SQLite 左连接计数（*）？答案

【问题标题】：SQLite LEFT JOIN count(*)?SQLite 左连接计数（*）？
【发布时间】：2015-05-31 22:33:35
【问题描述】：

我需要连接两个表（实际上是两个视图），以便对于左视图的每个选定行，右视图中的行数。这对我来说听起来像是 LEFT JOIN，但在 SQLite（此测试 database）和 LEFT JOIN 查询中：

SELECT TARGET.session_id session_id, TARGET.labeltype_id labeltype_id, TARGET.label_id label_id, count(SECONDARY.label_id) NOlabels 
FROM segment_extended TARGET LEFT JOIN segment_extended SECONDARY
WHERE TARGET.session_id = SECONDARY.session_id AND TARGET.lt_name= "Word" AND SECONDARY.lt_name ="Comments" 
AND ((SECONDARY.start <= TARGET.start AND TARGET.END <= SECONDARY.END) OR (TARGET.start <= SECONDARY.start AND SECONDARY.END <= TARGET.END))
AND TARGET.label != '' AND SECONDARY.label != '' 
GROUP BY TARGET.session_id,TARGET.labeltype_id, TARGET.label_id;

我只得到了我期望的一小部分：

2   3   3   1
2   3   9   1

更扩展的查询给出正确的结果：

SELECT session_id, labeltype_id, label_id, max(NOlabels) NOlabels 
FROM (SELECT TARGET.session_id session_id, TARGET.labeltype_id labeltype_id, TARGET.label_id label_id, count(SECONDARY.label_id) NOlabels 
    FROM segment_extended TARGET , segment_extended SECONDARY
    WHERE TARGET.session_id = SECONDARY.session_id AND TARGET.lt_name= "Word" AND SECONDARY.lt_name ="Comments" 
    AND ((SECONDARY.start <= TARGET.start AND TARGET.END <= SECONDARY.END) OR (TARGET.start <= SECONDARY.start AND SECONDARY.END <= TARGET.END))
    AND TARGET.label != '' AND SECONDARY.label != '' 
    GROUP BY TARGET.session_id,TARGET.labeltype_id, TARGET.label_id
    UNION
    SELECT TARGET.session_id session_id, TARGET.labeltype_id labeltype_id, TARGET.label_id label_id, 0 NOlabels 
    FROM segment_extended TARGET
    WHERE TARGET.lt_name= "Word"
    AND TARGET.label != '' 
    GROUP BY TARGET.session_id,TARGET.labeltype_id, TARGET.label_id)
GROUP BY session_id, labeltype_id, label_id
ORDER BY session_id,labeltype_id, label_id


session_id  labeltype_id    label_id    NOlabels
2   3   2   0
2   3   3   1
2   3   4   0
2   3   5   0
2   3   7   0
2   3   8   0
2   3   9   1
2   3   10  0

但它似乎不必要地复杂。左连接我做错了什么？

【问题讨论】：

标签： join sqlite count left-join

【解决方案1】：

在进行左连接时，您必须将左连接中的空值计为 0 条记录，但仍包括它们。您可以在内部查询中使用 CASE 构造来完成此操作，然后在外部 group-by 中使用 SUM 聚合函数。

SELECT session_id, labeltype_id, label_id, sum(has_label) NOlabels
FROM (
    SELECT TARGET.session_id session_id, TARGET.labeltype_id labeltype_id, TARGET.label_id label_id, CASE WHEN SECONDARY.label_id is NULL then 0 else 1 END has_label
    FROM
        segment_extended TARGET
        LEFT JOIN
            segment_extended SECONDARY on
                TARGET.session_id = SECONDARY.session_id
                AND SECONDARY.lt_name ="Comments"
                AND ((
                    SECONDARY.start <= TARGET.start AND TARGET.END <= SECONDARY.END)
                    OR (TARGET.start <= SECONDARY.start AND SECONDARY.END <= TARGET.END))
                AND SECONDARY.label != ''
    WHERE TARGET.lt_name= "Word" AND TARGET.label != '')
GROUP BY session_id, labeltype_id, label_id

【讨论】：

【解决方案2】：

您的联接不是左联接。

如果没有符合连接条件的行，左连接会为右表添加 NULL 值。但是，您的查询没有连接条件，并且 WHERE 条件不受 LEFT JOIN 子句的影响。

将 WHERE 替换为 ON。

【讨论】：