[英]MYSQL COUNT not returning 0 for Left Join and Group By
我只是想讓下面的代碼識別計數為零並返回0。
設置日期變量和星期數表
//set first of the year and today
$year = date("Y");
$start = "01/01/".$year;
$today = date("Y-m-d");
$first = $year."-01-01";
$start = $year."-01-01";
$date1 = new DateTime($first);
$date2 = new DateTime($today);
$interval = $date1->diff($date2);
$weeks = floor(($interval->days) / 7);
//create weeks table
$sql = "DROP TEMPORARY TABLE IF EXISTS weekstbl" ;
mysqli_query ($db, $sql ) or ( "Error " . mysqli_error () ) ;
$weekstbl = "
CREATE TEMPORARY TABLE weekstbl (
`weekNo` int NOT NULL,
`weekStart` Date,
`weekEnd` Date,
PRIMARY KEY(weekNo)
)
";
mysqli_query($db, $weekstbl) or die ("Sql error : ".mysqli_error());
for($i = 1; $i <= $weeks; $i++){
$week = $date1->format("W");
$date1->add(new DateInterval('P6D'));
$date1->format('Y-m-d');
$newDate1 = $date1->format('Y-m-d');
$statement = $db->prepare("INSERT INTO weekstbl (weekNo,weekStart,weekEnd) VALUES (?,?,?)");
$statement->bind_param('iss', $week,$start,$newDate1);
$statement->execute();
$date1->add(new DateInterval('P1D'));
$start = $date1->format('Y-m-d');
}
該代碼輸出當前年份到當前日期的所有星期數,星期的開始日期和星期的結束日期。 看起來如下:
------------------------------------
| weekNo | weekStart | weekEnd |
------------------------------------
| 52 | 2017-01-01 | 2017-01-07 |
| 1 | 2017-01-08 | 2017-01-14 |
| 2 | 2017-01-15 | 2017-01-21 |
從訪問系統的當前日期一直持續到現在。 然后,它將用於與其他一些有條件的其他表聯接。 這是我希望返回的結果,但是在下面列出的所有試驗中,我都無法完成。 我在當年有12個星期時運行此程序,這意味着名稱后應有12個數字。
[["Ryan Balcom",3,30,3,1,10,0,1,2,5,1,3,3],["Jared Beckwith",40,8,13,8,13,7,5,3,11,5,3,1],["Jim Roberts",0,32,8,7,2,9,4,2,8,4,4,10],["Jim Kelly",44,24,12,14,14,16,10,25,12,8,7,6],["Josh Bell",34,10,10,5,9,8,5,23,7,6,6,14],["Rick Zuniga",0,0,0,0,0,3,1,1,0,0,0,0],["Mike Horton",37,20,7,10,16,5,6,4,5,3,5,1],["Paul Schilthuis",31,11,9,2,8,6,4,0,5,9,8,4],["TJ Sutton",1,10,0,0,0,0,0,0,0,0,0,0]]
試驗1
$assignments = "
SELECT COUNT(j.leadid) AS leadcount, u.username
FROM weekstbl wk
LEFT JOIN jobbooktbl j
ON wk.weekNo=WEEK(j.leadcreated)
LEFT JOIN assignmentstbl a
ON j.leadid=a.custid
LEFT JOIN usertbl u
ON a.userid=u.userid
WHERE j.leadcreated >= '".$first."' AND j.leadcreated <= '".$today."' AND YEAR(j.leadcreated) = '".$year."'
GROUP BY WEEK(j.leadcreated), a.userid";
$assignmentsqry = mysqli_query($db,$assignments);
while ($row = mysqli_fetch_array($assignmentsqry)) {
$key = $row['username']; //unnecessary variable for demonstration purposes
if (!isset($volumeYTDsm[$key])) {
$volumeYTDsm[$key] = [$row['username']];
}
$float = floatval($row['leadcount']);
$volumeYTDsm[$key][] = $float;
}
$volumeYTDsm = array_values($volumeYTDsm);//removing keys
}
輸出上述內容,但不包括0周:
[["Ryan Balcom",3,30,3,1,10,1,2,5,1,3,3],["Jared Beckwith",40,8,13,8,13,7,5,3,11,5,3,1],["Jim Roberts",32,8,7,2,9,4,2,8,4,4,10],["Jim Kelly",44,24,12,14,14,16,10,25,12,8,7,6],["Josh Bell",34,10,10,5,9,8,5,23,7,6,6,14],["Rick Zuniga",3,1,1],["Mike Horton",37,20,7,10,16,5,6,4,5,3,5,1],["Paul Schilthuis",31,11,9,2,8,6,4,5,9,8,4],["TJ Sutton",1,10]]
試用2
$assignments = "
SELECT COUNT(*) AS leadcount, u.username
FROM weekstbl wk
LEFT OUTER JOIN jobbooktbl j
ON (wk.weekNo=WEEK(j.leadcreated)) AND j.leadcreated >= '".$first."' AND j.leadcreated <= '".$today."' AND YEAR(j.leadcreated) = '".$year."'
