Sum tids varighed efter stedsændring

Følgende ser ud til at gøre tricket:

SET @locationID=0,@ts=NULL,@changed=0;

SELECT
  MIN(assetID) AS id
  , MIN(locationID) AS location
  , SUM(secDiff) AS duration
FROM
  (SELECT
    assetID
    , locationID
    , @changed := IF(locationID <> previousLocationID, @changed + 1, @changed) AS changed
    , IFNULL(TIMESTAMPDIFF(SECOND,
                           previousTs,
                           ts
                           ),
             0
      ) AS secDiff
  FROM
    (SELECT
      assetID
      , locationID
      , @locationID AS previousLocationID
      , @locationID := locationID AS currentLocationID
      , ts
      , @ts AS previousTs
      , @ts := ts AS currentTs
    FROM Logs L1
    WHERE assetid = 1157    
    ORDER BY ts
    ) L2
  ORDER BY ts
  ) L3
GROUP BY changed
ORDER BY changed DESC
;

Se det i aktion:SQL Fiddle .

Opdatering:

Hvis du har brug for at deltage i yderligere tabeller, bør du faktisk JOIN og ikke undervalg. Da der er en GROUP BY på det aktuelt yderste niveau skal den eksisterende erklæring pakkes ind i et andet sæt parenteser - for at forhindre gruppering over faktatabellerne. Med nogle andre justeringer til det formål:

SET @locationID=0,@ts=NULL,@changed=0;

SELECT
  A.name
  , L4.assetID
  , L.name
  , L4.locationID
  , duration
FROM
  (SELECT
    MIN(assetID) AS assetID
    , MIN(locationID) AS locationID
    , SUM(secDiff) AS duration
    , changed
  FROM
    (
-- no change in here
    ) L3
  GROUP BY changed
  ) L4
JOIN Asset A
  ON L4.assetID = A.id
JOIN Location L
  ON L4.locationID = L.id
ORDER BY changed DESC
;

Udvidet SQL Fiddle .

Opdatering 2:

Den mest ligetil måde at adressere dubletter på bør være at DISTINCT dem væk som det allerførste skridt:

-- no change here
  (SELECT
    assetID
    , locationID
    , @locationID AS previousLocationID
    , @locationID := locationID AS currentLocationID
    , ts
    , @ts AS previousTs
    , @ts := ts AS currentTs
  FROM
    (SELECT DISTINCT
      assetID
      , locationID
      , ts
    FROM Logs
    WHERE assetid = 1157
    ) L1
  ORDER BY ts
  ) L2
-- no change here either

Denne SQL violin returnerer for de duplikerede logfiler data det samme resultatsæt som SQL Fiddle , hvor den tidligere forespørgsel kører mod data uden dubletter.

Kommenter venligst, hvis og da dette kræver justering / yderligere detaljer.