Python equivalent of Perl e flag with regular expression

[Friedman, Jason]

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;
}

How would I do this with Python?

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[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

[George Sakkis]

On Oct 2, 1:06 pm, "Friedman, Jason" <[e-mail]>
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;

}

Are you sure that this version returns the desired results ? How does
perl know not to much the keyword within the quotes ?

How would I do this with Python?

Leaving aside the fact that regexps are not the right tool to check
whether a string is within quotes or not, you can use re.sub() and
pass a callable instead of a replacement string:

>>import re
keywords = 'select from where as'.split()
regex = re.compile('|'.join(r'\b%s\b' % re.escape(k) for k in keywords), re.I)
sql = """

select column1, 'select' as type
from table
where column2 = 'foo'
"""

>>print regex.sub(lambda match: match.group().upper(), sql)

SELECT column1, 'SELECT' AS type
FROM table
WHERE column2 = 'foo'

George

[MRAB]

On Oct 2, 6:06*pm, "Friedman, Jason" <[e-mail]>
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.

[snip]

FYI, yhe replace method can take a third argument, which is the
maximum number of replacements to do:

>>query = "select column1, 'select' as type from table where column2 = 'foo'"
query.replace("select", "SELECT", 1)

"SELECT column1, 'select' as type from table where column2 = 'foo'"