Question

I’m having trouble with this SQL:

$sql = mysql_query("SELECT $menucompare ,
  (COUNT($menucompare ) * 100 / (SELECT COUNT( $menucompare )
  FROM data WHERE $ww = $button ))  AS percentday FROM data WHERE $ww >0  ");

• $menucompare is table fields names what ever field is selected and contains data bellow
• $button is the week number selected (lets say week ‘6’)
• $ww table field name with row who have the number of week ‘6’

For example, I have data in $menucompare like that:

123456bool
521478bool
122555heel
147788itoo

and I want to select those, who have same word in the last of the data and make percentage.

The output should be like that:

• bool — 50% (2 entries)
• heel — 25% (1 entry)
• itoo — 25% (1 entry)

Any clearness to my SQL will be very appreciated.
I didn’t find anything like that around.

Answer 1

Well, keeping data in such format probably not the best way, if possible, split the field into 2 separate ones.

First, you need to extract the string part from the end of the field.

• if the length of the string / numeric parts is fixed, then it’s quite easy;
• if not, you should use regular expressions which, unfortunately, are not there by default with MySQL. There’s a solution, check this question: How to do a regular expression replace in MySQL?

I’ll assume, that numeric part is fixed:

SELECT s.str, CAST(count(s.str) AS decimal) / t.cnt * 100 AS pct
  FROM (SELECT substr(entry, 7) AS str FROM data) AS s
  JOIN (SELECT count(*) AS cnt FROM data) AS t ON 1=1
 GROUP BY s.str, t.cnt;

If you’ll have regexp_replace function, then substr(entry, 7) should be replaced to regexp_replace(entry, '^[0-9]*', '') to achieve the required result.

Variant with substr can be tested here.