LEFT JOIN assignmentstbl a
ON j.leadid=a.custid
LEFT JOIN usertbl u
ON a.userid=u.userid
GROUP BY wk.weekNo, a.userid";
這將輸出以下內容...不確定使用空名稱輸出的內容:
[[null,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],["Ryan Balcom",3,30,3,1,10,1,2,5,1,3,3],["Jared Beckwith",40,8,13,8,13,7,5,3,11,5,3,1],["Jim Roberts",32,8,7,2,9,4,2,8,4,4,10],["Jim Kelly",44,24,12,14,14,16,10,25,12,8,7,6],["Josh Bell",34,10,10,5,9,8,5,23,7,6,6,14],["Rick Zuniga",3,1,1],["Mike Horton",37,20,7,10,16,5,6,4,5,3,5,1],["Paul Schilthuis",31,11,9,2,8,6,4,5,9,8,4],["TJ Sutton",1,10]]
使用這兩種方法,我都試圖包括COALESCE和IFNULL,但是都沒有改變任何一個查詢的結果:
COALESCE(COUNT(j.leadid),0) AS leadcount
IFNULL(COUNT(j.leadid),0) AS leadcount
我一直在努力解決這個問題一個月,但似乎沒有任何投入。 任何幫助或指示將不勝感激!
您需要將主表中的一列添加到分組依據,否則,您將按僅存在於您的左聯接之一中的值進行分組。
SELECT
COUNT(j.leadid) AS leadcount,
IF(u.username IS NULL,'NOT FOUND',u.username) AS username
FROM weekstbl wk
LEFT JOIN jobbooktbl j
ON wk.weekNo=WEEK(j.leadcreated)
LEFT JOIN assignmentstbl a
ON j.leadid=a.custid
LEFT JOIN usertbl u
ON a.userid=u.userid
WHERE (j.leadcreated >= '".$first."' AND
j.leadcreated <= '".$today."' AND
YEAR(j.leadcreated) = '".$year."') OR
j.leadcreated IS NULL
GROUP BY wk.weekno, WEEK(j.leadcreated), a.userid
ORDER BY wk.weekStart
更好的where子句將與weekstbl
的日期進行weekstbl
。
WHERE
的條件是將第一個LEFT JOIN
變成一個內部LEFT JOIN
。 因此,您似乎想要:
SELECT COUNT(j.leadid) AS leadcount, u.username
FROM weekstbl wk LEFT JOIN
jobbooktbl j
ON wk.weekNo = WEEK(j.leadcreated) AND
j.leadcreated >= '".$first."' AND j.leadcreated <= '".$today."' AND
YEAR(j.leadcreated) = '".$year."'LEFT JOIN
assignmentstbl a
ON j.leadid = a.custid LEFT JOIN
usertbl u
ON a.userid = u.userid
GROUP BY WEEK(j.leadcreated), a.userid;
但是,您按星期進行匯總,但不包括SELECT
的星期。 我懷疑您想要:
SELECT wk.weekNo, u.username, COUNT(j.leadid) AS leadcount
FROM weekstbl wk LEFT JOIN
jobbooktbl j
ON wk.weekNo = WEEK(j.leadcreated) AND
j.leadcreated >= '".$first."' AND j.leadcreated <= '".$today."' AND
YEAR(j.leadcreated) = '".$year."'LEFT JOIN
assignmentstbl a
ON j.leadid = a.custid LEFT JOIN
usertbl u
ON a.userid = u.userid
GROUP BY wk.weekNo, a.userid;
您不希望通過WEEK(j.leadcreated)
進行聚合,因為這可能為NULL
。
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