以编程方式访问<declare-styleable>资源

Chr*_*ock 6 android

是否有可能以编程方式接收由a [int]保存的资源-id而不引用资源类R?

<declare-styleable name="com_facebook_login_view">
    <attr name="confirm_logout" format="boolean"/>
    <attr name="fetch_user_info" format="boolean"/>
    <attr name="login_text" format="string"/>
    <attr name="logout_text" format="string"/>
</declare-styleable>
Run Code Online (Sandbox Code Playgroud)

问题是我无法解析定义的'declare-styleable'属性的ID - 始终返回0x00:

int id = context.getResources().getIdentifier( "com_facebook_login_view", "declare-styleable", context.getPackageName() ); 
int[] resourceIDs = context.getResources().getIntArray( id );
Run Code Online (Sandbox Code Playgroud)

Chr*_*ock 15

以下是以编程方式为标记child-<attr>-tags定义的资源ID的解决方案<declare-styleable>:

/*********************************************************************************
*   Returns the resource-IDs for all attributes specified in the
*   given <declare-styleable>-resource tag as an int array.
*
*   @param  context     The current application context.
*   @param  name        The name of the <declare-styleable>-resource-tag to pick.
*   @return             All resource-IDs of the child-attributes for the given
*                       <declare-styleable>-resource or <code>null</code> if
*                       this tag could not be found or an error occured.
*********************************************************************************/
public static final int[] getResourceDeclareStyleableIntArray( Context context, String name )
{
    try
    {
        //use reflection to access the resource class
        Field[] fields2 = Class.forName( context.getPackageName() + ".R$styleable" ).getFields();

        //browse all fields
        for ( Field f : fields2 )
        {
            //pick matching field
            if ( f.getName().equals( name ) )
            {
                //return as int array
                int[] ret = (int[])f.get( null );
                return ret;
            }
        }
    }
    catch ( Throwable t )
    {
    }

    return null;
}
Run Code Online (Sandbox Code Playgroud)

也许这有一天可以帮助某人.