根据选定的表参数显示列

Question

我正在做一份临时报告。用户将能够从数据库中选择一个表，然后选择他们想要显示的列。这是我拥有的代码和参数。

SELECT 
    name, TABLE_NAME, COLUMN_NAME  
FROM 
    sys.tables
INNER JOIN  
    INFORMATION_SCHEMA.COLUMNS ON TABLE_NAME = name
WHERE 
    TABLE_NAME = @TableName <--multiselect drop down
    AND COLUMN_NAME = @ColumnName <--multiselect drop down

我的问题是如何在报告中创建列，以便根据选择显示相应的列名称。有些表有 4 列，有些有 30 列。我已经看到如何根据选择隐藏列的可见性，但是我必须这样做 278 列，必须有更好的方法。

谢谢。

Answer 1

这可以通过两种方式完成。

方法一：传递表名作为参数，列名作为逗号分隔值

DECLARE @TableName NVARCHAR(100) = 'Tablename'
DECLARE @ColumnName NVARCHAR(100) = 'Column1,Column2'

SELECT 
    name, TABLE_NAME, COLUMN_NAME  
FROM 
    sys.tables
INNER JOIN  
    INFORMATION_SCHEMA.COLUMNS ON TABLE_NAME = name
WHERE 
TABLE_NAME = @TableName
AND COLUMN_NAME IN
(
    -- Convert Comma Separated Values to rows
    SELECT PARSENAME(REPLACE(Split.a.value('.', 'VARCHAR(100)'),'-','.'),1) 'Columns' 
    FROM  
    (
         SELECT CAST ('<M>' + REPLACE(@ColumnName, ',', '</M><M>') + '</M>' AS XML) AS Data              
    ) AS A 
    CROSS APPLY Data.nodes ('/M') AS Split(a)
)

方法2：将表名作为参数传递，并在临时表中传递列名

DECLARE @TableName NVARCHAR(100) = 'Tablename'    

SELECT 
    name, TABLE_NAME, COLUMN_NAME  
FROM 
    sys.tables
INNER JOIN  
    INFORMATION_SCHEMA.COLUMNS ON TABLE_NAME = name
WHERE 
TABLE_NAME = @TableName 
AND COLUMN_NAME IN
(
    -- Consider you are inserting the column names to a temporary table before executing this query
    SELECT COLS FROM TEMPTABLE
)