Question: I'm trying to figure out an easy query I can do to test if a large table has a list of entries that has at least ONE blank (NULL / empty) value in ANY column.

I need something like

```
SELECT * FROM table AS t WHERE ANY(t.* IS NULL)

```

I don't want to have to do

```
SELECT * FROM table AS t WHERE t.c1 = NULL OR t.c2 = NULL OR t.c3 = NULL

```

This would be a HUGE query.

Answer: 
An extension to @db2's answer with less (read:zero) hand-wrangling:

```
DECLARE @tb nvarchar(512) = N'dbo.[table]';

DECLARE @sql nvarchar(max) = N'SELECT * FROM ' + @tb
    + N' WHERE 1 = 0';

SELECT @sql += N' OR ' + QUOTENAME(name) + N' IS NULL'
    FROM sys.columns 
    WHERE [object_id] = OBJECT_ID(@tb)
    AND is_nullable = 1;

EXEC sys.sp_executesql @sql;

```
You should list out all the columns as per JNK's comment. 

```
WHERE c1 IS NULL OR c2 IS NULL OR c3 IS NULL

```

A somewhat less efficient approach that avoids this is below though.

```
;WITH xmlnamespaces('http://www.w3.org/2001/XMLSchema-instance' AS ns) 
SELECT * 
FROM   YourTable AS T1 
WHERE (
    SELECT T1.* 
    FOR XML PATH('row'), ELEMENTS XSINIL, TYPE
  ).exist('//*/@ns:nil') = 1 

```

[(Based on this SO answer)](https://stackoverflow.com/questions/9079037/how-to-count-in-sql-all-fields-with-null-values-in-one-record)