XQuery APIs -- how do you identify the XML source

Discussion in 'XML' started by Benjamin G. Jones, Jan 8, 2004.

  1. I am having a very basic problem with XQuery.

    I want to use an XQuery API in Java (either Saxon or Qexo), and I have
    an XQuery expression that works as expeced from the command line if I
    specify a filename before the xpath expression.

    What I want to know is this: If I want to use
    Qexo's gnu.xquery.lang.XQuery.eval() or
    Saxon's QueryExpression.evaluate(),
    how do I specifiy the input in the Xquery statement?
    For line 3 of the xquery, I want to say something analagous to:

    let $vars=string({xml string goes here})//ci/text()


    let $vars=input()//ci/text()

    to read whatever the argument of eval is.
    but neither of these seem to work.

    Please please help!


    (Expressions appear below).
    ------------ start xquery ---------------------------
    1 <varlist>
    2 {
    3 let $vars:=document("equation.mml")//ci/text()
    4 for $var in $vars
    5 return
    6 <var>{$var}</var>
    7 }
    8 </varlist>

    ------------ end xquery ---------------------------

    on the xml file:

    ------------ start xml "equation.mml" ---------------------------

    ------------ end xml equation.mml--------------------------

    to produce (as expected)

    -------------- begin output ------------------------------

    -------------- end output --------------------------------
    + Integre Technical Publishing, inc. +
    + http://www.integretechpub.com +
    + (505)889-8189 +
    Benjamin G. Jones, Jan 8, 2004
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