Timediff calucate where query

Question

我有一個查詢在哪裏我計算每個用戶的工作時間。

$sql = " 
SELECT actief 
    , userid 
    , DATE_FORMAT(datum, '%x - %V') 
    , username 
    , min_urenperweek 
    , SEC_TO_TIME(SUM(TIME_TO_SEC(tijd_tot) - TIME_TO_SEC(tijd_van))) timediff 
    FROM " . TBL_URENREGISTRATIE . " ur 
    JOIN " . TBL_CMS_USERS . " us 
    ON us.userid = ur.userid 
GROUP 
    BY YEARWEEK(datum) 
    , userid 
ORDER 
    BY us.username ASC 
    " ; 

我想要的是：如果min_urenperweek可以說其40小時和timediff更多的則是40小時，我不希望它表現出來，如果是小於40的話，我想表明它

我試圖Where min_urenperweek <= timediff，但它不工作

Answer 1

1. 你應該有比你SELECT條款不屬於聚合函數GROUP BY子句中相同的字段。

2. 您不能在WHERE子句中從SELECT子句中指定別名。

3. 如果你想在你的聚集體的結果進行篩選，你不能這樣做的WHERE子句中的HAVING條款存在爲此

嘗試是這樣的

$sql = " 
SELECT actief 
    , userid 
    , DATE_FORMAT(datum, '%x - %V') 
    , username 
    , min_urenperweek 
    , SEC_TO_TIME(SUM(TIME_TO_SEC(tijd_tot) - TIME_TO_SEC(tijd_van))) timediff 
    FROM " . TBL_URENREGISTRATIE . " ur 
    JOIN " . TBL_CMS_USERS . " us 
    ON us.userid = ur.userid 
GROUP 
    BY actief 
    , userid 
    , DATE_FORMAT(datum, '%x - %V') 
    , username 
    , min_urenperweek 
HAVING min_urenperweek <= timediff 
ORDER 
    BY us.username ASC 
    " ;