how do I write a SQL where statement that checks if a string contains some substring and a number. For example:
string: macsea01
where string like 'macsea' plus a number
Regex is the most obvious solution to this question. Without more detail about the specific format of the string, I can suggest the following, which will match a sequence of a letter in the alphabet followed immediately by a digit:
where column_name like '%[a-zA-Z][0-9]%'
If you're literally looking for macsea at the beginning of the string followed by a digit, it would be:
where column_name like 'macsea[0-9]%'
Regex seem to bee a little slippery here, depending on your needs you can for instance divide the string into several parts, first the text part, and take the rest of the string, try to convert it into a number.
Somthing like this (but I think this perticular code is broken
where substring(column_name, 1, 6) = 'macsea' and cast(substring(column_name, 7, 1000) as int) > 0
I have the following test table in SQL Server 2005:
CREATE TABLE [dbo].[TestTable]
(
[ID] [int] NOT NULL,
[TestField] [varchar](100) NOT NULL
)
Populated with:
INSERT INTO TestTable (ID, TestField) VALUES (1, 'A value'); -- Len = 7
INSERT INTO TestTable (ID, TestField) VALUES (2, 'Another value '); -- Len = 13 + 6 spaces
When I try to find the length of TestField with the SQL Server LEN() function it does not count the trailing spaces - e.g.:
-- Note: Also results the grid view of TestField do not show trailing spaces (SQL Server 2005).
SELECT
ID,
TestField,
LEN(TestField) As LenOfTestField, -- Does not include trailing spaces
FROM
TestTable
How do I include the trailing spaces in the length result?
This is clearly documented by Microsoft in MSDN at http://msdn.microsoft.com/en-us/library/ms190329(SQL.90).aspx, which states LEN "returns the number of characters of the specified string expression, excluding trailing blanks". It is, however, an easy detail on to miss if you're not wary.
You need to instead use the DATALENGTH function - see http://msdn.microsoft.com/en-us/library/ms173486(SQL.90).aspx - which "returns the number of bytes used to represent any expression".
Example:
SELECT
ID,
TestField,
LEN(TestField) As LenOfTestField, -- Does not include trailing spaces
DATALENGTH(TestField) As DataLengthOfTestField -- Shows the true length of data, including trailing spaces.
FROM
TestTable
You can use this trick:
LEN(Str + 'x') - 1
I use this method:
LEN(REPLACE(TestField, ' ', '.'))
I prefer this over DATALENGTH because this works with different data types, and I prefer it over adding a character to the end because you don't have to worry about the edge case where your string is already at the max length.
Note: I would test the performance before using it against a very large data set; though I just tested it against 2M rows and it was no slower than LEN without the REPLACE...
"How do I include the trailing spaces in the length result?"
You get someone to file a SQL Server enhancement request/bug report because nearly all the listed workarounds to this amazingly simple issue here have some deficiency or are inefficient. This still appears to be true in SQL Server 2012. The auto trimming feature may stem from ANSI/ISO SQL-92 but there seems to be some holes (or lack of counting them).
Please vote up "Add setting so LEN counts trailing whitespace" here:
https://feedback.azure.com/forums/908035-sql-server/suggestions/34673914-add-setting-so-len-counts-trailing-whitespace
Retired Connect link:
https://connect.microsoft.com/SQLServer/feedback/details/801381
There are problems with the two top voted answers. The answer recommending DATALENGTH is prone to programmer errors. The result of DATALENGTH must be divided by the 2 for NVARCHAR types, but not for VARCHAR types. This requires knowledge of the type you're getting the length of, and if that type changes, you have to diligently change the places you used DATALENGTH.
There is also a problem with the most upvoted answer (which I admit was my preferred way to do it until this problem bit me). If the thing you are getting the length of is of type NVARCHAR(4000), and it actually contains a string of 4000 characters, SQL will ignore the appended character rather than implicitly cast the result to NVARCHAR(MAX). The end result is an incorrect length. The same thing will happen with VARCHAR(8000).
What I've found works, is nearly as fast as plain old LEN, is faster than LEN(#s + 'x') - 1 for large strings, and does not assume the underlying character width is the following:
DATALENGTH(#s) / DATALENGTH(LEFT(LEFT(#s, 1) + 'x', 1))
This gets the datalength, and then divides by the datalength of a single character from the string. The append of 'x' covers the case where the string is empty (which would give a divide by zero in that case). This works whether #s is VARCHAR or NVARCHAR. Doing the LEFT of 1 character before the append shaves some time when the string is large. The problem with this though, is that it does not work correctly with strings containing surrogate pairs.
There is another way mentioned in a comment to the accepted answer, using REPLACE(#s,' ','x'). That technique gives the correct answer, but is a couple orders of magnitude slower than the other techniques when the string is large.
Given the problems introduced by surrogate pairs on any technique that uses DATALENGTH, I think the safest method that gives correct answers that I know of is the following:
LEN(CONVERT(NVARCHAR(MAX), #s) + 'x') - 1
This is faster than the REPLACE technique, and much faster with longer strings. Basically this technique is the LEN(#s + 'x') - 1 technique, but with protection for the edge case where the string has a length of 4000 (for nvarchar) or 8000 (for varchar), so that the correct answer is given even for that. It also should handle strings with surrogate pairs correctly.
LEN cuts trailing spaces by default, so I found this worked as you move them to the front
(LEN(REVERSE(TestField))
So if you wanted to, you could say
SELECT
t.TestField,
LEN(REVERSE(t.TestField)) AS [Reverse],
LEN(t.TestField) AS [Count]
FROM TestTable t
WHERE LEN(REVERSE(t.TestField)) <> LEN(t.TestField)
Don't use this for leading spaces of course.
