如何检测和删除仅包含空值的列？

Question

在我的表table1中有6列Locations,a,b,c,d,e.

Locations [a]   [b]   [c]  [d]   [e]

[1]       10.00 Null  Null 20.00 Null

[2]       Null  30.00 Null Null  Null

我需要像这样的结果

Locations [a]   [b]   [d]

[1]       10.00 Null  20.00

[2]       Null  30.00 Null

我的问题是如何使用sql查询检测和删除包含所有空值的列.可能吗？

如果是,请帮助并提供样品.

Answer 1

这是一个快速(和丑陋)的存储过程,它接受表的名称并打印(或者如果你想要的话,删除)充满空值的字段.

ALTER procedure mysp_DropEmptyColumns 
  @tableName nvarchar(max)
as begin
  declare @FieldName nvarchar(max)
  declare @SQL nvarchar(max)
  declare @CountDef nvarchar(max)
  declare @FieldCount int

  declare fieldNames cursor  local fast_forward for
    select c.name
      from syscolumns c 
        inner join sysobjects o on c.id=o.id
      where o.xtype='U'
        and o.Name=@tableName

  open fieldNames 
  fetch next from fieldNames into @FieldName
  while (@@fetch_status=0)
  begin
    set @SQL=N'select @Count=count(*) from "'+@TableName+'" where "'+@FieldName+'" is not null'
    SET @CountDef = N'@Count int output';
    exec sp_executeSQL @SQL, @CountDef, @Count = @FieldCount output
    if (@FieldCount=0)
    begin
      set @SQL = 'alter table '+@TableName+' drop column '+@FieldName
      /* exec sp_executeSQL @SQL */
      print @SQL
    end
    fetch next from fieldNames into @FieldName
  end

  close fieldNames
end

这使用了一个游标,并且有点慢和令人费解,但我怀疑这是一种你经常运行的程序

Answer 2

如何检测给定列是否只有值NULL：

SELECT 1  -- no GROUP BY therefore use a literal
  FROM Locations
HAVING COUNT(a) = 0 
       AND COUNT(*) > 0;

结果集将由零行（列a没有NULL值）或一行（列a只有NULL值）组成。FWIW 该代码是标准 SQL-92。