php - MYSQL COUNT 不为 Left Join 和 Group By 返回 0

Question

我只是想让下面的代码识别计数何时为零并返回 0。

设置日期变量和周数表

//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');
  }

此代码输出当年到今天的所有周数、一周的开始日期和一周的结束日期。看起来像下面这样:

------------------------------------
| weekNo |  weekStart |  weekEnd   |
------------------------------------
|   52   | 2017-01-01 | 2017-01-07 |
|    1   | 2017-01-08 | 2017-01-14 |
|    2   | 2017-01-15 | 2017-01-21 |

这一直持续到系统被访问的当前日期。然后，这将用于与其他几个有条件的表连接。这是我希望得到的返回，但在我下面列出的所有试验中，我无法做到这一点。我在当年有 12 个星期的时候运行了这个，这意味着名称后面应该有 12 个数字。

[["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]]

试验 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
    }

这会输出以上但不包括第 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]]

试验 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";

这会输出以下内容...不太确定这是用空名称输出的内容:

[[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]]

我尝试使用这两种方法来包含 COALESCE 和 IFNULL，但都没有改变任何一个查询的结果:

COALESCE(COUNT(j.leadid),0) AS leadcount
IFNULL(COUNT(j.leadid),0) AS leadcount

一个月来我一直在努力解决这个问题，但似乎没有任何效果。任何帮助或指导将不胜感激!

Answer 1

WHERE 中的条件是将第一个 LEFT JOIN 转换为内部联接。所以，你似乎想要:

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，您按周聚合，但不包括 SELECT 中的周。我怀疑你想要:

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;

您不想按 WEEK(j.leadcreated) 进行聚合，因为那可能是 NULL。