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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.
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I have two columns in a database I want to compare. One is email addresses and the other is username.
I'm looking to return results if the first part of username matches the first component of the email address (stuff before the # sign).
Sample data:
username emailaddress
badzzycshulzey9802 badzzycshulzey#gmail.com
trogleddg1919 trogleddg#gmail.com
Tried a variety of queries but I can't seem to get this one.... thanks!
You don't really need a regexp for this, strings functions can do it:
select *
from mytable
where username like concat(substr(emailaddress, 1, locate('#', emailaddress) - 1), '%')
locate('#', emailaddress) gives you the position of the arobas in the email address, and substr(emailaddress, 1, locate(...) - 1) extracts everything before that. Then, we can check if the username starts with that part of the email address using like with a wildcard on the right side.
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I am parsing an excel file and adding the values im getting into a database. But one of the columns I am parsing can be either a float or a string. Like such: 0,45 or Contact Support. What is the best way for me to add this as it is, to my database? Or more how should I format my database column for this?
use varchar datatype(in MySQL) for the column you want to parse and you can save both the integer and character values.
I recommend you create a float type column and a varchar column. Write your parser so that valid float values are filled to the float column and the rest to the varchar one. This will allow you to both sort your database by the value of the float type column and run queries on fields that are missing that value: SELECT * FROM mytable WHERE myfloat IS NULL
first am with Kaivosukeltaja answer.
but if you can't do that, make a varchar column to save the strings or doubles.
BTW when you get data from a database all the data comes in string format then you can cast it to any type you want.
if you are using PHP:
$x = $row['string_double_column'];
if(is_numeric($x))
return doubleval($x);
else
return $x;
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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