title: Cannot use the sqlite3 parameter substitution with PRAGMA?
body: <p>I'm trying to change the <code>user_version</code> of an SQLite database via Python 2.6, and I've tried to do the following:</p>
  
  <pre><code>cur.execute( "PRAGMA user_version = ?" , (version,) )
  </code></pre>
  
  <p>It fails with the following error:</p>
  
  <pre><code>    cur.execute( "PRAGMA user_version = ?" , (version,) )
  sqlite3.OperationalError: near "?": syntax error
  </code></pre>
  
  <p>I've tried the named style of substitution (instead of question marks) but that also fails with the same error.</p>
  
  <p>If I drop a number in there as part of the SQL string or using Python's string operations it all works fine, but I'd rather not do either of those.</p>
  
  <p>So why isn't this working?<br>
  And how do I safety insert a number from a variable in to this call?</p>