4

在下面的 sql 语句中：

 SELECT `keywords`.keyID, count(`keywords-occurencies`.keyID) as countOccurencies 
                    FROM `keywords-occurencies`  
                    LEFT JOIN `keywords` 
                    ON `keywords-occurencies`.keyID = `keywords`.keyID 
                    WHERE `keywords-occurencies`.`keyID` IN (1,2,3) AND date BETWEEN '2013/01/25' AND '2013/01/27'
                    GROUP BY `keywords`.`keyID`

如果 keyID 3 没有返回值，则不计为 0，也不包含在结果集中，并显示类似这样的结果

keyID countOccurencies
1       3
3       5

我想显示零结果，例如

keyID countOccurencies
1       3
2       0
3       5

用于测试的样本数据：

--
-- Table structure for table `keywords`
--

CREATE TABLE IF NOT EXISTS `keywords` (
  `keyID` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `keyName` varchar(40) NOT NULL,
  PRIMARY KEY (`keyID`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

--
-- Dumping data for table `keywords`
--

INSERT INTO `keywords` (`keyID`, `keyName`) VALUES
(1, 'testKey1'),
(2, 'testKey2');

-- --------------------------------------------------------

--
-- Table structure for table `keywords-occurencies`
--

CREATE TABLE IF NOT EXISTS `keywords-occurencies` (
  `occurencyID` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `keyID` int(10) unsigned NOT NULL,
  `date` date NOT NULL,
  PRIMARY KEY (`occurencyID`),
  KEY `keyID` (`keyID`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

--
-- Dumping data for table `keywords-occurencies`
--

INSERT INTO `keywords-occurencies` (`occurencyID`, `keyID`, `date`) VALUES
(1, 1, '2013-01-27'),
(2, 1, '2013-01-26');

--
-- Constraints for table `keywords-occurencies`
--
ALTER TABLE `keywords-occurencies`
  ADD CONSTRAINT `keywords@002doccurencies_ibfk_1` FOREIGN KEY (`keyID`) REFERENCES `keywords` (`keyID`) ON DELETE CASCADE ON UPDATE CASCADE;
4

4 回答 4

6

要做的事情

  • 你应该把它分组 GROUP BY keywords-occurencies.keyID
  • 并且您必须显示keywords-occurencies.keyID不是keywords.keyID
  • 数数keywords.keyID
  • （可选）使用ALIAS，这样您就可以摆脱除 tableNames 之外的反引号

询问，

SELECT  a.keyID,
        count(b.keyID) AS countOccurencies
FROM    `keywords - occurencies` a
        LEFT JOIN `keywords` b
            ON a.keyID = b.keyID
WHERE   a.keyID IN ( 1, 2, 3 ) AND 
        DATE BETWEEN '2013/01/25' AND '2013/01/27'
GROUP   BY a.keyID

更新 1

根据示例记录，您需要执行以下操作，

  • 交换表名
  • 把这个条件DATE BETWEEN '2013-01-25' AND '2013-01-27'放在ONjoin子句上。
  • （可选）使用ALIAS，这样您就可以摆脱除 tableNames 之外的反引号

询问，

SELECT  a.keyID,
        count(b.keyID) AS countOccurencies
FROM    `keywords` a
        LEFT JOIN `keywords-occurencies` b
            ON a.keyID = b.keyID AND
               b.DATE BETWEEN '2013-01-25' AND '2013-01-27'
WHERE   a.keyID IN ( 1, 2, 3 ) 
GROUP   BY a.keyID
于 2013-01-27T16:00:25.187 回答
2

date在::上打一个反勾号

SELECT 
`keywords`.keyID, 
count(`keywords-occurencies`.keyID) as countOccurencies 
FROM `keywords-occurencies`  
LEFT JOIN `keywords`  ON `keywords-occurencies`.keyID = `keywords`.keyID 
WHERE `keywords-occurencies`.`keyID` IN (1,2,3) AND `date` BETWEEN '2013/01/25' AND '2013/01/27'
GROUP BY `keywords-occurencies`.`keyID`
于 2013-01-27T15:57:58.173 回答
1

有两件事。您需要按左外连接第一部分的 id 进行分组。然后你需要计算第二面是什么。对于右外连接，顺序相反：

SELECT k.keyID, count(ko.keyID) as countOccurencies 
FROM `keywords-occurencies` ko
      RIGHT JOIN `keywords` k
      ON ko.keyID = k.keyID 
WHERE k.`keyID` IN (1,2,3) AND date BETWEEN '2013/01/25' AND '2013/01/27'
GROUP BY k.`keyID`

其原因与左外连接有关。即使没有匹配项，它也会将所有内容保留在第一个表中。所以，这就是你获得完整列表的地方。至于计数，你想计算匹配。如果你计算第一个表中的 id，你总是会得到至少 1。计算第二个表中的 id 可以得到 0。

请注意，我还在您的表中添加了别名，以使查询更具可读性。

于 2013-01-27T16:05:50.663 回答
0

将其更改为LEFT JOIN对我有用：

SELECT k.keyID, count(ko.keyID) as countOccurencies 
FROM `keywords-occurencies` ko
      LEFT JOIN `keywords` k
      ON ko.keyID = k.keyID 
WHERE k.`keyID` IN (1,2,3) AND date BETWEEN '2013/01/25' AND '2013/01/27'
GROUP BY k.`keyID`
于 2015-05-19T07:38:31.753 回答