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)

不能这样做没有别名，只是因为，你将如何引用一个字段在where子句，如果该字段存在于2或3个表，你要连接? 这将是不清楚的mysql，你试图引用哪一个。

我完全理解您关于重复字段名的问题。

我也需要它，直到我编写了自己的函数来解决它。如果您正在使用PHP，您可以使用它，或者如果您有以下设施，则可以使用您正在使用的语言编写代码。

这里的技巧是mysql_field_table()返回表名，mysql_field_name()返回结果中每一行的字段(如果它是使用mysql_num_fields()获得的)，这样您就可以将它们混合在一个新数组中。

这是所有列的前缀;)

问候,

function mysql_rows_with_columns($query) {
    $result = mysql_query($query);
    if (!$result) return false; // mysql_error() could be used outside
    $fields = mysql_num_fields($result);
    $rows = array();
    while ($row = mysql_fetch_row($result)) { 
        $newRow = array();
        for ($i=0; $i<$fields; $i++) {
            $table = mysql_field_table($result, $i);
            $name = mysql_field_name($result, $i);
            $newRow[$table . "." . $name] = $row[$i];
        }
        $rows[] = $newRow;
    }
    mysql_free_result($result);
    return $rows;
}

或者你可以使用Red Gate SQL Refactor或SQL Prompt，它通过单击Tab按钮将SELECT *展开为列列表

所以在你的例子中，如果你输入SELECT * FROM A JOIN B… 转到*的末尾，Tab键，瞧!你会看到 选择a . columnn1, A.column2， ....， B. columnn1, B.column2 FROM A JOIN

但它不是免费的

PHP 7.2 + MySQL/Mariadb

MySQL会给你发送多个相同名称的字段。甚至在终端客户端。但如果你想要一个关联数组，你必须自己创建键。

感谢@axelbrz的原创。我已经将它移植到更新的php，并对它进行了一些清理:

function mysqli_rows_with_columns($link, $query) {
    $result = mysqli_query($link, $query);
    if (!$result) {
        return mysqli_error($link);
    }
    $field_count = mysqli_num_fields($result);
    $fields = array();
    for ($i = 0; $i < $field_count; $i++) {
        $field = mysqli_fetch_field_direct($result, $i);
        $fields[] = $field->table . '.' . $field->name; # changed by AS
        #$fields[] = $field->orgtable . '.' . $field->orgname; # actual table/field names
    }
    $rows = array();
    while ($row = mysqli_fetch_row($result)) {
        $new_row = array();
        for ($i = 0; $i < $field_count; $i++) {
            $new_row[$fields[$i]] = $row[$i];
        }
        $rows[] = $new_row;
    }
    mysqli_free_result($result);
    return $rows;
}

$link = mysqli_connect('localhost', 'fixme', 'fixme', 'fixme');
print_r(mysqli_rows_with_columns($link, 'select foo.*, bar.* from foo, bar'));

从这个解决方案出发，我将如何处理这个问题:

首先创建一个所有AS语句的列表:

DECLARE @asStatements varchar(8000)

SELECT @asStatements = ISNULL(@asStatements + ', ','') + QUOTENAME(table_name) + '.' + QUOTENAME(column_name) + ' AS ' + '[' + table_name + '.' + column_name + ']'
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_NAME = 'TABLE_A' OR TABLE_NAME = 'TABLE_B'
ORDER BY ORDINAL_POSITION

然后在你的查询中使用它:

EXEC('SELECT ' + @asStatements + ' FROM TABLE_A a JOIN TABLE_B b USING (some_id)');

然而，这可能需要修改，因为类似的东西只在SQL Server中测试。但是这段代码在SQL Server中并不完全有效，因为不支持USING。

请评论，如果你可以测试/纠正这段代码，例如MySQL。