My site has urls like 'http://someRandomUsername.mysite.com'.
Sometimes users will try urls like
'http://www.someRandomeUsername.mysite.com'. I'd like to have some
logic in my url mappings to deal with this.
With the mappings below when I hit the page , with or without the
unneeded www, I get:
2012-03-01 14:52:16,014 [http-8080-5] ERROR [localhost].[/ambit] -
Unhandled exception occurred whilst decorating page
java.lang.IllegalArgumentException: URL mapping must either provide a
controller or view name to map to!
Any idea how to accomplish this? The mapping is below.
Thanks!
Jason
static mappings = {
name publicMap: "/$action?/$id?" {
def ret = UrlMappings.check(request)
controller = ret.controller
userName = ret.userName
}
}
static check =
{ request ->
def tokens = request?.serverName?.split(/\./) as List ?: []
def ret = [controller:'info']
if(tokens.size() > 3 && token[0] == 'www')
{
ret.userName = tokens[1]
ret.controller = 'redirect'
ret.action = 'removeWWW'
}
else if(tokens.size() == 3)
{
ret.userName = tokens[0]
ret.controller = 'info'
}
return ret
}
Honestly, like DmitryB said, the best way to do this is via the web server, whether it's IIS, Apache, or Tomcat.
Having said that, I feel the best way to accomplish this in Grails would be using filters.
You could create something like this in your ~/conf directory:
public class StripFilters {
def filters = {
stripWWWFilter(controller: '*', action: '*') {
before = {
def tokens = request.serverName.tokenize(/\./) ?: []
if(tokens.size() > 3 && tokens[0] == 'www') {
def url = request.request.requestURL.toString().replace('www.', '')
redirect([url:url, params: [userName: tokens[1]], permanent: true])
return false
}
}
}
}
}
This should do the trick.
Related
Our grails application uses ldap authentication, without any problems, now I need to prevent access, to the entire application, if a user has no specific ldap role.
I can see the role and use it in my Config.groovy annotations or secure the actions in the controllers, but instead I need a scenario/way to just show a "Denied ..." message and logout. (POST Forbidden 403).
def filters = {
loginFilter(controller:'login', action:'ajaxSuccessSproutcore') {
before = {
switch(Environment.current.name) {
case { it == 'development' || it == 'hrm'}:
if (springSecurityService.isLoggedIn() && grails.plugin.springsecurity.SpringSecurityUtils.ifAnyGranted("ROLE_ADMIN, ROLE_SEA_HRM_LOGIN")){
} else {
if (springSecurityService.isLoggedIn()) {
render ([msg:''] as JSON)
session.invalidate()
return false
}
}
break
default:
if (springSecurityService.isLoggedIn() && grails.plugin.springsecurity.SpringSecurityUtils.ifAnyGranted("ROLE_ADMIN , ROLE_USER")){
} else {
if (springSecurityService.isLoggedIn()) {
render ([msg:''] as JSON)
session.invalidate()
return false
}
}
break
}
}
after = { Map model ->
}
afterView = { Exception e ->
}
}
}
In grails 3 you can set up an Interceptor to check every request and take the appropriate action. In your case you'd want to add a check in the before block.
Edit: As Jeff Brown notes in the comments, grails 2 used Filters rather than interceptors.
Edit: Something like this in your logout logic:
...
else {
if (springSecurityService.isLoggedIn()) {
session.invalidate()
redirect action:'youShallNotPass'
return false
}
}
In my Grails 2.5.1 application , i was using a filter to use HTTPS with some controllers , everything was working fine but suddenly this filter is not working any more .
Filter :
def filters = {
all(controller:'checkout', action:'onlinePayment') {
before = {
if (!request.isSecure() /*&& !Environment.isDevelopmentMode()*/) {
def url = "https://" + request.serverName+':8443' + request.forwardURI
println "in filter"
redirect(url: url, permanent: true)
return false
}
}
after = { Map model ->
}
afterView = { Exception e ->
}
}
}
Here is the checkout page :
Also i found that no requests came to the filter as in filter was not printed out, is there something i need to check to fix this issue rather than this filter
Using the Grails OAuth plugin requires that an absolute callback URL be provided in Config.groovy. However I have different serverURLs for each environment.
