Spring Integration Java DSL: How to route with the channelMapping method to the channel which name is in the headers?
How to route with the channelMapping method to the channel which name is in the headers? So if I try this
@Bean
private IntegrationFlow postDataToChannelX()
return f -> f
...
.<String, Boolean> route(s -> s.equals(""), m -> m
.channelMapping(false, "headers['channelName']")
.channleMapping(true, ...);
there comes
Caused by: org.springframework.messaging.core.DestinationResolutionException: failed to look up MessageChannel with name 'headers['channelName']' in the BeanFactory.; nested exception is org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'headers['channelName']' available
spring-integration spring-integration-dsl
asked Nov 12 '18 at 5:59
MikeMike
12010
add a comment |
How to route with the channelMapping method to the channel which name is in the headers? So if I try this
@Bean
private IntegrationFlow postDataToChannelX()
return f -> f
...
.<String, Boolean> route(s -> s.equals(""), m -> m
.channelMapping(false, "headers['channelName']")
.channleMapping(true, ...);
there comes
Caused by: org.springframework.messaging.core.DestinationResolutionException: failed to look up MessageChannel with name 'headers['channelName']' in the BeanFactory.; nested exception is org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'headers['channelName']' available
spring-integration spring-integration-dsl
asked Nov 12 '18 at 5:59
MikeMike
12010
add a comment |
How to route with the channelMapping method to the channel which name is in the headers? So if I try this
@Bean
private IntegrationFlow postDataToChannelX()
return f -> f
...
.<String, Boolean> route(s -> s.equals(""), m -> m
.channelMapping(false, "headers['channelName']")
.channleMapping(true, ...);
there comes
Caused by: org.springframework.messaging.core.DestinationResolutionException: failed to look up MessageChannel with name 'headers['channelName']' in the BeanFactory.; nested exception is org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'headers['channelName']' available
spring-integration spring-integration-dsl
asked Nov 12 '18 at 5:59
MikeMike
12010
How to route with the channelMapping method to the channel which name is in the headers? So if I try this
@Bean
private IntegrationFlow postDataToChannelX()
return f -> f
...
.<String, Boolean> route(s -> s.equals(""), m -> m
.channelMapping(false, "headers['channelName']")
.channleMapping(true, ...);
there comes
Caused by: org.springframework.messaging.core.DestinationResolutionException: failed to look up MessageChannel with name 'headers['channelName']' in the BeanFactory.; nested exception is org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'headers['channelName']' available
spring-integration spring-integration-dsl
spring-integration spring-integration-dsl
asked Nov 12 '18 at 5:59
MikeMike
12010
asked Nov 12 '18 at 5:59
MikeMike
12010
edited Nov 13 '18 at 10:42
Mike
asked Nov 12 '18 at 5:59
MikeMike
12010
asked Nov 12 '18 at 5:59
MikeMike
12010
asked Nov 12 '18 at 5:59
MikeMike
12010
12010
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You can just do like this:
.route(Message.class, (m) -> m.getHeaders().get("channelName"))
So, you don't need any mapping at all since you resolve to the target channel directly in the routing function.
answered Nov 12 '18 at 14:28
Artem BilanArtem Bilan
64.3k84668
Edited the question. There was the route before thechannelMapping, so I think I need thechannelMapping. So how to route with the channelMapping method to the channel which name is in the headers?
– Mike
Nov 13 '18 at 10:46
Ok. You can include that into the routing function as well. I mean you do ternary operator and return the channel you need. The mapping is only for static channels
– Artem Bilan
Nov 13 '18 at 12:29
Thanks for the answer
– Mike
Nov 19 '18 at 11:54
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can just do like this:
.route(Message.class, (m) -> m.getHeaders().get("channelName"))
So, you don't need any mapping at all since you resolve to the target channel directly in the routing function.
answered Nov 12 '18 at 14:28
Artem BilanArtem Bilan
64.3k84668
Edited the question. There was the route before thechannelMapping, so I think I need thechannelMapping. So how to route with the channelMapping method to the channel which name is in the headers?
– Mike
Nov 13 '18 at 10:46
Ok. You can include that into the routing function as well. I mean you do ternary operator and return the channel you need. The mapping is only for static channels
– Artem Bilan
Nov 13 '18 at 12:29
Thanks for the answer
– Mike
Nov 19 '18 at 11:54
add a comment |
You can just do like this:
.route(Message.class, (m) -> m.getHeaders().get("channelName"))
So, you don't need any mapping at all since you resolve to the target channel directly in the routing function.
answered Nov 12 '18 at 14:28
Artem BilanArtem Bilan
64.3k84668
Edited the question. There was the route before thechannelMapping, so I think I need thechannelMapping. So how to route with the channelMapping method to the channel which name is in the headers?
– Mike
Nov 13 '18 at 10:46
Ok. You can include that into the routing function as well. I mean you do ternary operator and return the channel you need. The mapping is only for static channels
– Artem Bilan
Nov 13 '18 at 12:29
Thanks for the answer
– Mike
Nov 19 '18 at 11:54
add a comment |
You can just do like this:
.route(Message.class, (m) -> m.getHeaders().get("channelName"))
So, you don't need any mapping at all since you resolve to the target channel directly in the routing function.
answered Nov 12 '18 at 14:28
Artem BilanArtem Bilan
64.3k84668
You can just do like this:
.route(Message.class, (m) -> m.getHeaders().get("channelName"))
So, you don't need any mapping at all since you resolve to the target channel directly in the routing function.
answered Nov 12 '18 at 14:28
Artem BilanArtem Bilan
64.3k84668
answered Nov 12 '18 at 14:28
Artem BilanArtem Bilan
64.3k84668
answered Nov 12 '18 at 14:28
Artem BilanArtem Bilan
64.3k84668
answered Nov 12 '18 at 14:28
Artem BilanArtem Bilan
64.3k84668
64.3k84668
Edited the question. There was the route before thechannelMapping, so I think I need thechannelMapping. So how to route with the channelMapping method to the channel which name is in the headers?
– Mike
Nov 13 '18 at 10:46
Ok. You can include that into the routing function as well. I mean you do ternary operator and return the channel you need. The mapping is only for static channels
– Artem Bilan
Nov 13 '18 at 12:29
Thanks for the answer
– Mike
Nov 19 '18 at 11:54
add a comment |
Edited the question. There was the route before thechannelMapping, so I think I need thechannelMapping. So how to route with the channelMapping method to the channel which name is in the headers?
– Mike
Nov 13 '18 at 10:46
Ok. You can include that into the routing function as well. I mean you do ternary operator and return the channel you need. The mapping is only for static channels
– Artem Bilan
Nov 13 '18 at 12:29
Thanks for the answer
– Mike
Nov 19 '18 at 11:54
Edited the question. There was the route before the
channelMapping, so I think I need the channelMapping. So how to route with the channelMapping method to the channel which name is in the headers?– Mike
Nov 13 '18 at 10:46
Edited the question. There was the route before the
channelMapping, so I think I need the channelMapping. So how to route with the channelMapping method to the channel which name is in the headers?– Mike
Nov 13 '18 at 10:46
Ok. You can include that into the routing function as well. I mean you do ternary operator and return the channel you need. The mapping is only for static channels
– Artem Bilan
Nov 13 '18 at 12:29
Ok. You can include that into the routing function as well. I mean you do ternary operator and return the channel you need. The mapping is only for static channels
– Artem Bilan
Nov 13 '18 at 12:29
Thanks for the answer
– Mike
Nov 19 '18 at 11:54
Thanks for the answer
– Mike
Nov 19 '18 at 11:54
add a comment |
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