xpst0003 xquery 구문 오류

Question

x 쿼리 코드를 통해 webpage을 쿼리하고 싶습니다. 도와주세요. 그리고 다음 오류가 발생합니다 : XPST0003 : XQuery 구문 오류가 # ... // json // sentences // trans); # : 예상 "return", found ";".

<?xml version="1.0" encoding="UTF-8"?> 
<config charset="UTF-8"> 
<var-def name="scrappedContent"> 
<xquery> 
<xq-param name="doc"> 
    <html-to-xml outputtype="browser-compact" prunetags="yes"> 
     <http url="${url}"/> 
    </html-to-xml> 
    </xq-param> 

    <xq-expression><![CDATA[ 
    declare variable $doc as node() external; 
    let $transl := data($doc//query//results//json//sentences//trans); 
    let $translitl := data($doc//query//results//json//sentences//translit);  

    let $data := data($doc//div[@id="defId"]) 
    return 
     <myContent> 
      <transl>{$transl}</transl> 
      <translitl>{$translitl}</translitl> 
      <data>{$data}</data> 
     </myContent> 
    ]]> 
</xq-expression> 
</xquery> 
</var-def> 
</config> 

Answer 1

교체 :와

let $transl := data($doc//query//results//json//sentences//trans); 
let $translitl := data($doc//query//results//json//sentences//translit);  

:

let $transl := data($doc//query//results//json//sentences//trans) 
let $translitl := data($doc//query//results//json//sentences//translit) 

일을 개선해야한다 (하자 후 없음 세미콜론)

..

HTH!