Inner joins step 1. Counting newspaper subscribers


#1

Hello, have a problem with first exercise, about newspaper.
The “newspaper” table contains information about the newspaper subscribers. Count the number of subscribers who get a print newspaper.
My decision is:
SELECT COUNT(subscribers)
FROM newspaper
WHERE subscribers = ‘print newspaper’;

But it`s wrong, what a problem?

In the instruction to excersize we clearly see thar we need to count subcribers which have a type of “print newspaper” subscribe.


#2

It doesn’t say anywhere that ‘print newspaper’ is data existing in your table, nor does there exist a column named subscribers. (You might be getting an error telling you this)

You also chose column name/content to compare to without looking at the table (you got it out of thin air), seems unlikely that it’d match…

The instructions do however describe how the sought-for information is represented:

Some users subscribe to just the newspaper, some subscribe to just the online edition, and some subscribe to both.

The table newspaper contains information about the newspaper subscribers.

The table online contains information about the online subscribers.


#3

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