如何使用XQuery从XML将值列表提取到行中？

Question

I have an XQuery as under

DECLARE @x XML
SELECT @x = '<PartnerEmails>
<Email>a@xxxx.com</Email>
<Email>b@xxxx.com</Email>
</PartnerEmails>'
SELECT @x.query('data(PartnerEmails/Email)').value('.','varchar(100)') AS Val

Actual Output:

Val
a@xxxx.com b@xxxx.com

Expected Output

a@xxxx.com
b@xxxx.com

ie In two different rows.

How to do so?

Answer 1

Use this:

SELECT 
    node.value('.','varchar(100)') AS Val
FROM
    @x.nodes('/PartnerEmails/Email') AS PE(Node)    

Since you have multiple nodes inside <PartnerEmails> , you need to use the .nodes() function to create an "inline" table of XML fragments - each "row" in that table contains one <Email> node which you can then query on (and extract the contents of the XML node).

Answer 2

DECLARE @x XML
SELECT @x = '<PartnerEmails>
<Email>a@xxxx.com</Email>
<Email>b@xxxx.com</Email>
</PartnerEmails>'

SELECT   ColumnValue.value('.','varchar(1000)')  as Val            
FROM @x.nodes('/PartnerEmails/Email') as Table1(ColumnValue)