Python equivalent of Perl e flag with regular expression

[Peter Otten]

Friedman, Jason wrote:

> I have lines that look like this:
> select column1, 'select' as type
> from table
> where column2 = 'foo'
> 
> I want to return:
> SELECT column1, 'select' AS type
> FROM table
> WHERE column2 = 'foo'
> 
> This is SQL with the keywords converted to uppercase.  Note that the
> second "select" string is not a keyword and thus I do not want to
> convert it to uppercase.  Thus, I don't think the string.replace()
> method will work for me.
> 
> With Perl I might do something like this:
> $line =~ s/(select)/uc($1)/e;
> More generally:
> for $keyword in (@keyword) {
>   $line =~ s/($keyword)/uc($1)/e;
> }

I think your perl code is broken. It mechanically replaces the first
occurence of the keyword. This fails e. g. for

"select 'from' as x, b as y from table;"

> How would I do this with Python?

Here's my attempt, but you won't get 100% reliability without a real parser.

import re

sql = "select 'x' as t, 'y''' as u, selected, 'from' as fromage from temp;"


def fix_sql(sql):
    def sub(m):
        kw = m.group(3)
        if kw:
            return kw.upper()
        return m.group(1) or m.group(2)
    return re.compile("""('.*?')|(".*?")|\\b(select|as|from)\\b""").sub(sub,
sql)

print fix_sql(sql)

Peter