find the end point of a pattern in SQL server - sql

Question

There is a comma separated string in a column which looks like
test=1,value=2.2,system=321
I want to extract value out from the string. I can use select PatIndex('%value=%',columnName) then use left, but this only find the beginning of the patindex.
How to identify the end of pattern value=%, so we can extract the value out?

Answer 1

Chain a few SUBSTRING with CHARINDEX and your PATHINDEX.
DECLARE #text VARCHAR(100) = 'test=1,value=2.21954,system=321'
SELECT
Original = #text,
Parsed = SUBSTRING( -- Get a portion of the original value
#text,
PATINDEX('%value=%',#text) + 6, -- ... starting from the 'value=' (without the 'value=')
-1 + CHARINDEX( -- ... and get as many characters until the first comma
',',
SUBSTRING( -- ... (find the comma starting from the 'value=' onwards)
#text,
PATINDEX('%value=%',#text) + 6,
100)))
Result:
Original Parsed
test=1,value=2.2,system=321 2.2
Note that the CHARINDEX will fail if there is no comma after your value=. You can filter this with a WHERE.
I strongly suggest to store your values already split on a proper table and you wont have to deal with string nightmares like this.

Answer 2

You can use CHARINDEX with starting position to find the first comma after the pattern. CROSS APPLY is used to keep the query easier to read:
WITH tests(str) AS (
SELECT 'test=1,value=2.2,system=321'
)
SELECT str, substring(str, pos1, pos2 - pos1) AS match
FROM tests
CROSS APPLY (SELECT PATINDEX('%value=%', str) + 6) AS ca1(pos1)
CROSS APPLY (SELECT CHARINDEX(',', str, pos1 + 1)) AS ca2(pos2)
-- 2.2

Answer 3

First of all, don't store denormalized data in this way, if you want to query them. SQL, the language, isn't good at string manipulation. Parsing and splitting strings can't take advantage of indexes either, which means any query that tried to find eg all records that refer to system 321 would have to scan and parse all rows.
SQL Server 2016 and JSON
SQL Server 2016 added suppor for JSON and the STRING_SPLIT function. Earlier versions already provided the XML type. It's better to store complex values as JSON or XML instead of trying to parse the string.
One option is to convert the string into a JSON object and retrieve the value contents, eg :
DECLARE #text VARCHAR(100) = 'test=1,value=2.2,system=321'
select json_value('{"' + replace(replace(#text,',','","'),'=','":"') + '"}','$.value')
This returns 2.2.
The replacements converted the original string into
{"test":"1","value":"2.2","system":"321"}
JSON_VALUE(#json,'$.') will return the value property of that object
Earlier SQL Server versions
In earlier SQL Server version, you can convert that string into an XML element the same way and use XQuery :
DECLARE #text VARCHAR(100) = 'test=1,value=2.2,system=321';
declare #xml varchar(100)='<r ' + replace(replace(#text,',','" '),'=',' ="') + '" />';
select #xml
select cast(#xml as xml).value('(/r[1]/#value)','varchar(20)')
In this case #xml contains :
<r test ="1" value ="2.2" system ="321" />
The query result is 2.2

Answer 4

You can try like following.
DECLARE #xml AS XML
SELECT #xml = Cast(( '<X>' + Replace(txt, ',', '</X><X>') + '</X>' ) AS XML)
FROM (VALUES ('test=1,value=2.2,system=321')) v(txt)
SELECT LEFT(value, Charindex('=', value) - 1) AS LeftPart,
RIGHT(value, Charindex('=', Reverse(value)) - 1) AS RightPart
FROM (SELECT n.value('.', 'varchar(100)') AS value
FROM #xml.nodes('X') AS T(n))T
Online Demo
Output
+----------+-----------+
| LeftPart | RightPart |
+----------+-----------+
| test | 1 |
+----------+-----------+
| value | 2.2 |
+----------+-----------+
| system | 321 |
+----------+-----------+

Answer 5

You can try the below query if you are using SQL Server (2016 or above)
SELECT RIGHT(Value,CHARINDEX('=',REVERSE(Value))-1) FROM YourTableName
CROSS APPLY STRING_SPLIT ( ColumnName , ',' )
WHERE Value Like 'Value=%'