6

I get the above error for executing the below INSERT-statement. The database file ce.db is in the same directory as my code and I have successfully created the tables therein.

My sqlite version is 2.8.17 and I am confident that my db file exists as I can see it in my directory and have succeeded in creating tables therein.

import sqlite3

@app.route("/sign_up", methods=["GET", "POST"])
def sign_up():
  # [..other code..]

    conn = sqlite3.connect("ce.db")
    c = conn.cursor()
    result = c.execute("INSERT INTO users (name, hash) VALUES (:name, :hash)", {"name":request.form.get("username"), "hash":hashp})
    conn.commit()

Debugger shows "sqlite3.DatabaseError: file is not a database" error for the line starting with "result=...".

Improve this question
3
  • The file you're trying to open as a database is obviously not a sqlite database; it's some other file that happens to have the same name as what you're trying to open - maybe at a different path? Always a possibility when using a relative path to the database file instead of an absolute one. Commented Oct 28, 2019 at 8:23
  • 1
    Also, sqlite 2.8.17? Er.... why? That was the last sqlite2 release and dates from 2005. There is no reason to be using such an old obsolete version. Hmm. Maybe you're trying to open a sqlite3 database? Commented Oct 28, 2019 at 8:25
  • 1
    I figured it out - posting answer below Commented Oct 28, 2019 at 8:52

4 Answers 4

Reset to default
8

I also had this problem with creating a db in Python using sqlite3, what solved it for me was to remove .db from database name:

conn = sqlite3.connect("ce.db") --> conn = sqlite3.connect("ce")

(+ also have 2.6.0 DB-API version and my sqlite version is 3.31.1)

Improve this answer
Sign up to request clarification or add additional context in comments.

1 Comment

Thanks bro, now it works after i see this solution
4

Did some more digging and it turns out that 2.6.0 is the DB-API version (obtained via print(sqlite3.version) - not entirely sure what that is or what it's for) and my sqlite version (print(sqlite3.sqlite_version)) is 3.22.0. Also realised (running file ce.db in bash) that my ce.db was created on (or with?) 2.x... managed to remove and recreate it with sqlite3 ce.db statement.. seems to work now. All beginner problems, I know but figured it might helpt to share this for any future lost souls like me :)

Improve this answer

1 Comment

Just fell for this after having created a db through cli and trying to access it with sqlalchemy!
1

Anna, please, take a close look at your code. You are trying to connect to ce.db, but, as you say, your database is cd.db. You have typo mistake in your code, here is Sqlite trying you to say file ce.db is not a database file.

conn = sqlite3.connect("ce.db") # here is misspelled database name
Improve this answer

1 Comment

Apologies. It is ce.db. I mistyped in the text. Both the code and the filename are ce.db.
0

Sometimes the DB files are not checked out correctly which can lead to this error. Try doing a git-lfs pull so that it checked out the lfs file correctly as a binary file.

Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.