more pythonic way to reveal string from sqlite buffer

Question

I'm storing FB post ids in a sqlite table. Now I want to query other data in the FB graph using the post_id. The hack below works but it feels pretty hokey...

cur.execute("SELECT post_id FROM fbposts") 
data_all = cur.fetchall()
for x in data_all:
    y = "%s" % x

the value of y is now something like 96085205666_10153983162390667 and can be used to construct a new FB api call but there must be a more pythonic way

I may be missing something- x doesn't act like a string; type(x) returns <type 'tuple'> type(x[0]) returns <type 'buffer'> — Timothy Lombard
– Timothy Lombard, Commented Dec 20, 2016 at 7:17

Jean-François Fabre · Accepted Answer · 2016-12-20 07:17:12Z

4

With your request, data_all is an iterable on tuples of 1 element

`y = "%s" % x` properly converts a 1-element tuple to a string, but you're right, it's not the best way.

fetchall(): Fetches all (remaining) rows of a query result, returning a list. Note that the cursor’s arraysize attribute can affect the performance of this operation. An empty list is returned when no rows are available.

To take this element and create a list with the strings, do:

[x for (x,) in data_all]

this unpacks x inside the tuple and creates a list of strings (equivalent to [x[0] for x in data_all])

answered Dec 20, 2016 at 7:17

Jean-François Fabre♦

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2 Comments

David542 Over a year ago

That's interesting, I've never seen the x for (x,) pattern before. Nice to know now.

Jean-François Fabre Over a year ago

yeah that was the occasion to expose it :)

David542 · Accepted Answer · 2016-12-20 07:19:41Z

2

What about using the following pattern:

cur.execute("SELECT post_id FROM fbposts") 
data_all = [item[0] for item in cur.fetchall()]

Now your data_all will be the list of strings that you want. And if you want to make a call for each post_id, you can now do:

for post_id in data_all:
    fb.call(post_id)

answered Dec 20, 2016 at 7:19

David542

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1 Comment

Timothy Lombard Over a year ago

I made a small change to get a string... data_all = [str(item[0]) for item in cur.fetchall()]

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