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Check feature by permission #1813


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andmattia created

I try to check if a user has one permission depending on a feature. I try to explain; feature -- UseApplication1 -> featureDependency: new SimpleFeatureDependency("UseApplication1") -- UseApplication2 -> featureDependency: new SimpleFeatureDependency("UseApplication2")

Permission all permission depend on -- Document -> featureDependency: new SimpleFeatureDependency("UseApplication2") ---| Add -> featureDependency: new SimpleFeatureDependency("UseApplication2") ---| Update -> featureDependency: new SimpleFeatureDependency("UseApplication2") ... -- Application-> featureDependency: new SimpleFeatureDependency("UseApplication1") ---| Add -> featureDependency: new SimpleFeatureDependency("UseApplication1") ---| Update -> featureDependency: new SimpleFeatureDependency("UseApplication1")

1 user -- Has permission Document,Add,Update 2 user -- has no permission

If i get all feature and I check if is enable in tenant if(await FeatureChecker.IsEnabledAsync(feature.Name))

now I need to check if user has one(or more) permission enable depending on this feature ...

var user = await UserManager.GetUserByIdAsync(AbpSession.UserId.Value); var userPermission = await UserManager.GetGrantedPermissionsAsync(user); userPermission.Where(t=>t.FeatureDependency())

... But How can I check if exists 1 (or more) check permission depending on my feature mattia


1 Answer(s)
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    hikalkan created
    Support Team

    Hi,

    I suspect that you are using it a bit wrong :)

    Say that you have a "PermissionX" depends on "FeatureA". You should not have such a code

    if(tenant has FeatureA) { if(user has PermissionX) { //...do it } }

    Since PermissionX depends on FeatureA, this will be enough:

    if(user has PermissionX) { //...do it }

    So, most of times, no need to explictly check features.

    I haven't understood why you need to check "if exists 1 (or more) check permission depending on my feature", but to do it, use such a where condition:

    userPermission.Where(t=> (t.FeatureDependency is SimpleFeatureDependency) && t.FeatureDependency.As<SimpleFeatureDependency>().Features.Contains("MyFeatureName") );

    If that's not you want, please explain your real case for a better answer.