Question

我有两张桌子'all'和'jdetails'。我在所有表上都有一个现有的select查询。我想在jdetails表中添加一些额外的数据（如果有的话）。

所有表格：

judge, year, ...

jane doe, 2012

john doe, 2011

jdetails表：

name, designation,...

jane doe, level 1

jane doe, level 5

john doe, special

如何在下面更改我的查询以包含每个评判的“指定（来自jdetails）”（总共）？

我认为左连接是解决方案，但我有where子句要考虑。此外，我绝对必须在下面查询此查询的结果，但是如果存在，则添加来自jdetails表的数据。

此外，每个all.judge可以有多行（jdetails.name）指定，我希望将其列为单个值。例如，jane doe的指定值为“1级5级”。

我会加入all.judge = jdetails.name

当前查询：

$rows = $my->get_row("SELECT all.judge, `year`, `totlevel_avg`, `totlevel_count`,   `genrank`, `poprank`, `tlevel_avg`, `tlevel_count`, `1level_avg` as `onelevel_avg`, `1level_count` as `onelevel_count`, `2level_avg` as `twolevel_avg`, `2level_count` as `twolevel_count`, `3level_avg` as `threelevel_avg`, `3level_count` as `threelevel_count`, `4level_avg` as `fourlevel_avg`, `4level_count` as `fourlevel_count`, `PSGlevel_avg`, `PSGlevel_count`, `I1level_avg`, `I1level_count`, `I2level_avg`, `I2level_count`, `GPlevel_avg`, `GPlevel_count`, `states` from `all` where `id` ='{$term}'");

非常感谢任何帮助。

Answer 1

我没有在SELECT声明中包含您所拥有的所有内容，我只是用ams.*对其进行了总结。但是下面将all表链接到jdetails表，然后将指定分组到一个字段中。然后我将结果包装在外部查询中，该查询在all表（SQL Fiddle）中提取您需要的其余字段：

SELECT ams.*, am.Desigs
FROM 
(
  SELECT a.judge, GROUP_CONCAT(j.designation SEPARATOR ', ') AS Desigs
  FROM `all` AS a
  INNER JOIN jdetails AS j ON a.judge = j.name
  GROUP BY a.judge
) AS am 
INNER JOIN `all` AS ams ON am.judge = ams.judge

Answer 2

以下是一个可以帮助您的示例：

`SELECT table1.column_name(s),table2.column_name(s) FROM table1 
LEFT OUTER JOIN table2 ON table1.column_name=table2.column_name
 AND table1.column_name='Parameter'`

表1是全部，表2是jdetails

mySQL选择PHP - 左连接

2 个答案: