Passing input fileName along the complete route

["shekher.awasthi" <sh...@gmail.com>]

Hi All,

While working i came across a requirements where i have to name the final
output file based on the input file
e.g
if the name of the input file is a.xml and i need to convert it to csv at
the end than the output file should be named as
 customoutput_a.xml.csv
like in my case i have the following route

<route id="demo">  
	    
		   <from uri="{{demo.filePath}}"/>
		   
		   <choice>
       			<when>
           			<simple>${file:ext} ==  'csv'</simple>
           			<unmarshal>
	                <csv/>
	         </unmarshal>
	         
	       	<to uri="bean:converter?method=process" />
	       	<to uri="file://data/output?fileName=tmpXML.xml" />	
      		  </when>
		   </choice>
		  	   	  
	       <to uri="{{demo.tmpXMLtoStandardXML}}"/>
	       <to uri="file://data/output?fileName=3_StandardXML.xml" />
	       <to uri="{{demo.StandardXMLToImpexXslFile}}"/>	       
	       <to uri="{{demo.targetXMLFile}}"/>   
	    </route>  

now in such case in the last route point i have to name the file based on
the input file name, can any one help me how i can pass the input file name
throughout the route in camel so that i can extract the file name when
producing final output

Thanks in advance


--
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Re: Passing input fileName along the complete route

[Claus Ibsen <cl...@gmail.com>]

Hi

See the file language (its part of the simple language)
http://camel.apache.org/file-language.html

So in the to endpoint you can use a simple expression to define the file name

For example something like:
<to uri="file:dir?fileName=something-${file:onlyname}"/>



On Thu, Aug 25, 2011 at 9:42 AM, shekher.awasthi
<sh...@gmail.com> wrote:
> Hi All,
>
> While working i came across a requirements where i have to name the final
> output file based on the input file
> e.g
> if the name of the input file is a.xml and i need to convert it to csv at
> the end than the output file should be named as
>  customoutput_a.xml.csv
> like in my case i have the following route
>
> <route id="demo">
>
>                   <from uri="{{demo.filePath}}"/>
>
>                   <choice>
>                        <when>
>                                <simple>${file:ext} ==  'csv'</simple>
>                                <unmarshal>
>                        <csv/>
>                 </unmarshal>
>
>                <to uri="bean:converter?method=process" />
>                <to uri="file://data/output?fileName=tmpXML.xml" />
>                  </when>
>                   </choice>
>
>               <to uri="{{demo.tmpXMLtoStandardXML}}"/>
>               <to uri="file://data/output?fileName=3_StandardXML.xml" />
>               <to uri="{{demo.StandardXMLToImpexXslFile}}"/>
>               <to uri="{{demo.targetXMLFile}}"/>
>            </route>
>
> now in such case in the last route point i have to name the file based on
> the input file name, can any one help me how i can pass the input file name
> throughout the route in camel so that i can extract the file name when
> producing final output
>
> Thanks in advance
>
>
> --
> View this message in context: http://camel.465427.n5.nabble.com/Passing-input-fileName-along-the-complete-route-tp4733437p4733437.html
> Sent from the Camel - Users mailing list archive at Nabble.com.
>



-- 
Claus Ibsen
-----------------
FuseSource
Email: cibsen@fusesource.com
Web: http://fusesource.com
Twitter: davsclaus, fusenews
Blog: http://davsclaus.blogspot.com/
Author of Camel in Action: http://www.manning.com/ibsen/