This is sample code.
int myCodeVersion = 100;
My .py code as below.
fp = open(path, 'r').read()
myCodeVer = ''.join(map(str, re.findall('myCodeVersion'+ '\s=\s(.*?) ', fp)))
print 'My Code version: ' + myCodeVer
This is result of my .py code
My Code version: 100;
I want to print only digit as below.
My Code version: 100
How can I fix the my regex?
myCodeVer = ''.join(map(str, re.findall('myCodeVersion'+ '\s=\s(\d*)', fp)))
re.findall('myCodeVersion\s*=\s*(\d+)', fp)[0] is enough. In fact, \d* is wrong, must be at least \d+.With that particular file you can do it like so:
import re
with open(file, 'r') as f:
myCodeVer = re.findall('(?<=myCodeVersion)(?:\s*=\s*)(\d+)', f.read())
print 'My Code version: ' + myCodeVer[0]
Output:
My Code version: 100
Regex:
(?<=myCodeVersion) - Positive look behind of myCodeVersion
(?:\s*=\s*) - Non capturing group, 0 or more spaces on either side of =
(\d+) - Capture 1 or more digits
A slightly more generic approach will be to to capture groups. For example, we can capture variable type, name and value as groups using the pattern-matcher as follows:
s = 'int myCodeVersion = 100;'
import re
# compile the regex pattern, if you need to use multiple times, it will be faster
pattern = re.compile(r"(int|float|double)\s+([A-Za-z][A-Za-z0-9_]+)\s*=\s*(\d+)\s*;")
m = re.match(pattern, s)
print 'var type: ' + m.group(1)
print 'var name: ' + m.group(2)
print 'var value: ' + m.group(3)
#var type: int
#var name: myCodeVersion
#var value: 100
For the specific case you are interested in the code can be simplified to the following using the same approach:
pattern = re.compile(r"int\s+myCodeVersion\s*=\s*(\d+)\s*;")
print 'My Code version: ' + ''.join(map(str, re.match(pattern, s).group(1)))
# My Code version: 100
This could work:
.*\s*=\s*(.*?);
