Your expressions does not look correct jsonpath($(@.kind == 'full'). I
think that it should be .jsonpath($.@.kind == 'full')
On Thu, Aug 28, 2014 at 5:03 AM, Deven Phillips deven.phill...@gmail.com
wrote:
Hi all,
As an example, I was using a simple JSON document:
{
kind: full,
I tried that and I get:
com.jayway.jsonpath.PathNotFoundException: Path '@' not found in the
current context:
{kind:full}
I have tried a number of different permutations for that filter string, but
none have worked so far. Here are a few examples I have tried:
$(@.kind == 'full')
$.kind(@ eq
Raised a question to jsonpath google group as I don't know :
https://groups.google.com/forum/#!topic/jsonpath/MUNwbQ2UjTk
Unfortunately their code didn't cover this test case ;-)
On Thu, Aug 28, 2014 at 11:57 AM, Deven Phillips deven.phill...@gmail.com
wrote:
I tried that and I get:
Yeah, I looked for an example similar to my use case in their wiki, unit
tests, etc... Thanks for asking them though!
Deven
On Thu, Aug 28, 2014 at 7:04 AM, Charles Moulliard ch0...@gmail.com wrote:
Raised a question to jsonpath google group as I don't know :
That will work if you build your choice expression like this
from(websocket://0.0.0.0:8080/replication)
.choice()
.when()
.when(PredicateBuilder.isEqualTo(ExpressionBuilder.languageExpression(jsonpath,$.kind),ExpressionBuilder.constantExpression(full)))
where json content is
{
Hi all,
As an example, I was using a simple JSON document:
{
kind: full,
type: customer
}
I wanted to use a jsonpath predictate in a choice route as shown below:
from(websocket://0.0.0.0:8080/replication)
.choice()
.when()
.jsonpath($(@.kind == 'full')
