I'm just starting to learn XQuery and I want that it shows me the number of festivals with genre (genero) is the same as "metal". I can't get the total number of them, only separately.
Xquery
for $b in //festival
where $b/#genero="Metal"
return <prueba>{count ($b/#genero="Metal"), $b//nombre}</prueba>
XML
<Festivales>
<festival genero="Metal">
<informacion>
<nombre>Resurrection Fest</nombre>
<fecha_inicio>2020-07-01</fecha_inicio>
<fecha_fin>2020-07-04</fecha_fin>
</festival>
<festival genero="Rock-Heavy Metal">
<informacion>
<nombre>Rock the Night</nombre>
<fecha_inicio>2020-06-26</fecha_inicio>
<fecha_fin>2020-06-27</fecha_fin>
</festival>
<festival genero="Hardcore">
<informacion>
<nombre>Ieperfest</nombre>
<fecha_inicio>2020-07-03</fecha_inicio>
<fecha_fin>2020-07-05</fecha_fin>
</informacion>
</festival>
<festival genero="Metal">
<informacion>
<nombre>Download UK</nombre>
<fecha_inicio>2020-06-12</fecha_inicio>
<fecha_fin>2020-06-14</fecha_fin>
</informacion>
</festival>
</Festivales>
Result
<prueba>1<nombre>Resurrection Fest</nombre>
</prueba>
<prueba>1<nombre>Hellfest</nombre>
</prueba>
<prueba>1<nombre>Download UK</nombre>
</prueba>
Thanks!
for $b in //festival[#genero="Metal"]
let $n := $b/informacion/nombre/text()
return
<prueba>
{
<cnt>{count(//festival[#genero="Metal"]/informacion/nombre[. = $n])}</cnt>
, $b/informacion/nombre
}
</prueba>
Related
Hi I am trying to convert some xml to csv using xquery and found a previous post that helped me get to this point:
for $b in /root/Result
return
concat(escape-html-uri(string-join(($b/HolidayEndDate,
$b/HolidayType,
$b/FirstName,
$b/AllowanceRemainingDays,
$b/HolidayStartDate,
$b/EmployeeId,
$b/AllowanceDays,
$b/LastName,
$b/HolidayDurationDays
)
/normalize-space(),
",")
),
codepoints-to-string(10))
This returns all of the data as required but no Header row. Is there a simple addition to the above code that would also return the header row? Thanks. :)
Since your query returns a sequence of lines, you can just prepend another line before the FLWOR expression:
"HolidayEndDate,HolidayType,FirstName,AllowanceRemainingDays,HolidayStartDate,EmployeeId,AllowanceDays,LastName,HolidayDurationDays
",
for $b in /root/Result
return
concat(escape-html-uri(string-join(($b/HolidayEndDate,
$b/HolidayType,
$b/FirstName,
$b/AllowanceRemainingDays,
$b/HolidayStartDate,
$b/EmployeeId,
$b/AllowanceDays,
$b/LastName,
$b/HolidayDurationDays
)
/normalize-space(),
",")
),
codepoints-to-string(10))
Because nested sequences are flattened (i.e. concatenated) in XQuery, this results in one output sequence including the header. Note also that I used a character entity '
' for the newline character, which is much shorter than codepoints-to-string(10).
concat("HolidayEndDate,HolidayType,FirstName,AllowanceRemainingDays,HolidayStartDate,EmployeeId,AllowanceDays,LastName,HolidayDurationDays
",
string-join(
for $b in /root/Result
return
concat(escape-html-uri(string-join(($b/HolidayEndDate,
$b/HolidayType,
$b/FirstName,
$b/AllowanceRemainingDays,
$b/HolidayStartDate,
$b/EmployeeId,
$b/AllowanceDays,
$b/LastName,
$b/HolidayDurationDays
)
/normalize-space(),
",")
),
codepoints-to-string(10)), "")
)
"some" is not a special term which makes the googling seem to just ignore that search.
