[SOLVED] How to remove pragma busy_timeout value from sqlite3 command result in bash?

Issue

result=$(sqlite3 sample.db "SELECT count(*) FROM table WHERE value='sample';")
echo $result

Result is 1.
Because of "Database is locked" error I changed the code:

result=$(sqlite3 sample.db "pragma busy_timeout=20000; SELECT count(*) FROM table WHERE value='sample';")
echo $result

But now the result is:

20000 1

I don't want busy_timeout value in the result.
I want the result to be 1.
How can I do that by editing sqlit3 commands?

Solution

Thanks to @user4157124
Sqlite3 command must edit as below:
result=$(sqlite3 -cmd ".timeout 20000" sample.db "SELECT count(*) FROM table WHERE value='sample';")
It prevents Database is locked Error and the result varibale doesn't combine with the timeout value.

Answered By - Amin SZ

Answer Checked By - Pedro (WPSolving Volunteer)