Find value in comma-separated list - sql

Question

I have lists like this stored in a field in SQL:
Id MyListField
1 soy, meat, milk, wheat, nuts
2 soy, meat, nuts, milk, wheat
3 nuts
4 walnuts, nutshell, nuts
5 nuts, beans
6 walnuts, hazel nuts
7 nutshell, nutsize
This is not normalized and I can't change it. I now have to find nuts and replace them with insane (but keep e. g. hazel nuts and nutshell).
A Find is relatively easy, there are four cases and I have the placeholder character % at my disposal:
SELECT * FROM MyTable WHERE
MyListField LIKE 'nuts' OR
MyListField LIKE 'nuts, %' OR
MyListField LIKE '%, nuts' OR
MyListField LIKE '%, nuts, %'
But a Replace is hard, because I don't have placeholder or startofstring/endofstring characters:
UPDATE MyTable SET MyListField =
REPLACE(
REPLACE(
REPLACE(
REPLACE(
MyListField,
', nuts, ' ,
', insane, '
),
????,
????
),
????,
????
),
????,
????
)
WHERE
MyListField LIKE 'nuts' OR
MyListField LIKE 'nuts, %' OR
MyListField LIKE '%, nuts' OR
MyListField LIKE '%, nuts, %'
I can easily replace it in the middle of the string, but not at the start or the end. Or can I?
I am using SQL Server 2008, if that matters.

Answer 1

If you concatenate the field between , and _, then you only have one case to look for.
e.g.
update MyTable set MyFieldList = replace(', ' + MyFieldList + ',', ', nuts,', ', insane,')
update MyTable set MyFieldList = substring(MyFieldList, 3, len(MyFieldList) - 3)

Answer 2

By putting a comma at the start and end of the list, each element is wrapped in a leading ", " and a trailing ",".
Then your replace method becomes easy....
REPLACE(', ' + MyListField + ',',
', ' + #termToReplace + ',',
', ' + #replacement + ',')
Finally, strip the leading and trailing commas.

Answer 3

You could use something like
case
when Field = #val then #replaceVal
when Field like #val + ', %' then
#replaceVal + substring(Field, len(#val) + 1, len(Field))
when Field like '%, ' + #val then
left(Field, len(Field) - len(#val)) + #replaceVal
Just add all the remaining options, and you should be fine.

Answer 4

With all that help, it was easier than I thought it to be.
UPDATE MyTable
SET MyListField =
SUBSTRING(
REPLACE(
CONCAT(', ', MyListField, ', '),
', nuts, ',
', insane, '
),
3,
DATALENGTH(MyListField)+LEN('insane')-LEN('nuts')
)
WHERE CONCAT(', ',MyListField,', ') LIKE '%, nuts, %'
This works as long as there is one or zero occurrences in the string. If there can be more than one, it will fail, but this does not matter to me. If it matters to you, feel free to multiply the (LEN-LEN) part with How do you count the number of occurrences of a certain substring in a SQL varchar?