What I am asking is in my learning below:
b.collect:
Array[(Int, String)] = Array((3,dog), (6,salmon), (3,rat), (8,elephant))
d.collect:
Array[(Int, String)] = Array((3,dog), (3,cat), (6,salmon), (6,rabbit), (4,wolf), (7,penguin))
if I do some join and then collect the result, like b.join(d).collect, I will get the following:
Array[(Int, (String, String))] = Array((6,(salmon,salmon)), (6,(salmon,rabbit)), (3,(dog,dog)), (3,(dog,cat)), (3,(rat,dog)), (3,(rat,cat)))
which seems understandable, however, if I do: b.leftOuterJoin(d).collect, I will get:
Array[(Int, (String, Option[String]))] = Array((6,(salmon,Some(salmon))), (6,(salmon,Some(rabbit))), (3,(dog,Some(dog))), (3,(dog,Some(cat))), (3,(rat,Some(dog))), (3,(rat,Some(cat))), (8,(elephant,None)))
My question is why do I get results seems to be expressed differently, I mean why the second result contains "Some"? what's the difference between with "Some" and without "Some"? Can "Some" be removed? Does "Some" have any impact to any later operations as the content of RDD?
Thank you very much.
When you do the normal join as b.join(d).collect, you get Array[(Int, (String, String))]
This is because of only the same key with RDD b and RDD d so it is always guaranteed to have a value so it returns Array[(Int, (String, String))].
But when you use b.leftOuterJoin(d).collect the return type is Array[(Int, (String, Option[String]))] this is because to handle the null. In leftOuterJoin, there is no guarantee that all the keys of RDD b are available in RDD d, So it is returned as Option[String] which contains two values
Some(String) =>If the key is matched in both RDD
None If the key is present in b and not present in d
You can replace Some by getting the value from it and providing the value in case of None as below.
val z = b.leftOuterJoin(d).map(x => (x._1, (x._2._1, x._2._2.getOrElse("")))).collect
Now you should get Array[(Int, (String, String))] and output as
Array((6,(salmon,salmon)), (6,(salmon,rabbit)), (3,(dog,dog)), (3,(dog,cat)), (3,(rat,dog)), (3,(rat,Some(cat)), (8,(elephant,)))
Where you can replace "" with any other string as you require.
Hope this helps.
declare variable $fb := doc("factbook.xml")/mondial;
for $c in $fb//country
where ($c/encompassed/#continent = 'f0_119') and ($c/#population < 100000)
return concat('Country: ',$c/name, ', Population: ',$c/#population);
it returns:
Type Error: Type of value '
()
' does not match sequence type: xs:anyAtomicType?
At characters 11681-11698
At File "q2_3.xq", line 4, characters 13-67
At File "q2_3.xq", line 4, characters 13-67
At File "q2_3.xq", line 4, characters 13-67
however, if i do not do a concat return, just name or population it will work, and most strange thing is i have another program :
declare variable $fb := doc("factbook.xml")/mondial;
for $c in $fb//country
where $c/religions = 'Seventh-Day Adventist'
order by $c/name
return concat('Country: ',$c/name, ', Population: ',$c/#population);
The return syntax is exactly same, however, it works.
Why this happens?
Without seeing an example of your data it's impossible to say for sure, but if $c/name returns more than one value, then your error would make sense. Do you have any results where there are more than one name element?
I have a problem nesting the result tags in each other the right way.
The result should look like this:
aimed result
<categoryA>
<position>...</position>
<position>...</position>
...
</categoryA>
<categoryB>
<position>...</position>
<position>...</position>
...
</categoryB>
currently I have only managed to get the right results for the positions, the categoryA and B are 1 hierarchic layer higher than the positions. the positions should be nested in the categories. The categories can be referenced by let $y := $d/Bilanz/Aktiva/* (respectively $d$d/Bilanz/Aktiva/LangfristigesVermoegen and $d$d/Bilanz/Aktiva/KurzfristigesVermoegen).
