SQLite3 Python用列表更新整行

Question

System

• Python 2.7
• SQLite3

Info

I have table with the following structure:

c.execute("CREATE TABLE foo (id INTEGER, number INTEGER, number2 REAL, letters TEXT)")

I got the following list

list_foo = [9, 1, 2.3, 'abc']

edit: In the table I already have fields, so the table should in this case be updated where id=9

Question

How do I update a row with my list variable?

Solution idea

c.execute("UPDATE foo SET VALUES = ? WERE id = ?", (list_foo, listfoo[0]))

But of course VALUES doesn't work. How do I instantly update the row without typing:

c.execute("UPDATE foo SET id = ?, number = ?, number2 = ?, letters = ? WERE id = ?", (list_foo[1], list_foo[2], list_foo[3], list[0]))

My actual table has more entries than this, so it would be a pain to type it like this.

Edit 2: If it's not possible and I have to use my long SQLite3 code, is it at least possible to have:

(list_foo[1:], list_foo[0])

instead of

(list_foo[1], list_foo[2], list_foo[3], list[0]))

Answer 1

Update** (You can hack something by escaping the quotes, other than that I'm not sure if what you are looking for exists)

def quote_identifier(s, errors="strict"):
    encodable = s.encode("utf-8", errors).decode("utf-8")

    nul_index = encodable.find("\x00")

    if nul_index >= 0:
        error = UnicodeEncodeError("NUL-terminated utf-8", encodable,
                                   nul_index, nul_index + 1, "NUL not allowed")
        error_handler = codecs.lookup_error(errors)
        replacement, _ = error_handler(error)
        encodable = encodable.replace("\x00", replacement)

    return "\"" + encodable.replace("\"", "\"\"") + "\""


list_foo  = [2, 2, 2.5, 'abd']
tab_names = ('number','number2','letters')
dic = dict(zip(tab_names, list_foo[1:len(list_foo)]))

for x in dic:
  c.execute("UPDATE foo SET"+quote_identifier(x)+"=? WHERE rowid=?",[dic[x], list_foo[0]])
rec_con.commit()