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    <p>Try this one -</p> <p><strong>Solution #1:</strong></p> <pre><code>DECLARE @t VARCHAR(100) SELECT @t = 'A:B:C;J:K;P:L:J;' SELECT * FROM ( SELECT token = t.c.value('.', 'VARCHAR(100)') FROM ( SELECT xmls = CAST('&lt;t&gt;' + REPLACE( REPLACE(@t, ':', ' '), ';', '&lt;/t&gt;&lt;t&gt;') + '&lt;/t&gt;' AS XML) ) r CROSS APPLY xmls.nodes('/t') AS t(c) ) t WHERE t.token != '' </code></pre> <p><strong>Output:</strong></p> <pre><code>---------- A B C J K P L J </code></pre> <p><strong>Solution #2:</strong></p> <pre><code>DECLARE @t VARCHAR(100) SELECT @t = 'A:B:C;J:K;P:L:J;' PRINT REPLACE(REPLACE(@t, ':', ' '), ';', CHAR(13) + CHAR(13)) </code></pre> <p><strong>Output:</strong></p> <pre><code>A B C J K P L J </code></pre> <p><strong>Solution #3:</strong></p> <pre><code>DECLARE @t VARCHAR(100) SELECT @t = 'A:B:C;J:K;P:L:J;' SELECT [1], [2], [3] FROM ( SELECT t2.id , t2.name , rn2 = ROW_NUMBER() OVER (PARTITION BY t2.id ORDER BY 1/0) FROM ( SELECT id = t.c.value('@n', 'INT') , name = t.c.value('@s', 'CHAR(1)') FROM ( SELECT x = CAST('&lt;t s = "' + REPLACE(token + ':', ':', '" n = "' + CAST(rn AS VARCHAR(10)) + '" /&gt;&lt;t s = "') + '" /&gt;' AS XML) FROM ( SELECT token = t.c.value('.', 'VARCHAR(100)') , rn = ROW_NUMBER() OVER (ORDER BY 1/0) FROM ( SELECT x = CAST('&lt;t&gt;' + REPLACE(@t, ';', '&lt;/t&gt;&lt;t&gt;') + '&lt;/t&gt;' AS XML) ) r CROSS APPLY x.nodes('/t') t(c) ) t ) d CROSS APPLY x.nodes('/t') t(c) ) t2 WHERE t2.name != '' ) t3 PIVOT ( MAX(name) FOR rn2 IN ([1], [2], [3]) ) p </code></pre> <p><strong>Output:</strong></p> <pre><code>1 2 3 ---- ---- ---- A B C J K NULL P L J </code></pre>
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    1. CO@rahul aggarwal, can you please tell what's wrong in my answer?
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    2. COanswer is correct but in above solution the output is shown in 3 different columns. Thanks for the solution
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    3. CO@rahul aggarwal, please see the updated answer.
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