查询表中每列的填充行数(SQL Server)

时间:2019-05-01 18:35:29

标签: sql sql-server tsql

我试图获取表中每一列的非空值的计数。我研究了以下先前提出的SO问题,但没有找到满意的答案:

Query to list number of records in each table in a database

get a count of each value from every column in a table SQL Server

SQL Server count number of distinct values in each column of a table

我编写了以下代码,以尝试为我的表构建一个数据字典,以包括列的名称,每列中填充的行数,数据类型,长度以及它是否为主数据。密钥:

SELECT 
    c.name 'Column Name',
    p.rows 'Row_Count',
    t.Name 'Data type',
    c.max_length 'Max Length',
    ISNULL(i.is_primary_key, 0) 'Primary Key'
FROM    
    sys.columns c
INNER JOIN 
    sys.types t ON c.user_type_id = t.user_type_id
LEFT OUTER JOIN 
    sys.index_columns ic ON ic.object_id = c.object_id AND ic.column_id = c.column_id
LEFT OUTER JOIN 
    sys.indexes i ON ic.object_id = i.object_id AND ic.index_id = i.index_id
LEFT OUTER JOIN
sys.partitions p ON p.OBJECT_ID = i.OBJECT_ID and i.index_id = p.index_id
WHERE
    c.object_id = OBJECT_ID('my_table')

但是,Row_Count列返回所有Null。

我的预期结果如下:

Column_Name Row_Count Data_Type Max_Length Primary_Key
A              10       varchar   50            0
B              10       varchar   50            0
C              7        float     50            0
D              3        float     50            0
E              10       varchar   50            0

2 个答案:

答案 0 :(得分:5)

这是一种不使用动态SQL的选项。

全面披露,我怀疑DS的表现会更好。也就是说,这几乎可以在任何表,视图或查询上使用。我正在使用master..spt_values作为演示

示例

Select ColumnName      = Item
      ,B.column_ordinal
      ,Row_Count       = sum(1)
      ,B.system_type_name
      ,B.max_length
      ,Distinct_Values = count(DISTINCT Value)
 From  (
        Select C.*
         From  master..spt_values A
         Cross Apply ( values (cast((Select A.* for XML RAW) as xml))) B(XMLData)
         Cross Apply (
                        Select Item  = replace(xAttr.value('local-name(.)', 'varchar(100)'),'_x0020_',' ')
                              ,Value = xAttr.value('.','varchar(max)')
                         From  XMLData.nodes('//@*') xNode(xAttr)
                     ) C
       ) A
 Left Join  (
        Select * from sys.dm_exec_describe_first_result_set('Select * from master..spt_values',null,null )  
       ) B on A.Item=B.name
 Group By A.Item
         ,B.system_type_name
         ,B.max_length
         ,B.column_ordinal 
 Order By B.column_ordinal 

返回

enter image description here

编辑

正如Larnu所提到的,这将导致(var)binary和image失败。同样,这在大型桌子上效果不佳。我只是在发现阶段使用了这种方法。

答案 1 :(得分:3)

这太丑了...

DECLARE @SQL nvarchar(MAX);
DECLARE @Table sysname = 'SampleTable';
DECLARE @Schema sysname = 'dbo';

