Hi ashutoshanshu,
thanks a lot, your syntax works. I realy don't know why I haven't found this solution, maybe it was to late for me yesterday... ;-)
Do you have also a clue, how to do this in a xpath or xqerry?
If I try this like this:
replace(/*,"\\s+","")
then my XML structure is away. (in a Property Transfer Step)
input: <node>test1<subnode>test2</subnode>test3</node> output: test1 test2 test3
best regards,
Shamane2000
You should know that replace() or replaceAll() method takes only 2 arguments within quotes like this :- replace("", "") This is not the correct way of using this method :- replace(/*,"\\s+", "") and by the way, what is your input String and what is your expected output?
@ashutoshanshu wrote:...and by the way, what is your input String and what is your expected output?
I have a lot of properties, containing xml parts (with linebreaks) and I want to linearize them for using in a *.csv file.And save csv with aditionally linebreaks are ... :smileymad:
And because all of my tries with xpath and xqerry were unsucsessfully, I trieed it now with a grovy script.
Provide me a basic XML. I will try to linearize it.
for example:
<node>
test1
<subnode>
test2
</subnode>
test3
</node>what I want is:
<node>test1<subnode>test2</subnode>test3</node>
Thanks in advance,
Shamane2000