`transform=True` fails when you have an index on a column name with an underscore

I'm using this project (sqlite_minutils) as a part of a FastHTML web app and came across a bug. I used `python=3.12` and `sqlite-minutils==3.37.0.post3`. Here's a minimal reproduction:

```python
#script.py

from fasthtml.common import Database
from datetime import datetime

db = Database('test.db')

class Request:
id: str
time: str
# run with time_check and its index commented out first
time_check: str

requests = db.create(Request, if_not_exists=True, transform=True)
requests.create_index(('time',), unique=True, if_not_exists=True)
# first run with this index creation commented out
requests.create_index(('time_check',), unique=True, if_not_exists=True)

for i in range(5):
requests.insert(Request(time=datetime.now()))
```
To reproduce, simply first comment out `time_check: str` and the index on `time_check`, run `python script.py`. Then uncomment `time_check` and its index creation and re-run `python script.py`. You will get the following error, which is caused by the query returning nothing, and the index `[0]` trying to grab what isn't there.

```bash
File venv/lib/python3.12/site-packages/sqlite_minutils/db.py:1910, in Table.transform_sql(self, types, rename, drop, pk, not_null, defaults, drop_foreign_keys, add_foreign_keys, foreign_keys, column_order, tmp_suffix, keep_table)
1908 for index in self.indexes:
1909 if index.origin not in ("pk"):
-> 1910 index_sql = self.db.execute(
1911 """SELECT sql FROM sqlite_master WHERE type = 'index' AND name = :index_name;""",
1912 {"index_name": index.name},
1913 ).fetchall()[0][0]
1914 assert index_sql is not None, (
1915 f"Index '{index}' on table '{self.name}' does not have a "
1916 "CREATE INDEX statement. You must manually drop this index prior to running this "
1917 "transformation and manually recreate the new index after running this transformation."
1918 )
1919 if keep_table:

IndexError: list index out of range
```

It looks like the parameter substitution fails when there is an extra underscore from the column name. If I do the same query manually using the `?` substitution syntax it works, and if I do the query manually typing out the index name verbatim it also works.

Beginne in sqlite_minutils/db.py bei Table.transform_sql, insbesondere bei der sqlite_master-Suche um Zeile 1910, und führe die Reproduktion des Issues mit script.py aus. Prüfe die parametrisierte Abfrage des Indexnamens anhand von Namen, die Unterstriche enthalten; abgeschlossen ist die Aufgabe, wenn die Transformation ohne IndexError abgeschlossen wird und den betroffenen Index beibehält.

Vom Indexierungsmodell aus dem Issue-Text verfasst.