You need also to ensure that your data is actually saved with the trailing blanks. When ANSI PADDING is OFF (non-default):
Trailing blanks in character values
inserted into a varchar column are
trimmed.
You should define a CLR function that returns the String's Length field, if you dislike string concatination.
I use LEN('x' + #string + 'x') - 2 in my production use-cases.
If you dislike the DATALENGTH because of of n/varchar concerns, how about:
select DATALENGTH(#var)/isnull(nullif(DATALENGTH(left(#var,1)),0),1)
which is just
select DATALENGTH(#var)/DATALENGTH(left(#var,1))
wrapped with divide-by-zero protection.
By dividing by the DATALENGTH of a single char, we get the length normalised.
(Of course, still issues with surrogate-pairs if that's a concern.)
This is the best algorithm I've come up with which copes with the maximum length and variable byte count per character issues:
ISNULL(LEN(STUFF(#Input, 1, 1, '') + '.'), 0)
This is a variant of the LEN(#Input + '.') - 1 algorithm but by using STUFF to remove the first character we ensure that the modified string doesn't exceed maximum length and remove the need to subtract 1.
ISNULL(..., 0) is added to deal with the case where #Input = '' which causes STUFF to return NULL.
This does have the side effect that the result is also 0 when #Input is NULL which is inconsistent with LEN(NULL) which returns NULL, but this could be dealt with by logic outside this function if need be
Here are the results using LEN(#Input), LEN(#Input + '.') - 1, LEN(REPLACE(#Input, ' ', '.')) and the above STUFF variant, using a sample of #Input = CAST(' S' + SPACE(3998) AS NVARCHAR(4000)) over 1000 iterations
Algorithm
DataLength
ExpectedResult
Result
ms
LEN
8000
4000
2
14
+DOT-1
8000
4000
1
13
REPLACE
8000
4000
4000
514
STUFF+DOT
8000
4000
4000
0
In this case the STUFF algorithm is actually faster than LEN()!
I can only assume that internally SQL looks at the last character and if it is not a space then optimizes the calculation
But that's a good result eh?
Don't use the REPLACE option unless you know your strings are small - it's hugely inefficient
use
SELECT DATALENGTH('string ')
I am using ISNUMERIC to get all non numeric rows in my table - but all I get in return is the following example 1.437.230,61 or 3.511.980,00. I really dont know how to get these few rows converted to numeric! I have conveted about 2,5 mil rows without problem but I am getting about 9000 rows that are not numeric - but as displayed above they are numbers. I have tried to trim my coloumn with no luck!
You should always name the dbms you are using. Many dbms have problems converting proper numbers such as 1.437.230,61 due to the thousand separators. So isnumeric works fine here, but the conversion function doesn't.
Use a string replace function to remove the thousand separators from the string before converting. Such as:
to_number( replace(numstr, '.', '') )
I have a nvarchar field in my table, which contains all sorts of strings.
In case there are strings which contain a number following a non-number sign, I want to insert a space before that number.
That is - if a certain entry in that field is abc123, it should be turned into abc 123, or ab12.34 should become ab 12. 34.I want this to be done throughout the entire table.
What's the best way to achieve it?
You can try something like that:
select left(col,PATINDEX('%[0-9]%',col)-1 )+space(1)+
case
when PATINDEX('%[.]%',col)<>0
then substring(col,PATINDEX('%[0-9]%',col),len(col)+1-PATINDEX('%[.]%',col))
+space(1)+
substring(col,PATINDEX('%[.]%',col)+1,len(col)+1-PATINDEX('%[.]%',col))
else substring(col,PATINDEX('%[0-9]%',col),len(col)+1-PATINDEX('%[0-9]%',col))
end
from tab
It's not simply, but I hope it will help you.
SQL Fiddle
I used functions (link to MSDN):
LEFT, PATINDEX, SPACE, SUBSTRING, LEN
and regular expression.
I have a table with a field that denotes whether the data in that row is valid or not. This field contains a string of undetermined length. I need a query that will only pull out rows where all the characters in this field are N. Some possible examples of this field.
NNNNNNNNNNNNNNNNNNN
NNNNNNNNNNNNNNNNNNNNNNN
NNNNNEEEENNNNNNNNNNNN
NNNNNOOOOOEEEENNNNNNNNNNNN
Any suggestions on a postcard please.
Many thanks
This should do the trick:
SELECT Field
FROM YourTable
WHERE Field NOT LIKE '%[^N]%' AND Field <> ''
What it's doing is a wildcard search, broken down:
The LIKE will find records where the field contains characters other than N in the field. So, we apply a NOT to that as we're only interested in records that do not contain characters other than N. Plus a condition to filter out blank values.
SELECT *
FROM mytable
WHERE field NOT LIKE '%[^N]%'
I don't know which SQL dialect you are using. For example Oracle has several functions you may use. With oracle you could use condition like :
WHERE LTRIM(field, 'N') = ''
The idea is to trim out all N's and see if the result is empty string. If you don't have LTRIM, check if you have some kind of TRANSLATE or REPLACE function to do the same thing.
Another way to do it could be to pick length of your field and then construct comparator value by padding empty string with N. Perhaps something like:
WHERE field = RPAD('', field, 'N)
Oracle pads that empty string with N's and picks number of pad characters from length of the second argument. Perhaps this works too:
WHERE field = RPAD('', LENGTH(field), 'N)
I haven't tested those, but hopefully that give you some ideas how to solve your problem. I guess that many of these solutions have bad performance if you have lot of rows and you don't have other WHERE conditions to select proper index.