+6Hi FME'ers,
I am trying to update values in an XML documents but these values are conditional. For example:
IF <MethodCode> like 'FF%' THEN <MethodCode> = 'PM'
and
IF <MethodCode> = 'FF12' AND '<ItemDesc> = 'Basket' THEN <ItemDesc> = 'Wicker Basket'. ELSEIF <MethodCode> = 'FF10' AND '<ItemDesc> = 'Basket' THEN <ItemDesc> = 'Laundry Basket'
I think XQUERY will solve my issue but I don't have time to learn a new query language.
Many thanks,
David
Yes, you can do that with an XQuery expression (XMLXQueryUpdater). I think the following example contains every syntax you have to learn for the present.
Source XML
<items>
<item>
<code>X1</code>
<desc>aaa</desc>
</item>
<item>
<code>X2</code>
<desc>aaa</desc>
</item>
<item>
<code>X2</code>
<desc>bbb</desc>
</item>
<item>
<code>Z3</code>
<desc>aaa</desc>
</item>
</items>
XQuery Expression (assume that an attribute storing the source XML is set to the XML Attribute parameter in the XMLXQueryUpdater)
for $item in //item
let $code := data($item/code)
let $desc := data($item/desc)
return
if (matches($code, '^X.*')) then (
replace value of node $item/code with 'KK',
if ($code = 'X1' and $desc = 'aaa') then
replace value of node $item/desc with 'foo aaa'
else if ($code ='X2' and $desc = 'aaa') then
replace value of node $item/desc with 'bar aaa'
else ()
)
else ()
Result
<items>
<item>
<code>KK</code>
<desc>foo aaa</desc>
</item>
<item>
<code>KK</code>
<desc>bar aaa</desc>
</item>
<item>
<code>KK</code>
<desc>bbb</desc>
</item>
<item>
<code>Z3</code>
<desc>aaa</desc>
</item>
</items>
If you need more specific solution, please post a small (but valid) sample XML and your desired result corresponding to the sample.
+6Yes, you can do that with an XQuery expression (XMLXQueryUpdater). I think the following example contains every syntax you have to learn for the present.
Source XML
<items>
<item>
<code>X1</code>
<desc>aaa</desc>
</item>
<item>
<code>X2</code>
<desc>aaa</desc>
</item>
<item>
<code>X2</code>
<desc>bbb</desc>
</item>
<item>
<code>Z3</code>
<desc>aaa</desc>
</item>
</items>
XQuery Expression (assume that an attribute storing the source XML is set to the XML Attribute parameter in the XMLXQueryUpdater)
for $item in //item
let $code := data($item/code)
let $desc := data($item/desc)
return
if (matches($code, '^X.*')) then (
replace value of node $item/code with 'KK',
if ($code = 'X1' and $desc = 'aaa') then
replace value of node $item/desc with 'foo aaa'
else if ($code ='X2' and $desc = 'aaa') then
replace value of node $item/desc with 'bar aaa'
else ()
)
else ()
Result
<items>
<item>
<code>KK</code>
<desc>foo aaa</desc>
</item>
<item>
<code>KK</code>
<desc>bar aaa</desc>
</item>
<item>
<code>KK</code>
<desc>bbb</desc>
</item>
<item>
<code>Z3</code>
<desc>aaa</desc>
</item>
</items>
If you need more specific solution, please post a small (but valid) sample XML and your desired result corresponding to the sample.
In was very close with my own attempt. Sorry to throw this in but the xml has a namespace. Do I have to declare it and therefore prefix the nodes with //x: (for example)
Yes, you can do that with an XQuery expression (XMLXQueryUpdater). I think the following example contains every syntax you have to learn for the present.
Source XML
<items>
<item>
<code>X1</code>
<desc>aaa</desc>
</item>
<item>
<code>X2</code>
<desc>aaa</desc>
</item>
<item>
<code>X2</code>
<desc>bbb</desc>
</item>
<item>
<code>Z3</code>
<desc>aaa</desc>
</item>
</items>
XQuery Expression (assume that an attribute storing the source XML is set to the XML Attribute parameter in the XMLXQueryUpdater)
for $item in //item
let $code := data($item/code)
let $desc := data($item/desc)
return
if (matches($code, '^X.*')) then (
replace value of node $item/code with 'KK',
if ($code = 'X1' and $desc = 'aaa') then
replace value of node $item/desc with 'foo aaa'
else if ($code ='X2' and $desc = 'aaa') then
replace value of node $item/desc with 'bar aaa'
else ()
)
else ()
Result
<items>
<item>
<code>KK</code>
<desc>foo aaa</desc>
</item>
<item>
<code>KK</code>
<desc>bar aaa</desc>
</item>
<item>
<code>KK</code>
<desc>bbb</desc>
</item>
<item>
<code>Z3</code>
<desc>aaa</desc>
</item>
</items>
If you need more specific solution, please post a small (but valid) sample XML and your desired result corresponding to the sample.
declare namespace <prefix> = "<namespace name>";Then, in the expression body, qualify all element names belonging to the namespace by the prefix.
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