Question

考虑这个问题：

SELECT COUNT(*) AS pageviews 
FROM stats_hits 
WHERE pageID='1' AND entrytime>DATE_SUB(now(), INTERVAL 5 DAY) 
GROUP BY DATE(`entrytime`) LIMIT 5

返回过去五天的点击次数：类似于：

如何重写此查询以返回水平结果集？

11, 12, 18, 15, 5

谢谢！

编辑：作为一个副作用问题，有没有人知道如何调整此查询以显示0没有命中的天数？目前，如果某一天没有点击，只返回4个结果。

Answer 1

SELECT GROUP_CONCAT(s.pageviews) AS pageviews
FROM
  (
  SELECT COUNT(*) AS pageviews 
  FROM stats_hits 
  WHERE pageID='1' AND entrytime>DATE_SUB(now(), INTERVAL 5 DAY) 
  GROUP BY DATE(`entrytime`) LIMIT 5
  ) s;

请参阅：http://dev.mysql.com/doc/refman/5.1/en/group-by-functions.html#function_group-concat

如果你想要总是有5个结果，那么丑陋的黑客可能会有效。

SELECT GROUP_CONCAT(s2.pageviews) AS pageviews
FROM 
  (
  SELECT * FROM
    (
    SELECT COUNT(*) AS pageviews 
    FROM stats_hits 
    WHERE pageID='1' AND entrytime>DATE_SUB(now(), INTERVAL 5 DAY) 
    GROUP BY DATE(`entrytime`) LIMIT 5
    UNION ALL SELECT 0 AS pageviews
    UNION ALL SELECT 0 AS pageviews
    UNION ALL SELECT 0 AS pageviews
    UNION ALL SELECT 0 AS pageviews
    UNION ALL SELECT 0 AS pageviews
    ) s1
  ORDER BY pageviews = 0 ASC
  LIMIT 5) s2;

另一个选择是

CREATE TEMPORARY TABLE temp_table LIKE SELECT 0 as pageviews;
INSERT INTO temp_table VALUES (0),(0),(0),....,(0);
INSERT INTO temp_table 
  SELECT COUNT(*) AS pageviews 
  FROM stats_hits 
  WHERE pageID='1' AND entrytime>DATE_SUB(now(), INTERVAL x DAY) 
  GROUP BY DATE(`entrytime`) LIMIT x; //replace x with the number of days

SELECT GROUP_CONCAT(t.pageviews) AS pageviews
FROM temp_table t
ORDER BY t.pageview = 0 ASC 
LIMIT x;  //replace x with the number of days

简单的数据透视查询帮助

1 个答案: