Question

嘿我正在尝试使用一个计数来显示乐队所拥有的数字总数，到目前为止我有：

SELECT cds.cd_id,cds.position,bands.band_name
FROM cds
INNER JOIN releases
INNER join bands
ON cds.cd_id =releases.cd_id and releases.band_id = bands.band_id
Where band_name IS NOT NULL AND position = 1;

SELECT bands.band_name, COUNT(band_name) AS Total FROM bands
WHERE band_name ='Prodigy';

这会将表格与位置cd_id和band_name连接在一起，然后使用计数来计算乐队＆＃39; Prodigy＆＃39;已经

cd020   1   Prodigy
cd021   1   Prodigy
cd024   1   Kylie Minogue
cd028   1   Kylie Minogue
cd031   1   Kylie Minogue
cd049   1   Kylie Minogue
cd065   1   Robbie Williams
cd068   1   Robbie Williams
cd070   1   Robbie Williams

正如你所看到的，Prodigy应该有2个1号位，我的计数只显示它有，我认为这是计算band_name出现次数的最简单方法，而不是每个频段的位置数。你能给予的任何帮助或建议都会非常有帮助

Answer 1

聚合查询需要GROUP BY语句。像这样：

SELECT bands.band_name, COUNT(band_name) AS Total FROM bands
WHERE band_name ='Prodigy'
GROUP BY bands.band_name;

要获得结果，请使用以下内容：

SELECT
    bands.band_name,
    COUNT(*)
FROM
    cds
        INNER JOIN
    releases ON (cds.cd_id =releases.cd_id)
        INNER JOIN
    bands ON (releases.band_id = bands.band_id)
WHERE
    bands.band_name IS NOT NULL
    AND cds.position = 1
GROUP BY bands.band_name;

Answer 2

这应该适合你。

SELECT cds.position, bands.band_name, COUNT(cds.cd_id) as COUNT
FROM cds
    INNER JOIN releases
    INNER join bands
        ON cds.cd_id =releases.cd_id and releases.band_id = bands.band_id
WHERE band_name IS NOT NULL AND position = 1
GROUP BY bands.band_name, cds.position;

计算Band_Name的数字1

2 个答案: