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?error=true don't want additional suffix on incorrect login.

akash251998
Star Contributor
Star Contributor

While doing the login if correct credentials is given then the url in the header is http://127.0.0.1:8080/share/page/user/admin/dashboard

While giving incorrect credentials it will give error message Your authentication details haven't been recognized or Alfresco Content Services may not be available at this time. 

And the url come in header is as below:

http://127.0.0.1:8080/share/page/?error=true

While giving incorrect User Name and password i want only error message not the extra suffix  (?error=true).

Is it possible to eliminate the ?error=true in the url.

Regards

Akash

1 ACCEPTED ANSWER

It can be done i found the solution.

Go to the location:

installed_folder\tomcat\webapps\share\WEB-INF\classes\alfresco\site-webscripts\org\alfresco\components\guest

There you will find an file named login.get.js.  Go to line number 23 .

Replace 

               model.failureUrl = successUrl + (successUrl.indexOf("?") != -1 ? "&" : "?") + "error=true";

                                               by

                     model.failureUrl = successUrl ;

Regards 

Akash

View answer in original post

5 REPLIES 5

sanjaybandhaniya
Elite Collaborator
Elite Collaborator

Hi,

You can change/remove but it will not display error message.

In share side there is a webscript called guest login(dont remember exact file name),you can change param in that file.

If you can specify the location i will be very thankful to you. Because i am searching but not getting.

Thanks in Advance

Can anyone suggest me the solution . As the above one might be correct but how and where to do it is not .

It can be done i found the solution.

Go to the location:

installed_folder\tomcat\webapps\share\WEB-INF\classes\alfresco\site-webscripts\org\alfresco\components\guest

There you will find an file named login.get.js.  Go to line number 23 .

Replace 

               model.failureUrl = successUrl + (successUrl.indexOf("?") != -1 ? "&" : "?") + "error=true";

                                               by

                     model.failureUrl = successUrl ;

Regards 

Akash

EddieMay
World-Class Innovator
World-Class Innovator

Hi @akash251998,

Great that you found a solution & thanks for updating us on how you did it.

Cheers, 

Digital Community Manager, Alfresco Software.
Problem solved? Click Accept as Solution!