SQL Server : find all stored procedures containing Left(column,5) - sql-server

Question

I'm using SQL Server 2014 and I need to find all stored procedures that use a Left(Column, 5) function
For example, it should find
left(jobno, 5)
but not
left(jobno, 50)
it should find
left(ZIP, 5)
I initially thought I could use
Like '%Left(%,5)%'
but it returns no records
Thanks
mark

Answer 1

This is a bit costly approach, but is serves the purpose. It will work when Length to LEFT function is specified as Number, but fails when Length is specified using variable or some other expressions. Hope this helps.
DECLARE #Search VARCHAR(10) = '5'
;WITH CTE AS
(
SELECT 1 AS Lvl, o.name, o.type, o.Object_ID
,NULLIF(CHARINDEX('left(',definition),0) AS LeftPos
,CHARINDEX(')', definition, (NULLIF(CHARINDEX('left(',definition),0))) AS LastIndex
,LTRIM(RTRIM(SUBSTRING(definition, CHARINDEX(',', definition, (NULLIF(CHARINDEX('left(',definition),0))) + 1, CHARINDEX(')', definition, (NULLIF(CHARINDEX('left(',definition),0))) - CHARINDEX(',', definition, (NULLIF(CHARINDEX('left(',definition),0))) -1 ))) AS LeftLen
FROM sys.all_sql_modules m
INNER JOIN sys.objects o on o.Object_ID = m.Object_ID
WHERE CHARINDEX(')', definition, (NULLIF(CHARINDEX('left(',definition),0))) - CHARINDEX(',', definition, (NULLIF(CHARINDEX('left(',definition),0))) -1 > 0
UNION ALL -- get all occurunces or Left
SELECT 2 AS Lvl, o.name, o.type, o.Object_ID
,NULLIF(CHARINDEX('left(',definition,LastIndex),0) AS LeftPos
,CHARINDEX(')', definition, (NULLIF(CHARINDEX('left(',definition,LastIndex),0))) AS LastIndex
,LTRIM(RTRIM(SUBSTRING(definition, CHARINDEX(',', definition, (NULLIF(CHARINDEX('left(',definition,LastIndex),0))) + 1, CHARINDEX(')', definition, (NULLIF(CHARINDEX('left(',definition,LastIndex),0))) - CHARINDEX(',', definition, (NULLIF(CHARINDEX('left(',definition,LastIndex),0))) -1 ))) AS LeftLen
FROM sys.all_sql_modules m
INNER JOIN sys.objects o ON o.Object_ID = m.Object_ID
INNER JOIN CTE c ON c.Object_ID = m.Object_ID
AND c.LastIndex IS NOT NULL AND ISNULL(c.LeftLen,#Search) <> #Search
WHERE CHARINDEX(')', definition, (NULLIF(CHARINDEX('left(',definition,LastIndex),0))) - CHARINDEX(',', definition, (NULLIF(CHARINDEX('left(',definition,LastIndex),0))) -1 > 0
)
SELECT * FROM CTE
WHERE LeftLen = #Search
OPTION (MAXRECURSION 10000);

Answer 2

SQL unfortunately cannot have wild cards in the middle of a string. Try this
(column name) like '%left(%' and (column name) like '%,5)%'
That should give you the desired result

Answer 3

This is one of the reasons why we've implemented regex using the SQL CLR. However, you can do a couple other things to solve your issue.
You could do something like this ...
select *
from sys.objects
where object_definition(object_id) like '%left(%,5)%';
Or something like this, which seems a bit cleaner, however, sometimes I use the above method, because I need some other stuff that spans from sys.objects ...
select *
from sys.sql_modules
where definition like '%left(%,5)%';
Another alternative is to come up with a list of possible column names. If you don't have any cross database worries, then you may want to use sys.columns, sys.all_columns or even sys.parameters as your pick list. To do this, you'd do something like this ...
select *
from sys.sql_modules m
cross join (
select distinct name
from sys.all_columns) c
where definition like '%left(' + c.name + ',5)%';
One other thing to consider is whitespace, which, again, is another reason to implement regex using SQL CLR. For instance, in the above case, somebody may put a space between the "," after the column name and the number "5". But, with that being said, the above queries should get you really close.
Hope this helps.

Answer 4

You could use something like
LIKE '%LEFT([a-z],5)%'