solving this whole problem is easier without using a long regex, so feel free proposed non regex solutions.
To solve the main problem simply change the central section of your regex to not accept empty lines:
.*\n -> .+\n
Then add a newline anchor \n before the central section to avoid the skip of the AM|PM section:
So your regex will be:
((1[0-2]|[0-9]):[0-5][0-9](\s?(AM|PM))?)\n(?:(.+\n)+)(correctsecurity)
^ ^
Changes --------------------------------|------|
Optional Optimization
You may remove many unneeded groups and add a generic multi-os regex for the newline (?:\r\n?|\n):
((?:1[0-2]|[0-9]):[0-5][0-9](?: [AP]M)?)(?:\r\n?|\n)(?:[^\r\n]+(?:\r\n?|\n))+(correctsecurity)
You can solve this with negative lookahead :
a((?!a).)*correctsecurity
Where a is the pattern that you want to start the match at and don't want to encouter in the middle of the match.
Applied to your specif needs :
\d*:\d* [AP]M((?!\d*:\d* [AP]M).)*correctsecurity
Don't forget to let the dot match line breaks.
I assume VBA uses thee VBScript regex dialect which requires the following modification:
\d*:\d* [AP]M((?!\d*:\d* [AP]M)[\s\S])*correctsecurity
I am trying to use PSQL, specifically AWS Redshift to parse a line. Sample data follows
{"c.1.mcc":"250","appId":"sx-calllog","b.level":59,"c.1.mnc":"01"}
{"appId":"sx-voice-call","b.level":76,"foreground":9}
I am trying the following regex in order to to extract the appId field, but my query is returning empty fields.
'appId\":\"[\w*]\",'
Query
SELECT app_params,
regexp_substr(app_params, 'appId\":\"[\w*]\",')
FROM sample;
You can do that as follows:
(\"appId\":\"[^"]*\")(?:,)
Demo: http://regex101.com/r/xP0hW3
The first extracted group is what you want.
Your regex was not matching because \w does not match -
Adding this here despite this being an old question since it may help someone viewing this down the road...
If your lines of data are valid json, you can use Redshift's JSON_EXTRACT_PATH_TEXT function to extract the value a given key. Emphasis on the json being valid, as it will fail if even one line cannot be parsed and Redshift will throw a JSON parsing error.
Example using given data:
select json_extract_path_text('{"c.1.mcc":"250","appId":"sx-calllog","b.level":59,"c.1.mnc":"01"}','appId');
returns sx-calllog
This is especially useful since Redshift does not support lookahead/lookbehind (it is POSIX regex) & extract groups.
You can try using some lookahead and look behinds to isolate just the text inside the quotes for the appid. (?<=appId\":\")(?=.*\",)[^\"]*. I tested this out a bit using your examples you provided here.
To explain the regex a bit more: (?<=appId\":\")(?=.*\",)[^\"]*
(?<=appId\":\"): positive look behind for appid":". Since you don't want the appid text itself being returned (just the value), you can preface the regex with a look behind to say "find me the following regex, but only when it is following the look behind text.
(?=.*\",): positive look ahead for the ending ",. You don't want quotes to be returned in your match, but as with number 1 you want your regex to be bounded a bit and a look ahead does that.
[^\"]*: The actual matching portion. You want to find the string of chars that are NOT ". This will match the entire value and stop matching right before the closing ".
EDIT: Changed the 3rd step a little bit, removed the , from that last piece, it is not needed and would break the match if the value were to actually contain a ,.
I need to build a RegEx expression which gets its text strings from the Title field of my Database. I.e. the complete strings being searched are: Mr. or Ms. or Dr. or Sr. etc.
Unfortunately this field was a free field and anything could be written into it. e.g.: M. ; A ; CFO etc.
The expression needs to match on everything except: Mr. ; Ms. ; Dr. ; Sr. (NOTE: The list is a bit longer but for simplicity I keep it short.)
WHAT I HAVE TRIED SO FAR:
This is what I am using successfully on on another field:
^(?!(VIP)$).* (This will match every string except "VIP")
I rewrote that expression to look like this:
^(?!(Mr.|Ms.|Dr.|Sr.)$).*
Unfortunately this did not work. I assume this is because because of the "." (dot) is a reserved symbol in RegEx and needs special handling.
I also tried:
^(?!(Mr\.|Ms\.|Dr\.|Sr\.)$).*
But no luck as well.
I looked around in the forum and tested some other solutions but could not find any which works for me.
I would like to know how I can build my formula to search the complete (short) string and matches everything except "Mr." etc. Any help is appreciated!
Note: My Question might seem unusual and seems to have many open ends and possible errors. However the rest of my application is handling those open ends. Please trust me with this.
If you want your string simply to not start with one of those prefixes, then do this:
^(?!([MDS]r|Ms)\.).*$
The above simply ensures that the beginning of the string (^) is not followed by one of your listed prefixes. (You shouldn't even need the .*$ but this is in case you're using some engine that requires a complete match.)
If you want your string to not have those prefixes anywhere, then do:
^(.(?!([MDS]r|Ms)\.))*$
The above ensures that every character (.) is not followed by one of your listed prefixes, to the end (so the $ is necessary in this one).
I just read that your list of prefixes may be longer, so let me expand for you to add:
^(.(?!(Mr|Ms|Dr|Sr)\.))*$
You say entirely of the prefixes? Then just do this:
^(?!Mr|Ms|Dr|Sr)\.$
And if you want to make the dot conditional:
^(?!Mr|Ms|Dr|Sr)\.?$
^
Through this | we can define any number prefix pattern which we gonna match with string.
var pattern = /^(Mrs.|Mr.|Ms.|Dr.|Er.).?[A-z]$/;
var str = "Mrs.Panchal";
console.log(str.match(pattern));
this may do it
/(?!.*?(?:^|\W)(?:(?:Dr|Mr|Mrs|Ms|Sr|Jr)\.?|Miss|Phd|\+|&)(?:\W|$))^.*$/i
from that page I mentioned
Rather than trying to construct a regex that matches anything except Mr., Ms., etc., it would be easier (if your application allows it) to write a regex that matches only those strings:
/^(Mr|Ms|Dr|Sr)\.$/
and just swap the logic for handling matching vs non-matching strings.
re.sub(r'^([MmDdSs][RSrs]{1,2}|[Mm]iss)\.{0,1} ','',name)
I am trying to extract some information from facebook using Regex. Here is a link with an example:
https://graph.facebook.com/210989592315921
I was interested in what would the regular expression be in order to extract just the number of likes from this string.
I have tried for example this expression:
"likes":\s[0-9]$
Thank you in advance for any advice regarding this matter,
Mark
You should follow "#Hope I helped" comment and use a json parser. You can't be sure the text is going to be formatted always the same way:
Are you always going to have a single space between : and the number ?
By the way, here is the error you are looking for, your current regex matches a single figure, not a multiple digit number, you should use something like: [0-9]+ and probably remove the $ which is not correct in your example, as you have a comma after the number.