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I got a string representing users data.
What is the proper regex to extract domain in this string?
I know that I have to find all strings with 2 characters matching the condition that it comes after the last "." after a "#".
However I still failed to implement it.
import re
regex = r"#.+\.([a-z]{2}),"
your_string = ("001,Francisca,Dr Jhonaci,jhonadr#abc.com,32yearsold,120.238.225.0\n"
"002,Lavenda,Bocina,lavenboci#banck.ac.uk,50yearsold,121.186.221.182\n"
"003,Laura,Eglington,elinton#python.co.jp,26yearsold,36.55.173.63\n"
"004,Timo,Baum,timobaum#tennis.co.cn,22yearsold,121.121.110.10")
matches = re.finditer(regex, your_string, re.MULTILINE)
for match in matches:
result = match.group(1)
print(result)
The comma seems to be the delimiter in the string.
To not cross-matching a comma (to prevent matching too much), and also not cross-matching a second # char you can use a negated character class starting with [^
If the entry can also be at the end of the string, you can assert either a , or the end of the string.
#[^#,]*\.([A-Za-z]{2})(?=,|$)
Regex demo
import re
regex = r"#[^#,]*\.([A-Za-z]{2})(?=,|$)"
s = ("001,Francisca,Dr Jhonaci,jhonadr#abc.com,32yearsold,120.238.225.0\n"
"002,Lavenda,Bocina,lavenboci#banck.ac.uk,50yearsold,121.186.221.182\n"
"003,Laura,Eglington,elinton#python.co.jp,26yearsold,36.55.173.63\n"
"004,Timo,Baum,timobaum#tennis.co.cn,22yearsold,121.121.110.10")
print(re.findall(regex, s, re.M))
Output
['uk', 'jp', 'cn']
Use the comma after the email instead of the last point.
Using this regex
#.+\.(\w+)(?<!com),
the capturing group will contain the info that you want.
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 2 years ago.
The question is to 'query the list of CITY names from STATION that either do not start with vowels or do not end with vowels.' But the not here should work as "and" instead of "or"?
The problem is stated as "NOT(starting with a vowel) OR NOT(ending with a vowel)". The sql expression is saying "NOT(starting with a vowel AND ending with a vowel)", which is equivalent: https://en.wikipedia.org/wiki/De_Morgan%27s_laws
...except that the sql expression will produce the wrong result for city names that are just a single vowel character, because the regex will only match values when there are at least two characters.
You regex should be
(^[aeiou])|([aeiou]$)
Right now regex will only match if it start and ends with a vowel, while you want to say, start with vowel and it doesn't matter what is after, or it doesn't matter what is at the begining as long as it ends with a vowel.
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I saw this question and tested the answers but noticed that executing SELECT ... WHERE column LIKE "%string%" OR string LIKE CONCAT("%", column, "%")
string LIKE CONCAT("%", column, "%") is not secure if the value of the column contains % and secondly if the column is null it returning true which is not correct since the column contains nothing.
You can just escape the percent signs, if any exist:
SELECT column1
FROM table
WHERE (
column2 LIKE "%string%"
OR string LIKE CONCAT("%", REPLACE(column2, '%', '|%'), "%") ESCAPE |
)
AND column2 IS NOT NULL;
The default escape character is a backslash but this is not ANSI compliant and can be a pain to work with if you're building a query in another language. So I use the LIKE ... ESCAPE syntax to specify my own escape.
CONCAT() returns NULL if any of its arguments are null, so if you're concerned about that, just check for it.
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I need to find all entries that contain more than one colon (:) character.
However when I do LIKE %:% it shows the entire table because of http://. How can I find more than one colon?
SELECT *
FROM `downloads`
WHERE `url` LIKE '%:%'
LIMIT 0 , 30
If you want to find a colon that occurs after the scheme of your URL, then change your LIKE clause accordingly:
SELECT *
FROM `downloads`
WHERE `url` LIKE '%:%:%'
LIMIT 0 , 30
The first colon will be in your scheme, and the second will be somewhere else in the Url after the scheme.
A word of caution, however - it is completely valid to have a colon in the Url when a port number is specified, e.g.: http://localhost:8080
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MySQL does not support the Negative Lookahead. How can I find the result not containing a string using REGEXP.
I am using 'NOT REGEXP' but the result is unexpected.
there is a 'Content' column in my table,i want to find the rows which the Content column contains '' label,but i still want some src to be excluded.
here is the sql:
Content REGEXP '.' AND Content NOT REGEXP '.(test.mywebsite1.com/|img.mywebsite.com/face/|test.mywebsite.com/phoneIcon.jpg).*'
but when the Content contain both and it works unexpected;
Test your REGEXP on a known set, get that working, and verify it is working.
Then add the NOT to get the boolean inverse.
Note that a MySQL boolean expression will return one of three possible values: TRUE, FALSE and NULL.
And note that NOT expr will also return one of three possible values: TRUE, FALSE and NULL.
When expr returns NULL, then NOT expr will also return NULL.
It's not really productive to attempt to provide any other assistance, absent an actual question and more details of what you are attempting to do.