RE: Changing the python sqlite3 version to the latest

[Jose Isaias Cabrera <jicman@outlook.com>]

On February 14, 2023 8:46 PM, Jose Isaias Cabrera expressed:
> 
> 
> On Tuesday, February 14, 2023 5:58 PM, Marco Atzeri expressed:
> >
> > On 11.02.2023 22:37, Jose Isaias Cabrera via Cygwin wrote:
> > >
> > > Greetings.
> > >
> > > If I install python39 and thus,
> > > $ python
> > > Python 3.9.10 (main, Jan 20 2022, 21:37:52) [GCC 11.2.0] on cygwin
> > > Type "help", "copyright", "credits" or "license" for more information.
> > >>>> import sqlite3
> > >>>> sqlite3.sqlite_version
> > > '3.34.0'
> > Last version packaged for Cygwin is 3.34.0
> >
> > Until Jan (or someone else) is packaging a more recent version that is
> > what is available
> >
> > Regards
> > Marco
> 
> Yes, but the question is how can I update the SQLite version? I have been
> able to do it with 2 PCs, but there is one that is not able. Some of the
> sqlite folks said that is a DLL issue.

In case anyone needs the answer, These steps worked for me:

-- Downloaded the Pre-release Snapshots
$ wget https://sqlite.org/snapshot/sqlite-snapshot-202302131932.tar.gz

-- untared the snapshot
$ tar xvf sqlite-snapshot-202302131932.tar.gz

-- cd to the untared directory
$ cd sqlite-snapshot-202302131932

$ ./configure --prefix=/usr

$ make install

And this process has set the python SQLite version to the sqlite-snapshot version. After that, you can download the trunk and follow the same procedure, and the version of the trunk will be changed also.

$ ./SQLiteVersion.py
3.41.0
['/usr/lib/python3.9/sqlite3']
3.41.0
2023-02-13 19:32:40 ecdeef43b27412b0b0b09e09a62ad3a03836a3fc80f2070268090e7ca8f02712

I hope this helps.
[JIC] 
This script may be useful...

$ cat SQLiteVersion.py
#!/usr/bin/python3

import sqlite3

def ConnectToSharedDB(sdb):
    return sqlite3.connect(sdb)

print(sqlite3.sqlite_version)
print(sqlite3.__path__)
SharedDB = ":memory:"
con = ConnectToSharedDB(SharedDB)
cur = con.cursor()
cur.execute("SELECT sqlite_version(),sqlite_source_id();")
for row in cur:
    print(row[0] + '\r\n' + row[1])

con.close()