How to get this output with XQuery? [duplicate]

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Closed 9 years ago.

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XQuery returning an error..?

Below is the XML file -

<Continents>
  <Continent name="Asia">
    <Country name="Japan">
    <City name="Tokyo"><Desc>Tokoyo is a most populated City</Desc></City>
    <City name="Horishima"><Desc>What to say.. Faced the destruction due to Atom Bomb</Desc></City>
    </Country>
    <Country name="India">
    <City name="New Delhi"><Desc>Capital of India</Desc></City>
    <City name="Mumbai"><Desc>Financial Capital of India</Desc></City>
    <City name="Lucknow"><Desc>City of Nawabs</Desc></City>
    </Country>
  </Continent>  
</Continents>

I want to list Cities for Country="India"

My XQuery FLWOR code is -

for $x in doc("Continent")/Continents/Continent
  where $x/Country/@name='India'
  return $x/Country/City/@name

I am wishing output as -

 name="New Delhi" name="Mumbai" name="Lucknow"

but getting output as -

 name="Tokyo" name="Horishima" name="New Delhi" name="Mumbai" name="Lucknow"

Can anybody help me to get correct output? Also how to get it on separate line?

Answer 1

You asked almost the exact same question here and the same answer applies here, too. What you want is:

doc("Continent")/Continents/Continent/Country[@name = 'India']/City/@name

If you want each result on its own line, try this:

string-join(
  doc("Continent")/Continents/Continent/Country[@name = 'India']/City/@name,
  '
'
)

This should result in:

New Delhi
Mumbai
Lucknow

Answer 2

You could also rewrite your FLWOR a bit to get the wanted results:

for $x in doc("Continent")/Continents/Continent/Country
where $x/@name='India'
return $x/City/@name

(Note that I move the /Country part)

If necessary you can wrap this FLWOR in a string-join as suggested by Leo the same way he wraps his XPath expression alternative..

Answer 3

/*/*/Country[@name='India']/City/concat('name="', @name, '" ')

Note: This also happens to be a pure XPath 2.0 expression.