Question

我有2个表，一个包含peopleID和名字的人员表，以及一个他们所做的承诺表，其中有一个pledgeID（1到6），一个personID来说明它来自谁，以及一个金额字段。

我们需要有一个查询，提供一个明确的人员列表，他们总共承诺了多少，以及他们认为正确的承诺数量（pledgeIDs 1,3和5将被认为是正确的）

所以我们需要知道，例如约翰史密斯总共认捐了500英镑，并且他认可了2个承诺（因为他已经承诺了1,3和6），而萨莉詹姆斯承诺2000英镑并承诺在ids 1,3和5上，因此匹配3

我希望这很清楚。我真的很感激这个帮助。

非常感谢

戴夫

Answer 1

听起来像这样的东西会起作用：

SELECT PersonID,
    SUM(Amount) AS TotalPledged,
    SUM(CASE WHEN PledgeID IN (1,3,5) THEN 1 ELSE 0 END) AS CorrectPledges
FROM PersonPledges
GROUP BY PersonID

Answer 2

如果您使用主表作为承诺，您可以像这样一次性获取所有信息：

SELECT People.PersonName, pledgetotals.PledgeTotal, pledgecounts.PledgesCorrect FROM People
  LEFT OUTER JOIN (SELECT PeoplePledges.peopleID, SUM(PeoplePledges.pledgeAmount) AS PledgeTotal FROM PeoplePledges GROUP BY PeoplePledges.peopleID) pledgetotals ON People.peopleID = pledgetotals.peopleID
  LEFT OUTER JOIN (SELECT PeoplePledges.peopleID, COUNT(DISTINCT PeoplePledges.pledgeID) AS PledgesCorrect FROM PeoplePledges JOIN Pledge ON PeoplePledges.pledgeID = Pledge.pledgeID WHERE Pledge.correct = 1 GROUP BY PeoplePledges.peopleID) pledgecounts ON People.peopleID = pledgecounts.peopleID

希望它不太清楚;每个子查询都需要进行汇总（汇总承诺金额和分别计算正确的承诺）;如果您使用左外连接以这种方式排列查询，您可以列出所有相关人员，无论他们是否真的有任何承诺。

编辑：这就是我所承诺的'主表'的意思：

CREATE TABLE Pledge (INT pledgeID INT PRIMARY KEY, correct BOOLEAN NOT NULL);
INSERT INTO Pledge (pledgeID, correct) VALUES (1, 1);
INSERT INTO Pledge (pledgeID, correct) VALUES (2, 0);
INSERT INTO Pledge (pledgeID, correct) VALUES (3, 1);
INSERT INTO Pledge (pledgeID, correct) VALUES (4, 0);
INSERT INTO Pledge (pledgeID, correct) VALUES (5, 1);
INSERT INTO Pledge (pledgeID, correct) VALUES (6, 0);

编辑：如果您无法添加主承诺表，那么您必须在代码中使用“魔术数字”，但结构非常相似：

SELECT People.PersonName, pledgetotals.PledgeTotal, pledgecounts.PledgesCorrect FROM People
    JOIN (SELECT PeoplePledges.peopleID, SUM(PeoplePledges.pledgeAmount) AS PledgeTotal FROM PeoplePledges GROUP BY PeoplePledges.peopleID) pledgetotals ON People.peopleID = pledgetotals.peopleID
    JOIN (SELECT PeoplePledges.peopleID, COUNT(DISTINCT PeoplePledges.pledgeID) AS PledgesCorrect FROM PeoplePledges WHERE PeoplePledges.pledgeID IN (1,3,5) GROUP BY PeoplePledges.peopleID) pledgecounts ON People.peopleID = pledgecounts.peopleID

Answer 3

最简单的方法是：

1-从你人民的桌子上获取所有人，例如：

SELECT peopleId, name FROM people

2-遍历每一行并获得每个人的承诺。在PHP中你会做这样的事情：

foreach ($people as $k=>$person)
{
  $sql = "SELECT SUM(amount) as total FROM pledges
            WHERE peopleId = '$person[peopleId]'";
  $result = mysql_query($sql);
  if (mysql_num_rows($result) <= 0)
     $total = '0';
  else
  {
     $row = mysql_fetch_assoc($result);
     $total = $row['total'];
  }
  $people[$k]['total'] = $total;
}

具有匹配计数的多个标准

3 个答案: