How can i log out from my own created gsp view with spring security installed in grails app?

I have installed spring security plug in to gra开发者_如何学Goils project.i have made my default action as auth. and when i login i get gsp view created by me.now how can i log out from there..

Thanks, LAxmi

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After changes in Spring Security Plugin 2.0-XX, you can use a POST request using the remoteLink tag like in the following snippet:

<g:remoteLink class="logout" controller="logout" method="post" asynchronous="false" onSuccess="location.reload()">Logout</g:remoteLink>

You can customize asynchronous attribute or onSuccess event handler to adapt your usage case.

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Add a link to /logout - LogoutController will log you out.

For example: <g:link controller='logout'>Logout</g:link>

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If you want to use a link to logout, first you must disable de logout.postOnly property as described in https://grails-plugins.github.io/grails-spring-security-core/3.2.x/index.html#configGroovy

in your application.groovy:

grails.plugin.springsecurity.logout.postOnly = false

then just call it as a link in any html tag

<a href="/logoff">Salir</a>

Hope this helps

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Add in Config grails.plugin.springsecurity.logout.postOnly = false

uses <g:link controller='logout'>Logout</g:link>

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Can be done with a form too:

<g:form controller="logout">                        
    <g:submitButton name="logout" value="Logout" />
</g:form>

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add a logout link to url: /logout

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As other users mentioned that you have to send the request as POST request. Another way to do that in an elegant way using a hidden form and JavaScript to submit it.

//jQuery 
 $(function(){
     $('#logout-btn').on('click',function () {
         $('.logout-form').submit()
        })
    })

Your GSP should have a hidden form and an anchor.

 //GSP
<a class="dropdown-item" href="#" id="logout-btn">Logout</a>
<g:form class="form-inline d-none logout-form" controller="logout"></g:form>

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To logout from application, you need to call the LogoutController like:

<g:link controller="logout" action="index">Exit</g:link>

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By default, the action index of LogoutController only accept 'POST' method.

You need to build a form to call it or set the next property in the application.yml:

grails.plugin.springsecurity.logout.postOnly = false

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If you want to keep that option with its default value, this is how should the form look like:

<g:form controller="logout" action="index" method="POST">
    <g:submitButton name="Exit"/>
</g:form>

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INFO: action key is not needed (optional), specifying only the controller will get you to the index action.