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

我根据答案实现了一个解决方案,建议在节点中使用虚拟或哨兵列。你可以通过生成SQL来使用它:

select 
    s.*
  , '' as _prefix__creator_
  , u.*
  , '' as _prefix__speaker_
  , p.*
from statements s 
  left join users u on s.creator_user_id = u.user_id
  left join persons p on s.speaker_person_id = p.person_id

然后对从数据库驱动程序返回的行进行后处理,比如addPrefixes(row)。

实现(基于我的驱动程序返回的字段/行,但应该很容易为其他DB驱动程序更改):

const PREFIX_INDICATOR = '_prefix__'
const STOP_PREFIX_INDICATOR = '_stop_prefix'

/** Adds a <prefix> to all properties that follow a property with the name: PREFIX_INDICATOR<prefix> */
function addPrefixes(fields, row) {
  let prefix = null
  for (const field of fields) {
    const key = field.name
    if (key.startsWith(PREFIX_INDICATOR)) {
      if (row[key] !== '') {
        throw new Error(`PREFIX_INDICATOR ${PREFIX_INDICATOR} must not appear with a value, but had value: ${row[key]}`)
      }
      prefix = key.substr(PREFIX_INDICATOR.length)
      delete row[key]
    } else if (key === STOP_PREFIX_INDICATOR) {
      if (row[key] !== '') {
        throw new Error(`STOP_PREFIX_INDICATOR ${STOP_PREFIX_INDICATOR} must not appear with a value, but had value: ${row[key]}`)
      }
      prefix = null
      delete row[key]
    } else if (prefix) {
      const prefixedKey = prefix + key
      row[prefixedKey] = row[key]
      delete row[key]
    }
  }
  return row
}

测试:

const {
  addPrefixes,
  PREFIX_INDICATOR,
  STOP_PREFIX_INDICATOR,
} = require('./BaseDao')

describe('addPrefixes', () => {
  test('adds prefixes', () => {
    const fields = [
      {name: 'id'},
      {name: PREFIX_INDICATOR + 'my_prefix_'},
      {name: 'foo'},
      {name: STOP_PREFIX_INDICATOR},
      {name: 'baz'},
    ]
    const row = {
      id: 1,
      [PREFIX_INDICATOR + 'my_prefix_']: '',
      foo: 'bar',
      [STOP_PREFIX_INDICATOR]: '',
      baz: 'spaz'
    }
    const expected = {
      id: 1,
      my_prefix_foo: 'bar',
      baz: 'spaz',
    }
    expect(addPrefixes(fields, row)).toEqual(expected)
  })
})

其他回答

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)

我可以想到两种方法以可重用的方式实现这一点。一种方法是为所有列重命名,为它们所在的表加上前缀。我已经看过很多次了,但我真的不喜欢。我发现它是多余的,会导致大量的输入,而且当您需要覆盖列名来源不明确的情况时,您总是可以使用别名。

另一种方法是为每个表创建别名表名的视图,这也是我建议您在自己的情况下使用的方法。然后你们就会站在这些观点的对立面,而不是桌子的对立面。这样,如果您愿意,您可以自由地使用*,如果您愿意,可以自由地使用带有原始列名的原始表,而且它还使编写任何后续查询更容易,因为您已经在视图中完成了重命名工作。

最后,我不清楚为什么需要知道每个列来自哪个表。这重要吗?最终重要的是它们所包含的数据。UserID来自User表还是UserQuestion表并不重要。当然,当您需要更新它时,这很重要,但在这一点上,您应该已经足够了解您的模式以确定这一点。

与非常好的“PHP (Wordpress)函数”相同的响应,但为CakePHP 4.3编码。 放在src/Controller/Component/MyUtilsComponent.php中

<?php

namespace App\Controller\Component;

use Cake\Controller\Component;
use Cake\Datasource\ConnectionManager;

class MyUtilsComponent extends Component
{
    public static function prefixedTableFieldsWildcard(string $table, string $alias, string $connexion = 'default'): string
    {
        $c = ConnectionManager::get($connexion);
        $columns = $c->execute("SHOW COLUMNS FROM $table");
        $field_names = [];
        foreach ($columns as $column) {
            $field_names[] = $column['Field'];
        }

        $prefixed = [];
        foreach ($field_names as $field_name) {
            $prefixed[] = "`{$alias}`.`{$field_name}` AS `{$alias}.{$field_name}`";
        }
        return implode(', ', $prefixed);
    }
}

测试和使用

    function testPrefixedTableFieldsWildcard(): void
    {
        $fields = MyUtilsComponent::prefixedTableFieldsWildcard('metas', 'u', 'test');
        $this->assertEquals('`u`.`id` AS `u.id`, `u`.`meta_key` AS `u.meta_key`, `u`.`meta_value` AS `u.meta_value`, `u`.`meta_default` AS `u.meta_default`, `u`.`meta_desc` AS `u.meta_desc`', $fields,);
    }

这将创建具有给定前缀的字段列表

select
    name + ' as prefix.' + name + ','
from sys.columns where object_id = object_id('mytable')
order by column_id

通过重命名相关表中的字段,我解决了自己的一个类似问题。是的,我有这样做的特权,我知道可能不是每个人都有。我在表示表名的表中的每个字段中添加了前缀。因此,OP发布的SQL将保持不变

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

并且仍然给出了预期的结果——易于识别输出字段属于哪个表。