SET @SQL = N'WITH Counts AS (' + NCHAR(13) + NCHAR(10) + 
           N'    SELECT @Schema AS SchemaName,' + NCHAR(13) + NCHAR(10) +
           N'           @Table AS TableName,' +
           STUFF((SELECT N',' + NCHAR(13) + NCHAR(10) + 
                         N'           COUNT(' + QUOTENAME(C.COLUMN_NAME) + N') AS ' + QUOTENAME(COLUMN_NAME)
                  FROM INFORMATION_SCHEMA.COLUMNS C
                  WHERE C.TABLE_SCHEMA = @Schema
                    AND C.TABLE_NAME = @Table
                  FOR XML PATH(N''),TYPE).value('.','nvarchar(MAX)'),1,14,N'') + NCHAR(13) + NCHAR(10) + 
           N'    FROM ' + QUOTENAME(@Table) + N')' + NCHAR(13) + NCHAR(10) + 
           N'SELECT V.ColumnName,' + NCHAR(13) + NCHAR(10) + 
           N'       V.NonNullCount,' + NCHAR(13) + NCHAR(10) + 
           N'       ISC.DATA_TYPE + ISNULL(DT.S,'''') AS Datatype,' + NCHAR(13) + NCHAR(10) +
           N'       ISNULL(PK.IsPrimaryKey,''No'') AS PrimaryKey' + NCHAR(13) + NCHAR(10) +
           N'FROM Counts C' + NCHAR(13) + NCHAR(10) + 
           N'     CROSS APPLY(VALUES' + STUFF((SELECT N',' + NCHAR(13) + NCHAR(10) + 
                                                      N'                       (N' + QUOTENAME(C.COLUMN_NAME,'''') + N',C.' + QUOTENAME(C.COLUMN_NAME) + N')'
                                               FROM INFORMATION_SCHEMA.COLUMNS C
                                               WHERE C.TABLE_NAME = @Table
                                               FOR XML PATH(N''),TYPE).value('.','nvarchar(MAX)'),1,26,N'') + N')V(ColumnName,NonNullCount)' + NCHAR(13) + NCHAR(10) +
           N'     JOIN INFORMATION_SCHEMA.COLUMNS ISC ON C.SchemaName = ISC.TABLE_SCHEMA' + NCHAR(13) + NCHAR(10) +
           N'                                        AND C.TableName = ISC.TABLE_NAME' + NCHAR(13) + NCHAR(10) +
           N'                                        AND V.ColumnName = ISC.COLUMN_NAME' + NCHAR(13) + NCHAR(10) + 
           N'     CROSS APPLY (VALUES(''('' + STUFF(CONCAT('','' + CASE ISC.CHARACTER_MAXIMUM_LENGTH WHEN -1 THEN ''MAX'' ELSE CONVERT(varchar(4),ISC.CHARACTER_MAXIMUM_LENGTH) END,' + NCHAR(13) + NCHAR(10)+
           N'                                            '','' + CASE WHEN ISC.DATA_TYPE NOT LIKE ''%int'' THEN CONVERT(varchar(4),ISC.NUMERIC_PRECISION) END,' + NCHAR(13) + NCHAR(10) +
           N'                                            '','' + CASE WHEN ISC.DATA_TYPE NOT LIKE ''%int'' THEN CONVERT(varchar(4),ISC.NUMERIC_SCALE) END,' + NCHAR(13) + NCHAR(10) +
           N'                                            '','' + CONVERT(varchar(4),ISC.DATETIME_PRECISION)),1,1,'''') + '')'')) DT(S)' + NCHAR(13) + NCHAR(10) +
           N'     OUTER APPLY(SELECT ''Yes'' AS IsPrimaryKey ' + NCHAR(13) + NCHAR(10) + 
           N'                 FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS TC' + NCHAR(13) + NCHAR(10) + 
           N'                      JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE KCU ON TC.TABLE_SCHEMA = KCU.TABLE_SCHEMA' + NCHAR(13) + NCHAR(10) + 
           N'                                                                  AND TC.TABLE_NAME = KCU.TABLE_NAME' + NCHAR(13) + NCHAR(10) + 
           N'                                                                  AND TC.CONSTRAINT_NAME = KCU.CONSTRAINT_NAME' + NCHAR(13) + NCHAR(10) + 
           N'                 WHERE TC.CONSTRAINT_TYPE = ''PRIMARY KEY''' + NCHAR(13) + NCHAR(10) + 
           N'                   AND KCU.COLUMN_NAME = V.ColumnName' + NCHAR(13) + NCHAR(10) + 
           N'                   AND TC.TABLE_SCHEMA = ISC.TABLE_SCHEMA' + NCHAR(13) + NCHAR(10) + 
           N'                   AND TC.TABLE_NAME = ISC.TABLE_NAME) PK;';

PRINT @SQL; --Might need to use SELECT here
--SELECT @SQL;
EXEC sp_executesql @SQL, N'@Schema sysname,@Table sysname',@Schema = @Schema, @Table = @Table;

db<>fiddle

老实说,这里发生了很多事情。如果需要解释,我会尝试,但是会花费一些时间,因此(没有违法行为),如果没有人想要/不需要知道,我不会花大力气。

请注意,我有点懒,没有加入SCHEMA_NAME。如果您使用具有相同命名对象的多个架构,则会出现问题,需要解决。

编辑:显然,我是受惩罚的glut嘴。修复了模式“问题”,并增加了关于ints的逻辑

New Fiddle