Ignore uppercase in a query, Python

Question

I have this code

query = request.GET.get("q")
    if query:
        if query == "song" or query == "songs":
            return render(request, 'music/cloud.html', {
                'songs': song_results,
                'generic_songs': song_results_generic,
            })

The thing is, I want the "if" statement to ignor uppercases. For example, when the user search by "SoNg" or "SonG" or.. it will be converted to "song" before testing

But when you are using lower() make sure that what you are comparing is also ".lower()" because python IS case sensitive. — Alex, Oct 11 '17 at 21:52
the fact that this text comes from a `request.get` seems irrelevant here, it's just case insensitive string matching. — Adam Smith, Oct 11 '17 at 21:58

score 2 · Accepted Answer · answered Oct 11 '17 at 21:51

2

Use .lower()

if query.lower() in ['song', 'songs']:

answered Oct 11 '17 at 21:51

kloddant

1,026
12
19

score 1 · Answer 2 · edited Jun 20 '20 at 09:12

In Python3.3+, you should prefer str.casefold .

if query.casefold() in ['song'.casefold(), 'songs'.casefold()]:

From the docs:

Return a casefolded copy of the string. Casefolded strings may be used for caseless matching.

Casefolding is similar to lowercasing but more aggressive because it is intended to remove all case distinctions in a string. For example, the German lowercase letter 'ß' is equivalent to "ss". Since it is already lowercase, lower() would do nothing to 'ß'; casefold() converts it to "ss".

The casefolding algorithm is described in section 3.13 of the Unicode Standard.

Ignore uppercase in a query, Python

2 Answers2