I want to query a webpage through following x query code. please help me. And it gives me following errors: XPST0003: XQuery syntax error in #...//json//sentences//trans); let#: expected "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>
Replace:
let $transl := data($doc//query//results//json//sentences//trans);
let $translitl := data($doc//query//results//json//sentences//translit);
With:
let $transl := data($doc//query//results//json//sentences//trans)
let $translitl := data($doc//query//results//json//sentences//translit)
(No semi-colon after let's)
That should improve things..
HTH!