08-11-2020 01:02 AM
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
08-12-2020 02:12 AM
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
08-11-2020 02:41 AM
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.
08-11-2020 04:31 AM
If you can specify the location i will be very thankful to you. Because i am searching but not getting.
Thanks in Advance
08-12-2020 12:33 AM
Can anyone suggest me the solution . As the above one might be correct but how and where to do it is not .
08-12-2020 02:12 AM
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
08-12-2020 08:07 AM
Hi @akash251998,
Great that you found a solution & thanks for updating us on how you did it.
Cheers,
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