Here is my query:
query
let $d := doc('http://etutor.dke.uni-linz.ac.at/etutor/XML?id=5001')/Bilanzen
let $a02 := $d/Bilanz[#jahr='2002']/Aktiva/*
let $a03 := $d/Bilanz[#jahr='2003']/Aktiva/*
for $n02 in $a02//* , $n03 in $a03//*
(:
where name($n02) = name($n03)
where node-name($n02) = node-name($n03)
:)
where name($n02) = name($n03)
return <position name="{node-name($n02)}">
<j2002>{data($n02/#summe)}</j2002>
<j2003>{data($n03/#summe)}</j2003>
<diff>{data($n03/#summe) - data($n02/#summe)}</diff>
</position>
xml
<Bilanzen>
<Bilanz jahr="2002">
<Aktiva>
<LangfristigesVermoegen>
<Sachanlagen summe="1486575.8"/>
<ImmateriellesVermoegen summe="67767.2"/>
<AssoziierteUnternehmen summe="190826.3"/>
<AndereBeteiligungen summe="507692.7"/>
<Uebrige summe="92916.4"/>
</LangfristigesVermoegen>
<KurzfristigesVermoegen>
<Vorraete summe="78830.9"/>
<Forderungen summe="198210.3"/>
<Finanzmittel summe="181102.0"/>
</KurzfristigesVermoegen>
</Aktiva>
<Passiva>
<Eigenkapital>
<Grundkapital summe="91072.4"/>
<Kapitalruecklagen summe="186789.5"/>
<Gewinnruecklagen summe="798176.2"/>
<Bewertungsruecklagen summe="-34922.4"/>
<Waehrungsumrechnung summe="0"/>
<EigeneAktien summe="0"/>
</Eigenkapital>
<AnteileGesellschafter summe="23613.1"/>
<LangfristigeVerb>
<Finanzverbindlichkeiten summe="680007.1"/>
<Steuern summe="36555.8"/>
<Rueckstellungen summe="429286.1"/>
<Baukostenzuschuesse summe="169246.0"/>
<Uebrige summe="36166.9"/>
</LangfristigeVerb>
<KurzfristigeVerb>
<Finanzverbindlichkeiten summe="14614.6"/>
<Steuern summe="65247.6"/>
<Lieferanten summe="94939.2"/>
<Rueckstellungen summe="123664.8"/>
<Uebrige summe="89464.8"/>
</KurzfristigeVerb>
</Passiva>
</Bilanz>
<Bilanz jahr="2003">
<Aktiva>
<LangfristigesVermoegen>
<Sachanlagen summe="1590313.7"/>
<ImmateriellesVermoegen summe="69693.2"/>
<AssoziierteUnternehmen summe="198224.7"/>
<AndereBeteiligungen summe="418489.3"/>
<Uebrige summe="104566.7"/>
</LangfristigesVermoegen>
<KurzfristigesVermoegen>
<Vorraete summe="20609.8"/>
<Forderungen summe="289458.5"/>
<Finanzmittel summe="302445.9"/>
</KurzfristigesVermoegen>
</Aktiva>
<Passiva>
<Eigenkapital>
<Grundkapital summe="91072.4"/>
<Kapitalruecklagen summe="186789.5"/>
<Gewinnruecklagen summe="875723.4"/>
<Bewertungsruecklagen summe="-15459.5"/>
<Waehrungsumrechnung summe="-633.7"/>
<EigeneAktien summe="0"/>
</Eigenkapital>
<AnteileGesellschafter summe="22669.8"/>
<LangfristigeVerb>
<Finanzverbindlichkeiten summe="733990.2"/>
<Steuern summe="68156.8"/>
<Rueckstellungen summe="395997.2"/>
<Baukostenzuschuesse summe="177338.5"/>
<Uebrige summe="38064.9"/>
</LangfristigeVerb>
<KurzfristigeVerb>
<Finanzverbindlichkeiten summe="6634.7"/>
<Steuern summe="97119.1"/>
<Lieferanten summe="89606.0"/>
<Rueckstellungen summe="128237.5"/>
<Uebrige summe="98495.2"/>
</KurzfristigeVerb>
</Passiva>
</Bilanz>
</Bilanzen>
I would really appreciate some help, i have no clue at all. Thank you.
