CODEWITHSUNDEEP
xpathxquerymarklogic
Aswanikumar
Aswanikumar

Reputation: 115

how to iterate dynamically over child elements using for in xquery

I have a XML file which contains multiple XML documents. In large XML document each document has a root node and it is common for all the XML documents in that file. However the child elements inside the root name vary. I have to write a for loop to iterate dynamically each and every child node and have to show it on a table format. Here I have pasted my XML format below. I have multiple xml files in that way but the root node is common for every file. I want to map each child element as table row and data as table column. child elements numbers would vary

<source>
  <empname>john </empname>
  <empid>25825   </empid>
  <salary> 20000 </salary>
  <dob> 12-08-1993</dob>
</source>
<source>
  <empname>joe</empname>
  <empid>25826</empid>
  <salary>20000</salary>
  <dob>12-07-1993</dob>
<source>

<source>
  <emptype>developer</emptype>
  <address>3155 </address>
  <mobile>58258365</mobile>
</source>
<source>
  <emptype>analyst</emptype>
  <address>3155 </address>
  <mobile>58258365</mobile>
</source>

<table>
  <th>  empname    empid   salary   dob </th>
  <td>     john       25825  60000    date</td>
  <td>     joe        25826  70000    date</td>
</table>

Every time the <th> and <td> is changed and the common thing is root element source code. Can anyone help me out to iterate those elements over loop and represent those data over table format?

Upvotes: 0

Views: 5268

Answers (2)

grtjn
grtjn

Reputation: 20414

You could try this, but the approach of distinct-values doesn't scale well. You best determine the columns upfront, and hardcode like the solution from Tavolo:

let $xml := 
  <sources>
    <source>
      <empname>john </empname>
      <empid>25825   </empid>
      <salary> 20000 </salary>
      <dob> 12-08-1993</dob>
    </source>
    <source>
      <empname>joe</empname>
      <empid>25826</empid>
      <salary>20000</salary>
      <dob>12-07-1993</dob>
    </source>
    <source>
      <emptype>developer</emptype>
      <address>3155 </address>
      <mobile>58258365</mobile>
    </source>
    <source>
      <emptype>analyst</emptype>
      <address>3155 </address>
      <mobile>58258365</mobile>
    </source>
  </sources>
let $sources := $xml/source
let $labels := fn:distinct-values($sources/*/fn:node-name(.))
return
  <table>
    <tr>{
      for $label in $labels
      return <th>{ $label }</th>
    }</tr>
    {
      for $source in $sources
      return <tr>{    
        for $label in $labels
        return <th>{ fn:data($source/*[fn:node-name() = $label]) }</th>
      }</tr>
    }
  </table>

HTH!

Upvotes: 1

TavoloPerUno
TavoloPerUno

Reputation: 579

I do not fully understand your statement. Do you want to be able to combine all such XML files? How about this?

<table>
<tr><th>empname</th><th>empid</th><th>salary</th><th>dob</th>   <th>emptype</th></tr>
{for $child in doc("your-file")/source) 
return <tr>{    
return <td>{$child/empname}</td><td>{$child/empid}</td><td>{$child/salary}</td><td>{$child/dob}</td><td>{$child/emptype}</td>
}</tr></table>

Upvotes: 1

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