How to print exception payload within mule message flow

Question

I have a mule flow within which I am logging the entire payload in String format by following code snippet

<logger level="ERROR" message="#[payload:java.lang.String]"/>

Now if the error occurs there really is no need to print the entire payload. The payload object in the message is null and the exception payload is populated with the relevant exception in it.

If I can print just the exception payload in String format, that would suffice. Does any one know how to log the exception payload from the message ?

you have received two answers below, care to accept one? If not, comment why. — David Dossot, Feb 27 '15 at 15:02

score 8 · Accepted Answer · edited Apr 26 '17 at 16:16

8

    <logger level="ERROR" message=" 
#[groovy:message.getExceptionPayload().getRootException().getMessage()]" />

The above code extracts the message related to the cause of exception.

edited Apr 26 '17 at 16:16

Chanandler Bong

403
1
11
23

answered Feb 09 '14 at 10:14

tortoise

613
1
7
19

score 3 · Answer 2 · answered Feb 06 '14 at 20:47

3

Quite simply:

<logger level="ERROR" message="#[exception]"/>

answered Feb 06 '14 at 20:47

Anton Kupias

3,945
3
16
20

score 2 · Answer 3 · edited Jul 19 '16 at 13:41

2

small correction in the expression use:

[groovy:message.getExceptionPayload().getRootException().getMessage()]

edited Jul 19 '16 at 13:41

user987339

10,519
8
40
45

answered Apr 04 '16 at 13:23

Anil

111
6

score 0 · Answer 4 · answered Jan 24 '19 at 16:29

If you want to save error into payload you need to transform it:

%dw 2.0
output application/json
---
{
    error: error.detailedDescription,
    errorType: (error.errorType.namespace default '') ++ ":" ++ (error.errorType.identifier default '') ,
    recoverable: false
}

So, into payload you have your error

How to print exception payload within mule message flow

4 Answers4