Question

I have a MySQL table named “content”containing (a.o.) the fields “_date” and “text”, for example:

_date      text
---------------------------------------------------------
2011-02-18 I'm afraid my car won't start tomorrow
2011-02-18 I hope I'm going to pass my exams
2011-02-18 Exams coming up - I'm not afraid :P
2011-02-19 Not a single f was given this day
2011-02-20 I still hope I passed, but I'm afraid I didn't
2011-02-20 On my way to school :)

I’m looking for a query to count the number of times the words “hope” and “afraid” are being used per day. In other words, the output would have to be something like:

_date      word   count
-----------------------
2011-02-18 hope   1
2011-02-18 afraid 2
2011-02-19 hope   0
2011-02-19 afraid 0
2011-02-20 hope   1
2011-02-20 afraid 1

Is there an easy way to do this or should I just write I different query per term? I now have this, but I don’t know what to put instead of “?”

SELECT COUNT(?) FROM content WHERE text LIKE '%hope' GROUP BY _date

Can somebody help met with the correct query for this?

Answer 1

I think the most easy and redable way is to make subquerys:

 Select 
    _date, 'hope' as word, 
    sum( case when `text` like '%hope%' then 1 else 0 end) as n
 from content
 group by _date
 UNION
 Select 
    _date, 'afraid' as word, 
    sum( case when `text` like '%afraid%' then 1 else 0 end) as n
 from content
 group by _date

This approach has not the best performace. If you are looking for performance you should grouping in subquery by day, also this like condition is a performance killer. This is a solution if you only execute the query in batch mode time by time. Explain your performance requeriments for an accurate solution.

EDITED TO MATCH LAST OP REQUERIMENT