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 Posted Thursday, May 20, 2010 12:52 AM
 Old Hand Group: General Forum Members Last Login: Saturday, November 27, 2010 11:14 PM Points: 356, Visits: 99
 Hi,I. SELECT COUNT_BIG(*)II. SELECT COUNT_BIG(column_2)III. SELECT COUNT_BIG(ALL column_2)IV. SELECT COUNT_BIG(DISTINCT column_2)I Answered this question selecting II,III and IV (which actually is a correct answer) guess what i was indicated as INCORRECT! Pls let me know if I am so.coz all the three (2,3,4) have same output.
Post #924850
 Posted Thursday, May 20, 2010 3:09 AM
 SSCrazy Eights Group: General Forum Members Last Login: Friday, November 11, 2016 6:39 AM Points: 9,932, Visits: 11,346
 Avaneesh -388582 (5/20/2010)Hi,I. SELECT COUNT_BIG(*)II. SELECT COUNT_BIG(column_2)III. SELECT COUNT_BIG(ALL column_2)IV. SELECT COUNT_BIG(DISTINCT column_2)I Answered this question selecting II,III and IV (which actually is a correct answer) guess what i was indicated as INCORRECT! Pls let me know if I am so.coz all the three (2,3,4) have same output.Wow! That's amazing! What code did you use to come to that conclusion? I used this:`DECLARE @QotDTABLE ( Column_1 TINYINT NOT NULL, Column_2 TINYINT NULL ); INSERT @QotD VALUES (1, 1);INSERT @QotD VALUES (2, NULL);INSERT @QotD VALUES (3, 2);INSERT @QotD VALUES (4, 2);INSERT @QotD VALUES (5, NULL);INSERT @QotD VALUES (6, 3);INSERT @QotD VALUES (7, 4);INSERT @QotD VALUES (8, 5);INSERT @QotD VALUES (9, 6);INSERT @QotD VALUES (10, 7);--I.SELECT COUNT_BIG(*) FROM @QotD;--II.SELECT COUNT_BIG(Column_2) FROM @QotD;--III.SELECT COUNT_BIG(ALL Column_2) FROM @QotD;--IV.SELECT COUNT_BIG(DISTINCT Column_2) FROM @QotD;`I Answered this question selecting II,III (which actually is a correct answer) guess what i was indicated as CORRECT! Pls let me know if I am so. coz all the two (2,3) have same output. Paul WhiteSQLPerformance.comSQLblog.com@SQL_Kiwi
Post #924926
 Posted Monday, September 27, 2010 5:22 PM
 SSChasing Mays Group: General Forum Members Last Login: Sunday, November 17, 2013 11:53 AM Points: 623, Visits: 237
 All is by default so II & III are same expressions
Post #994119
 Posted Friday, April 29, 2011 7:52 AM
 SSChasing Mays Group: General Forum Members Last Login: Sunday, March 4, 2012 4:02 AM Points: 660, Visits: 134
 I think this question could have been worded better and is incomplete.
Post #1100880
 Posted Thursday, August 30, 2012 1:24 AM
 Grasshopper Group: General Forum Members Last Login: Friday, June 14, 2013 7:31 AM Points: 15, Visits: 14
 explination given here is not fully correct-->"SELECT COUNT_BIG(ALL COlumn_2) and SELECT COUNT_BIG(column_2). COUNT_BIG(all column_2) will count all rows from the table (including duplicate & null rows). COUNT_BIG(DISTINCT column_2) will evaluate distinct non-null values. because ---> "COUNT_BIG(all column_2) will count all rows from the table including duplicate & Excluding null rows).andCOUNT_BIG(*) will count all rows from the table (including duplicate & null rows).
Post #1352036
 Posted Thursday, August 30, 2012 1:29 AM
 Grasshopper Group: General Forum Members Last Login: Friday, June 14, 2013 7:31 AM Points: 15, Visits: 14
 explanation given here is not fully correct-->"SELECT COUNT_BIG(ALL COlumn_2) and SELECT COUNT_BIG(column_2). COUNT_BIG(all column_2) will count all rows from the table (including duplicate & null rows). COUNT_BIG(DISTINCT column_2) will evaluate distinct non-null values. because ---> "COUNT_BIG(all column_2) will count all rows from the table including duplicate & Excluding null rows).andCOUNT_BIG(*) will count all rows from the table (including duplicate & null rows).
Post #1352038

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