Is there a way to get the current environment from inside Config.groovy, here's an example of what I want to do:
def devServerUrl = 'http://dev.example.com'
def prodServerUrl = 'http://prod.example.com'
def currentServerUrl = grailsApplication.metadata.environment == 'development' ? devServerUrl : prodServerUrl;
environments {
development {
grails {
serverURL = devServerUrl
}
}
production {
grails {
serverURL = prodServerUrl
}
}
}
oauth {
providers {
runkeeper {
api = RunKeeperApi
key = 'key'
secret = 'secret'
callback = currentServerUrl + '/oauth/runkeeper/callback'
}
}
}
Any ideas? Thanks!
Try this:
def currentServerUrl = Environment.current.name == 'development' ? devServerUrl : prodServerUrl;
I think it is cleaner to set different grails.serverURL for each environment and then do:
callback = "${grails.serverURL}/oauth/runkeeper/callback"
How can I detect the language of the browser and automatically display the correctly localized version of my grails website depending on that value.
I put this in to Index action
Locale locale = new Locale(params.lang)
cookieLocaleResolver.setLocale(request, response, (Locale)
session.getAttribute('locale'))
{
render controller: "home", action: "index"
return
}
And I got exception--
Error 500: Executing action [index] of controller [socialworking.HomeController] caused exception: null
Servlet: grails
URI: /socialworking/grails/home.dispatch
Exception Message:
Caused by:
Class: Unknown
First off, you should put that in a filter in grails-app/conf directory. Create a filter if you don't already have one.
MyFilters.groovy
class MyFilters {
def filters = {
setLocale(controller:'*', action:'*') {
before = {
// Your logic here
}
}
}
}
Your logic here could look in many ways, but here is a try:
String langToSet = 'en';
if ( params.lang && params.lang in validLanguages )
langToSet = params.lang;
else if ( session.lang ) {
langToSet = session.lang;
}
else if ( ... ) // Cookie lang is set ( User might have accessed the site before and you have stored their preferred lang )
// Get cookie lang
Locale locale = new Locale( langToUse)
org.springframework.web.servlet.support.RequestContextUtils.getLocaleResolver(request).setLocale(request, response, locale);
// Set the cookie lang
...
// We set the session lang
session.lang = langToSet
Note that the above is not a complete implementation but it is almost. The cookie stuff and validLanguages you should be able to figure out what they do.
I hope that helps!
how can i redirect to the same state more than one time using web flow
ex:
on('submit'){
def destinationInstance = Destination.get(params.destination)
def destinationGroupsInstance = DestinationGroup.get(params.destinationGroups)
def h = destinationInstance.addToDestinationGroups(destinationGroupsInstance)
}.to('flowList')
what i need is how to enter to this state more than one time until destinations ends
thx
on('submit'){
def destinationInstance = Destination.get(params.destination)
def destinationGroupsInstance = DestinationGroup.get(params.destinationGroups)
def h = destinationInstance.addToDestinationGroups(destinationGroupsInstance)
}.to{
(condition or while loop or for loop)
if success then
return "<state name>"
else
return "flowList"
}
Reference:
http://livesnippets.cloudfoundry.com/docs/guide/2.%20Grails%20webflow%20plugin.html
Well, you'd probably have something like the following code, which is untested but may give you a general idea.
def destinationFlow = {
initialize {
action {
flow.destination = Destination.get(params.id)
}
on('success').to 'destinationList'
}
destinationList {
render(view: 'destinationList')
on('addDestination') {
def destinationGroup = DestinationGroup.get(params.destinationGroupId)
flow.destination.addToDestinationGroups(destinationGroup)
}.to 'destinationList'
on('finish').to 'done'
}
done {
flow.destination.save()
redirect(...) // out of the flow
}
}
You'll need buttons on your destinationList view that invoke the 'addDestination' or 'finish' actions. See the WebFlow documentation and Reference Guide.