php mysql check if vendor has 3 low consecutive ratings - php

Question

on my ratings table for my software i have 4 fields.
id autoincrement
rvid vendor id
ratedate date of rating
rating the actual numeric rating
I have done alot with it over the last few months but this time im stumped and i cant get a clear picture in my head of the best way to do this. What i am trying to do is find out if the vendor has had 3 low 'consecutive' ratings. If their last three ratings have been < 3 then i want to flag them.
I have been playing with this for a few hours now so i thought i would ask (not for the answer) but for some path direction just to push me forward, im stuck in thought going in circles here.
I have tried GROUP BY and several ORDER BY but those attempts did not go well and so i am wondering if this is not a mysql answer but a php answer. In other words maybe i just need to take what i have so far and just move to the php side of things via usort and the like and do it that way.
Here is what i have so far i did select id as well at first thinking that was the best way to get the last consective but then i had a small breakthrough that if they have had 3 in a row the id does not matter, so i took it out of the query.
$sql = "SELECT `rvid`, `rating` FROM `vendor_ratings_archive` WHERE `rating` <= '3' ORDER BY `rvid` DESC";
which give me this
Array
(
[0] => Array
(
[rvid] => 7
[rating] => 2
)
[1] => Array
(
[rvid] => 5
[rating] => 1
)
[2] => Array
(
[rvid] => 5
[rating] => 0
)
[3] => Array
(
[rvid] => 5
[rating] => 3
)
)
this is just just samples i tossed in the fields, and there are only 4 rows here where as in live it will be tons of rows. But basically this tells me that these are the vendors that have low ratings in the table. And that is where i get stumpted. I can only do one sort in the query so that is why i am thinking that i need to take this and move to the php side to finish it off.
I think i need to sort the elements by rvid with php first i think, and then see if three elements in a row are the same vender (rvid).
Hope that makes sense. My brain hurts lol...
update - here is all of the table data using *
Array
(
[0] => Array
(
[id] => 7
[rvid] => 7
[ratedate] => 2016-05-01
[rating] => 2
)
[1] => Array
(
[id] => 8
[rvid] => 5
[ratedate] => 2016-05-01
[rating] => 1
)
[2] => Array
(
[id] => 6
[rvid] => 5
[ratedate] => 2016-05-01
[rating] => 0
)
[3] => Array
(
[id] => 5
[rvid] => 5
[ratedate] => 2016-05-01
[rating] => 3
)
)

Answer 1

Here's one way you can begin approaching this - completely in SQL:
Get the last rating for the vendor. ORDER BY date DESC, limit 1.
Get the second to last rating for the vendor. ORDER BY date DESC, limit 1, OFFSET 1.
Then write a query that does a LEFT join of the first two tables. You will have a dataset that has three columns:
vendor id
last rating
second to last rating
Then you can write an expression that says "if column1 is <3 and column2 < 3, then this new column is true"
You should be able to extend this to three columns relatively easily.

Answer 2

Here is what a came up with to solve this riddle. I think explaining it on here helped as well as Alex also helped as he keyed my brain on using the date. I first started looking at using if statment inside of the query and actually that got my brain out of the box and then it hit me what to do.
It is not perfect and certainly could use some trimming to reduce the code, but i understand it and it seems to work, so that is par for me on this course.
the query...
$sql = "SELECT `rvid`, `ratedate`,`rating` FROM `vendor_ratings_archive` WHERE `rating` <= '3' ORDER BY `ratedate`, `rvid` DESC";
which gives me this
Array
(
[0] => Array
(
[rvid] => 7
[ratedate] => 2016-05-01
[rating] => 2
)
[1] => Array
(
[rvid] => 5
[ratedate] => 2016-05-01
[rating] => 1
)
[2] => Array
(
[rvid] => 5
[ratedate] => 2016-05-01
[rating] => 0
)
[3] => Array
(
[rvid] => 5
[ratedate] => 2016-05-01
[rating] => 3
)
)
notice how vendor (rvid) 5 is grouped together which is an added plus.
next a simple foreach to load a new array
foreach($results as $yield)
{
$rvidarray[] = $yield['rvid'];
}//close foreach
which gives me this
Array
(
[0] => 7
[1] => 5
[2] => 5
[3] => 5
)
then we count the array values to group dups
$rvidcounter = array_count_values($rvidarray);
which results in this
Array(
[7] => 1
[5] => 3
)
so now vender 7 as 1 low score and vendor 5 has 3 low scores and since they were already sorted by date i know that its consecutive. Well it sounds good anyway lol ")
then we create our final array with another foreach
foreach($rvidcounter as $key => $value)
{
//anything 3 or over is your watchlist
if($value > 2)
{
$watchlist[] = $key; //rvid number stored
}
}//close foreach
which gives me this
Array
(
[0] => 5
)
this was all done in a service function. So the final deal is everyone in this array has over 3 consecutive low ratings and then i just use a returned array back in my normal php process file and grab the name of each vender by id and pass that to the html and print out the list.
done...
please feel free to improve on this if you like. I may or may not use it because the above code makes sense to me. Something more complicated may not make sense to me 6 mos from now lol But it would be interesting to see what someone comes up with to shorten the process a bit.
Thanks so much and Happy Coding !!!!!
Dave :)

Answer 3

You could do it in SQL like that:
SET #rvid = -1;
SELECT DISTINCT rvid FROM
(
SELECT
rvid,
#neg := rating<3, /* 0 or 1 for addition in next line */
#count := IF(#rvid <> rvid , #neg, #count+#neg) AS `count`, /* set or add */
#rvid := rvid /* remember last row */
FROM
testdb.venrate
ORDER BY
rvid, datetime desc
) subq
WHERE count>=3
;
You set a variable to a non existing id. In each chronologically sorted row you check if rating is too low, that results in 1 (too low) or 0 (ok). If rvid is not equal to the last rvid, it means a new vender section is beginning. On begin of section set the value 0 or 1, else add this value. Finally store the current row's rvid for comparison in next row process.
The code above is looking for 3 consecutive low ratings (low means a value less than 3) over all the time.
A small modification checks if all the latest 3 ratings has been equal to or less than 3:
SET #rvid = -1;
SELECT DISTINCT
rvid
FROM
(
SELECT
rvid,
#high_found := rating>3 OR (#rvid = rvid AND #high_found) unflag,
#count := IF(#rvid <> rvid , 1, #count+1) AS `count`,
#rvid := rvid /* remember last row */
FROM
testdb.venrate
ORDER BY
rvid, datetime desc
) subq
WHERE count=3 AND NOT unflag
;