python -- for with an if statement

Question

I dont understand why, when i run my code the for each loop under the if statement isn't run. Even when the number of found is greater than 0!

def findpattern(commit_msg):
    pattern = re.compile("\w\w*-\d\d*")
    group = pattern.finditer(commit_msg)
    found = getIterLength(group)
    print found
    if found > 0:
        issues = 0
        for match in group:
                print " print matched issues:"
                auth = soap.login(jirauser,passwd)

                print match.group(0)
                getIssue(auth,match.group(0))
                issues = issues + 1
    else: 
        sys.exit("No issue patterns found.")

print "Retrieved issues: " + str(issues)  

Any help would be appreciated, I have been banging my head on this for an hour.

Answer 1

Check your code formatting it looks like under the for you have a double tab instead of a single tab, remember python is very picky about indentation

Answer 2

Your getIterLength() function is finding the length by exhausting the iterator returned by finditer(). You would then need a new iterator instance for the for loop. Instead, I would restructure your code like this:

def findpattern(commit_msg):
    pattern = re.compile("\w\w*-\d\d*")
    group = pattern.finditer(commit_msg)

    found = 0
    issues = 0
    for match in group:
        print " print matched issues:"
        auth = soap.login(jirauser,passwd)

        print match.group(0)
        getIssue(auth,match.group(0))
        issues = issues + 1
        found += 1
    if found == 0:
        sys.exit("No issue patterns found.")


    print "Retrieved issues: " + str(issues)  

OR, you could use the findall() method instead of finditer() to give you a list (which is an iterable, not an iterator) on which you can run len(group) to get the size and then use it to iterate over in your for loop.