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

edited Nov 13 '18 at 10:42

asked Nov 12 '18 at 5:59

Mike

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

edited Nov 13 '18 at 10:42

asked Nov 12 '18 at 5:59

Mike

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

edited Nov 13 '18 at 10:42

asked Nov 12 '18 at 5:59

Mike

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

edited Nov 13 '18 at 10:42

asked Nov 12 '18 at 5:59

Mike

12010

edited Nov 13 '18 at 10:42

asked Nov 12 '18 at 5:59

Mike

12010

edited Nov 13 '18 at 10:42

asked Nov 12 '18 at 5:59

Mike

12010

asked Nov 12 '18 at 5:59

Mike

12010

asked Nov 12 '18 at 5:59

Mike

12010

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 Bilan

64.3k84668

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

Thanks for the answer
– Mike
Nov 19 '18 at 11:54

add a comment |

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53256596%2fspring-integration-java-dsl-how-to-route-with-the-channelmapping-method-to-the%23new-answer', 'question_page');

);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

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 Bilan

64.3k84668

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

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 Bilan

64.3k84668

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

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 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 Bilan

64.3k84668

answered Nov 12 '18 at 14:28

Artem Bilan

64.3k84668

answered Nov 12 '18 at 14:28

Artem Bilan

64.3k84668

answered Nov 12 '18 at 14:28

Artem Bilan

64.3k84668

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

Thanks for the answer
– Mike
Nov 19 '18 at 11:54

add a comment |

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

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

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 |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Pfthb