Sqlite3 - Update table using Python code - syntax error near %s

Question

This is my Python code -

cursor.execute("""UPDATE tasks SET task_owner=%s,task_remaining_hours=%s,                      task_impediments=%s,task_notes=%s WHERE task_id=%s""",                      (new_task_owner,new_task_remaining_hours,new_task_impediments,
                      new_task_notes,task_id))

This is the SQL statement I try in SQLite3 manager (Firefox extension)

UPDATE tasks SET task_owner=%s,task_remaining_hours=%d,task_impediments=%s,task_notes=%s WHERE task_id=%d,("sumod",10,"none","test",1)

The error I get is -

sqlite3.OperationalError: near "%": syntax error

I have tried many web searches including SO, tutorials and self-troubleshooting, but this error does not go away. What exactly I am doing wrong here.

Michael Berkowski · Accepted Answer · 2012-03-24 02:54:32Z

31

I believe Python's SQLite implementation uses ? placeholders, unlike MySQLdb's %s. Review the documentation.

cursor.execute("""UPDATE tasks SET task_owner = ? ,task_remaining_hours = ?,task_impediments = ?,task_notes = ? WHERE task_id= ? """,
  (new_task_owner,new_task_remaining_hours,new_task_impediments,new_task_notes,task_id))

edited Mar 24, 2012 at 2:54

answered Mar 24, 2012 at 2:46

Michael Berkowski

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Comments

DevB2F · Accepted Answer · 2019-05-10 00:36:59Z

0

It is also possible to use %s:

cursor.execute("UPDATE tasks SET task_owner='%s', task_remaining_hours='%s', 
task_impediments = '%s', task_notes = '%s' WHERE task_id= '%s' " % 
(new_task_owner,new_task_remaining_hours,new_task_impediments,new_task_notes,task_id))

answered May 10, 2019 at 0:36

DevB2F

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1 Comment

prhmma Over a year ago

it is possible, but it is very dangerous and strongly advised against it.

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