Provide a link to the catalog admin from the api_admin approval page.
ECOM-4441
This commit is contained in:
@@ -1,5 +1,7 @@
|
||||
"""Admin views for API managment."""
|
||||
from django.contrib import admin
|
||||
from django.core.urlresolvers import reverse
|
||||
from django.utils.translation import ugettext as _
|
||||
|
||||
from config_models.admin import ConfigurationModelAdmin
|
||||
from openedx.core.djangoapps.api_admin.models import ApiAccessRequest, ApiAccessConfig
|
||||
@@ -15,4 +17,21 @@ class ApiAccessRequestAdmin(admin.ModelAdmin):
|
||||
readonly_fields = ('user', 'website', 'reason', 'company_name', 'company_address', 'contacted', )
|
||||
exclude = ('site',)
|
||||
|
||||
def get_fieldsets(self, request, obj=None):
|
||||
return (
|
||||
(None, {
|
||||
'fields': (
|
||||
'user', 'website', 'reason', 'company_name', 'company_address',
|
||||
)
|
||||
},),
|
||||
('Status', {
|
||||
'description': _(
|
||||
'Once you have approved this request, go to {catalog_admin_url} to set up a catalog for this user.'
|
||||
).format(
|
||||
catalog_admin_url='<a href="{0}">{0}</a>'.format(reverse('api_admin:catalog-search'))
|
||||
),
|
||||
'fields': ('status',),
|
||||
}),
|
||||
)
|
||||
|
||||
admin.site.register(ApiAccessConfig, ConfigurationModelAdmin)
|
||||
|
||||
Reference in New Issue
Block a user