Microsoft SQL - Count varchar where max len = n

Posted on 2014-02-06
Medium Priority
Last Modified: 2014-02-06
Hello -

Hopefully an easy one for you.  I have a varchar(7) field in which the majority of the values are 7 digit numbers ('#######').  

I'm trying to count the number of records that have a "valid" 7 digit number value in them.  

I tried:

SELECT COUNT(myField) FROM myTable HAVING MAX(LEN(myField)) ='7'
That returned the count of all records in my table.  I know there are some blanks and commas and things in there so I know that's not right - Maybe I should use rtrim/ltrim?  

Then after a little research I found "DATALENGTH" and tried:

That returned a count smaller than the total record count (more than half).  This seems realistic but I'm not sure if I'm getting what I'm asking for.  

What would be the best way to go about this?

Question by:ttist25
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LVL 16

Assisted Solution

by:Surendra Nath
Surendra Nath earned 500 total points
ID: 39839447
ok you can try the below one

select count(1) from myTable where len(myField) = 7

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LVL 25

Assisted Solution

by:Lee Savidge
Lee Savidge earned 500 total points
ID: 39839469
SELECT COUNT(myField) FROM myTable where LEN(myField) =7

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LVL 66

Expert Comment

by:Jim Horn
ID: 39839511
>I'm trying to count the number of records that have a "valid" 7 digit number value in them.
Define 'valid'.  Seven characters, all numbers?
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LVL 75

Assisted Solution

by:Aneesh Retnakaran
Aneesh Retnakaran earned 500 total points
ID: 39839520
SELECT COUNT(myField) FROM myTable WHERE myfield like '[0-9][0-9][0-9][0-9][0-9][0-9][0-9]'
LVL 66

Accepted Solution

Jim Horn earned 500 total points
ID: 39839534
CREATE TABLE #foo (myField varchar(7) )

INSERT INTO #foo (myField) 
   ('1234567'), ('123,567'), ('  12345'), ('12345  '), ('1234567'), ('(23456)'),
   ('42'), ('one'), (NULL), ('BR549'), ('OU812'), ('foo')

-- Here's your seven numbers
SELECT * FROM #foo WHERE myField LIKE '[0-9][0-9][0-9][0-9][0-9][0-9][0-9]'

-- Here's everything else
SELECT * FROM #foo WHERE myField NOT LIKE '[0-9][0-9][0-9][0-9][0-9][0-9][0-9]'

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Author Closing Comment

ID: 39839643
Thanks for the responses everyone.  I always forget how powerful "LIKE" can be.  

Anyway - I ended up using a combination of LEN(myField) = 7 and NOT LIKE '[0-9] ...' which may redefine what I consider a "valid" value to be (this is all discovery stage analysis).  

Thanks again everyone.  Always good to know you're there!

Author Comment

ID: 39839645
Jim!  VH fan????  :D
LVL 66

Expert Comment

by:Jim Horn
ID: 39839669
Yes, and I have a couple of sample data sets that use as customers Sammy Hagar, Eddie Van Halen, Joe Satriani, etc.

Author Comment

ID: 39839723
Hah!  That's great.  I'm a player myself and have recently started building pedals for fun.  My standard test for the sound quality of a prototype pedal has become the opening bars from Drop Dead Legs.  

Ok - that's probably enough off-topic garbage but I had to acknowledge the connection.
LVL 66

Expert Comment

by:Jim Horn
ID: 39839739
In that case check out this question but do me a favor and don't answer it right away, as I want to see how the other yahoos do.

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