'Need query to join table on column with comma separated
Table1
| ID | Notes | ReasonID |
|---|---|---|
| 1 | Test1 | [11,12] |
| 2 | Test2 | [13,14] |
Table 2
| Reasonid | Name |
|---|---|
| 11 | Other1 |
| 12 | Other2 |
| 13 | Other3 |
| 14 | Other4 |
Result should look like this, where Notes column from Table1 should concat with Name column from Table2.
| ID | Final_Notes |
|---|---|
| 1 | Test1,Other1,Other2 |
| 2 | Test2,Other3,Other4 |
Solution 1:[1]
If you use SQL Server 2017+, you may try to parse the ReasonID column as JSON, use an appropriate JOIN and then aggregate with STRING_AGG().
Sample data:
SELECT *
INTO Table1
FROM (VALUES
(1, 'Test1', '[11,12]'),
(2, 'Test2', '[13,14]')
) t (ID, Notes, ReasonID)
SELECT *
INTO Table2
FROM (VALUES
(11, 'Other1'),
(12, 'Other2'),
(13, 'Other3'),
(14, 'Other4')
) t (ReasonID, Name)
Statement:
SELECT
ID,
FinalNotes = CONCAT(
Notes,
',',
(
SELECT STRING_AGG(t2.Name, ',') WITHIN GROUP (ORDER BY CONVERT(int, j.[key]))
FROM OPENJSON(ReasonID) j
-- Important, JOIN with possible implicit conversion
JOIN Table2 t2 ON j.[value] = t2.ReasonID
)
)
FROM Table1
Result:
| ID | FinalNotes |
|---|---|
| 1 | Test1,Other1,Other2 |
| 2 | Test2,Other3,Other4 |
Solution 2:[2]
Please try the following solution.
It will work starting from SQL Server 2012 onwards.
It is using the following:
- XML/XQuery to tokenize comma separated list of values.
FOR XML PATHto compose FinalNotes comma separated list.
SQL
-- DDL and sample data population, start
DECLARE @Table1 TABLE(ID INT, Notes VARCHAR(60), ReasonID VARCHAR(60));
INSERT INTO @Table1(ID, Notes, ReasonID) VALUES
(1, 'Test1', '[11,12]'),
(2, 'Test2', '[13,14]');
DECLARE @Table2 TABLE(Reasonid INT, Name VARCHAR(60));
INSERT INTO @Table2(Reasonid, Name) VALUES
(11, 'Other1'),
(12, 'Other2'),
(13, 'Other3'),
(14, 'Other4');
-- DDL and sample data population, end
DECLARE @separator CHAR(1) = ',';
;WITH rs AS
(
SELECT ID, Notes, Name
FROM @Table1 AS t
CROSS APPLY (SELECT TRY_CAST('<root><r><![CDATA[' +
REPLACE(REPLACE(REPLACE(ReasonID,'[',''),']',''), @separator, ']]></r><r><![CDATA[') +
']]></r></root>' AS XML)) AS t1(c)
CROSS APPLY c.nodes('/root/r/text()') AS t2(x)
INNER JOIN @Table2 AS t3 ON t3.Reasonid = x.value('.', 'INT')
)
SELECT ID, CONCAT(Notes
, (SELECT @separator + c.Name AS [text()]
FROM rs AS c
WHERE c.ID = p.ID
FOR XML PATH(''))) AS FinalNotes
FROM rs AS p
GROUP BY ID, Notes;
Output
+----+---------------------+
| ID | FinalNotes |
+----+---------------------+
| 1 | Test1,Other1,Other2 |
| 2 | Test2,Other3,Other4 |
+----+---------------------+
Solution 3:[3]
use SUBSTRING(string, 2, LEN(string)-2) for deleting [] and Parsename to split based on comma and join and concat as follows
Your data
DECLARE @Table1 TABLE(
ID INTEGER NOT NULL,
Notes VARCHAR(60) NOT NULL,
ReasonID VARCHAR(60) NOT NULL
);
INSERT INTO @Table1(ID, Notes, ReasonID)
VALUES
(1, 'Test1', '[11,12]'),
(2, 'Test2', '[13,14]');
DECLARE @Table2 TABLE(
Reasonid INTEGER NOT NULL,
Name VARCHAR(60) NOT NULL
);
INSERT INTO @Table2(Reasonid, Name)
VALUES
(11, 'Other1'),
(12, 'Other2'),
(13, 'Other3'),
(14, 'Other4');
your query
SELECT id,
Concat(notes, ',', T2.name, ',', T3.name) FinalNotes
FROM (SELECT id,
notes,
Parsename(Replace(SUBSTRING(ReasonID, 2, LEN(ReasonID)-2), ',', '.'), 2) R1,
Parsename(Replace(SUBSTRING(ReasonID, 2, LEN(ReasonID)-2), ',', '.'), 1) R2
FROM @table1) T1
join @table2 T2
ON T1.R1 = T2.reasonid
join @table2 T3
ON T1.R2 = T3.reasonid
by using XML
DROP TABLE IF EXISTS #t -- temporary table
select t1.ID,t1.Notes, Name into #t -- temporary table
from
(
SELECT A.ID,a.Notes,
Split.a.value('.', 'VARCHAR(100)') AS String
FROM (SELECT ID, Notes,
CAST ('<M>' + REPLACE(SUBSTRING(ReasonID, 2, LEN(ReasonID)-2) , ',', '</M><M>') + '</M>' AS XML) AS String
FROM @Table1) AS A CROSS APPLY String.nodes ('/M') AS Split(a)) t1
join @Table2 t2 on t1.String=t2.Reasonid
---XML Path
SELECT ID,concat(notes,',',
STUFF((SELECT ', ' + CAST(name AS VARCHAR(10)) [text()]
FROM #t t1
WHERE t1.ID = t.ID
FOR XML PATH(''), TYPE)
.value('.','NVARCHAR(MAX)'),1,2,' ')) FinalNotes
FROM #t t
GROUP BY ID,notes
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Charlieface |
| Solution 2 | |
| Solution 3 |
