how to find all "1401/07/29 19:00:00" pattern in a text with regex in python
Question:
i have a text that contain "1401/07/29 19:00:00" pattern.
how find all top pattern in text with re.findall method in python
Answers:
if string = "1401/07/29 19:00:00"
re.findall(r'd{4}/d{2}/d{2} d{2}:d{2}:d{2}', string)
if string = '1401/07/29 19:00:00'
string = '1401/07/29 19:00:00'
#variant 1
print(re.findall(r'd{4}\/d{2}\/d{2} d{2}:d{2}:d{2}', string))
#output: ['1401\/07\/29 19:00:00']
#variant 2
print(re.findall(r'd{4}/d{2}/d{2} d{2}:d{2}:d{2}', string.replace("\",'' )))
#output: ['1401/07/29 19:00:00']
i have a text that contain "1401/07/29 19:00:00" pattern.
how find all top pattern in text with re.findall method in python
if string = "1401/07/29 19:00:00"
re.findall(r'd{4}/d{2}/d{2} d{2}:d{2}:d{2}', string)
if string = '1401/07/29 19:00:00'
string = '1401/07/29 19:00:00'
#variant 1
print(re.findall(r'd{4}\/d{2}\/d{2} d{2}:d{2}:d{2}', string))
#output: ['1401\/07\/29 19:00:00']
#variant 2
print(re.findall(r'd{4}/d{2}/d{2} d{2}:d{2}:d{2}', string.replace("\",'' )))
#output: ['1401/07/29 19:00:00']