我想知道这在SQL中是否可行。假设你有两个表A和B,你在表A上做一个选择,在表B上做一个连接:

SELECT a.*, b.* FROM TABLE_A a JOIN TABLE_B b USING (some_id);

如果表A有“a_id”、“name”、“some_id”列,表B有“b_id”、“name”、“some_id”列,查询将返回“a_id”、“name”、“some_id”、“b_id”、“name”、“some_id”列。有什么方法可以为表B的列名加上前缀而不单独列出每一列吗?等价于这个:

SELECT a.*, b.b_id as 'b.b_id', b.name as 'b.name', b.some_id as 'b.some_id'
FROM TABLE_A a JOIN TABLE_B b USING (some_id);

但是,如前所述,没有列出每一列,所以像这样:

SELECT a.*, b.* as 'b.*'
FROM TABLE_A a JOIN TABLE_B b USING (some_id);

基本上就是说,“用‘something’为b.*返回的每一列添加前缀”。这可能吗,还是我运气不好?

编辑

关于不使用SELECT *等的建议是有效的建议,但与我的上下文无关,因此请关注眼前的问题——是否可以在连接中为表的所有列名添加前缀(SQL查询中指定的常量)?

我的最终目标是能够通过连接对两个表执行SELECT *操作,并且能够从结果集中获得的列的名称中分辨出哪些列来自表a,哪些列来自表b。同样,我不想单独列出列,我需要能够执行SELECT *操作。


当前回答

我在PostgreSQL 13中使用to_jsonb函数来获得连接表中的所有字段作为一列。

select
  TABLE_A.*,
  to_jsonb(TABLE_B.*) as b,
  to_jsonb(TABLE_C.*) as c
from TABLE_A
left join TABLE_B on TABLE_B.a_id=TABLE_A.id
left join TABLE_C on TABLE_C.a_id=TABLE_A.id
where TABLE_A.id=1

结果你会得到TABLE_A列加上b和c列的数量:

id name some_other_col b c
1 Some name Some other value {"id":1,"a_id":1,"prop":"value"} {"id":1,"a_id":1,"prop":"value"}
1 Some other name Another value {"id":1,"a_id":1,"prop":"value"} {"id":1,"a_id":1,"prop":"value"}

您只需要解析b和c列以将它们转换为对象。

其他回答

Select *通常会导致糟糕的代码,因为往往会添加新列,或者表中列的顺序经常改变,这通常会以非常微妙的方式破坏Select *。所以列出列是正确的解决方案。

至于如何进行查询,mysql不确定,但在sqlserver中,您可以从syscolumns中选择列名,并动态构建select子句。

不同的数据库产品会给你不同的答案;但如果你走得太远,你是在自讨苦吃。您最好选择您想要的列,并为它们提供自己的别名,以便每个列的标识非常清晰,并且可以在结果中区分它们。

对此没有SQL标准。

然而,通过代码生成(在表创建或修改或运行时按需生成),你可以很容易地做到这一点:

CREATE TABLE [dbo].[stackoverflow_329931_a](
    [id] [int] IDENTITY(1,1) NOT NULL,
    [col2] [nchar](10) NULL,
    [col3] [nchar](10) NULL,
    [col4] [nchar](10) NULL,
 CONSTRAINT [PK_stackoverflow_329931_a] PRIMARY KEY CLUSTERED 
(
    [id] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

CREATE TABLE [dbo].[stackoverflow_329931_b](
    [id] [int] IDENTITY(1,1) NOT NULL,
    [col2] [nchar](10) NULL,
    [col3] [nchar](10) NULL,
    [col4] [nchar](10) NULL,
 CONSTRAINT [PK_stackoverflow_329931_b] PRIMARY KEY CLUSTERED 
(
    [id] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

DECLARE @table1_name AS varchar(255)
DECLARE @table1_prefix AS varchar(255)
DECLARE @table2_name AS varchar(255)
DECLARE @table2_prefix AS varchar(255)
DECLARE @join_condition AS varchar(255)
SET @table1_name = 'stackoverflow_329931_a'
SET @table1_prefix = 'a_'
SET @table2_name = 'stackoverflow_329931_b'
SET @table2_prefix = 'b_'
SET @join_condition = 'a.[id] = b.[id]'

DECLARE @CRLF AS varchar(2)
SET @CRLF = CHAR(13) + CHAR(10)

DECLARE @a_columnlist AS varchar(MAX)
DECLARE @b_columnlist AS varchar(MAX)
DECLARE @sql AS varchar(MAX)

SELECT @a_columnlist = COALESCE(@a_columnlist + @CRLF + ',', '') + 'a.[' + COLUMN_NAME + '] AS [' + @table1_prefix + COLUMN_NAME + ']'
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME = @table1_name
ORDER BY ORDINAL_POSITION

SELECT @b_columnlist = COALESCE(@b_columnlist + @CRLF + ',', '') + 'b.[' + COLUMN_NAME + '] AS [' + @table2_prefix + COLUMN_NAME + ']'
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME = @table2_name
ORDER BY ORDINAL_POSITION

SET @sql = 'SELECT ' + @a_columnlist + '
,' + @b_columnlist + '
FROM [' + @table1_name + '] AS a
INNER JOIN [' + @table2_name + '] AS b
ON (' + @join_condition + ')'

PRINT @sql
-- EXEC (@sql)

I see two possible situations here. First, you want to know if there is a SQL standard for this, that you can use in general regardless of the database. No, there is not. Second, you want to know with regard to a specific dbms product. Then you need to identify it. But I imagine the most likely answer is that you'll get back something like "a.id, b.id" since that's how you'd need to identify the columns in your SQL expression. And the easiest way to find out what the default is, is just to submit such a query and see what you get back. If you want to specify what prefix comes before the dot, you can use "SELECT * FROM a AS my_alias", for instance.

最近在NodeJS和Postgres中遇到了这个问题。

ES6方法

我知道没有任何RDBMS特性提供这种功能,所以我创建了一个包含我所有字段的对象,例如:

const schema = { columns: ['id','another_column','yet_another_column'] }

定义了一个reducer将字符串与表名连接在一起:

const prefix = (table, columns) => columns.reduce((previous, column) => {
  previous.push(table + '.' + column + ' AS ' + table + '_' + column);
  return previous;
}, []);

这将返回一个字符串数组。为每个表调用它并合并结果:

const columns_joined = [...prefix('tab1',schema.columns), ...prefix('tab2',schema.columns)];

输出最后的SQL语句:

console.log('SELECT ' + columns_joined.join(',') + ' FROM tab1, tab2 WHERE tab1.id = tab2.id');