问题
Below is the XML part of my data.
<A>
<a><Type>Fruit</Type><Name>Banana</Name></a>
<a><Type>Fruit</Type><Name>Orange</Name></a>
<a><Type>Fruit</Type><Name>Apple</Name></a>
<a><Type>Fruit</Type><Name>Lemon</Name></a>
<a><Type>Cars</Type><Name>Toyota</Name></a>
<a><Type>Cars</Type><Name>Lamborghini</Name></a>
<a><Type>Cars</Type><Name>Renault</Name></a>
</A>
Out put as -
<a>Fruits-Banana,Orange,Apple,Lemon</a>
<a>Cars-Toyota,Lamborghini,Renault</a>
I tried to get the required output by all in vain. I tried 'group by` clause too, but getting errors.
any help?
回答1:
let $x:=
<A>
<a><Type>Fruit</Type><Name>Banana</Name></a>
<a><Type>Fruit</Type><Name>Orange</Name></a>
<a><Type>Fruit</Type><Name>Apple</Name></a>
<a><Type>Fruit</Type><Name>Lemon</Name></a>
<a><Type>Cars</Type><Name>Toyota</Name></a>
<a><Type>Cars</Type><Name>Lamborghini</Name></a>
<a><Type>Cars</Type><Name>Renault</Name></a>
</A>
for $z in distinct-values($x//a/Type)
let $c := $x//a[Type=$z]/Name
return
<a>{concat($z, "-", string-join($c, ","))}</a>
First for
is taking the distinct values of the tag Type
, then for each distinct value of this, the respective values of all the Name
tags are derived.
Then using the concat
function I have concatenated the Type
text with the string generated by string-join
, used to add/append the Name
and ,
(comma).
HTH :)
来源:https://stackoverflow.com/questions/21284293/xquery-group-by-on-2-tags