If I understand you correctly, you want the information about LangfristigesVermoegen (and its children) to be grouped in the output under element categoryA, and the information about Kurzfristigesvermoegen to be grouped under categoryB.
So you will want first of all to do something to generate the categoryA and categoryB elements. For example,
let $d := doc(...)/Bilanzen
return (
<categoryA>{ ... children of category A here ... }</categoryA>,
<categoryB>{ ... children of category B here ... }</categoryB>
)
The positions in each category can be generated using code similar to what you've now got, except that instead of iterating over
for $n02 in $a02//* , $n03 in $a03//*
you will need to iterate over $a02[self::LangfristigesVermoegen]/* for category A, and over $a02[self::KurzfristigesVermoegen]/* for category B (and similarly, of course, for $n02 and $n03).
If the set of categories is not static and you just want to group things in the output using the same grouping elements present in the input, then you'll want an outer structure something like this:
for $assetclass1 in $anno2002/*
let $assetclass2 := $anno2003/*[name() = name($assetclass1)]
return
(element {name($assetclass1)} {
for $old in $assetclass1/*,
$new in $assetclass2/*
where name($old) eq name($new)
return <position name="{node-name($old)}">
<j2002>{data($old/#summe)}</j2002>
<j2003>{data($new/#summe)}</j2003>
<diff>{data($new/#summe) - data($old/#summe)}</diff>
</position>
})
Sorry for the title, I didnt know what's the best title for this question
I'm having difficulties with XQUERY. The result is not what I expected
here's the xml
<Equipes>
<Equipe>
<equipeId>1</equipeId>
<equipeNom>Equipe A</equipeNom>
<JoueurEquipe>
<dateDebut>2011-01-01</dateDebut>
<dateFin>2013-01-01</dateFin>
<numero>1</numero>
<joueurId>1</joueurId>
</JoueurEquipe>
<JoueurEquipe>
<dateDebut>2010-01-01</dateDebut>
<dateFin>2012-01-01</dateFin>
<numero>2</numero>
<joueurId>2</joueurId>
</JoueurEquipe>
</Equipe>
<Equipe>
<equipeId>2</equipeId>
<equipeNom>Equipe B</equipeNom>
<JoueurEquipe>
<dateDebut>2009-01-01</dateDebut>
<dateFin>2012-01-01</dateFin>
<numero>1</numero>
<joueurId>3</joueurId>
</JoueurEquipe>
<JoueurEquipe>
<dateDebut>2010-01-01</dateDebut>
<dateFin>2014-01-01</dateFin>
<numero>2</numero>
<joueurId>4</joueurId>
</JoueurEquipe>
</Equipe>
</Equipes>
and here's the query
for $b in doc("ligue.xml")/ligue/Equipes
return
<Equipes>
{
$b/Equipe/equipeId,
$b/Equipe/equipeNom
}
</Equipes>
the result is
<equipes>
<equipeId>1</equipeId>
<equipeId>2</equipeId>
<equipeNom>Equipe A</equipeNom>
<equipeNom>Equipe B</equipeNom>
</equipes>
what I need is
<equipes>
<equipeId>1</equipeId>
<equipeNom>Equipe A</equipeNom>
<equipeId>2</equipeId>
<equipeNom>Equipe B</equipeNom>
</equipes>
I don't know what i'm missing
Thank you
Iterate over the things you actually want your output to iterate over:
<Equipes>{
for $b in doc("ligue.xml")/ligue/Equipes/Equipe[equipeId][equipeNom]
return ($b/equipeId, $b/equipeNom)
}</Equipes>
Now, let's explore why your prior query behaved as it did. This:
for $b in doc("ligue.xml")/ligue/Equipes
return
<Equipes>
{
$b/Equipe/equipeId,
$b/Equipe/equipeNom
}
</Equipes>
...finds the Equipes element, of which there's only one (making the for loop useless). It then searches for $b/Equipe/equipeId and $b/Equipe/equipeNom, both of which evaluate to lists containing multiple elements, and concatenates those two lists to get the output.