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#  
# ============LICENSE_START========================================== 
# org.onap.vvp/engagementmgr
# ===================================================================
# Copyright © 2017 AT&T Intellectual Property. All rights reserved.
# ===================================================================
#
# Unless otherwise specified, all software contained herein is licensed
# under the Apache License, Version 2.0 (the “License”);
# you may not use this software except in compliance with the License.
# You may obtain a copy of the License at
#
#             http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
#
#
#
# Unless otherwise specified, all documentation contained herein is licensed
# under the Creative Commons License, Attribution 4.0 Intl. (the “License”);
# you may not use this documentation except in compliance with the License.
# You may obtain a copy of the License at
#
#             https://creativecommons.org/licenses/by/4.0/
#
# Unless required by applicable law or agreed to in writing, documentation
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
#
# ============LICENSE_END============================================
#
# ECOMP is a trademark and service mark of AT&T Intellectual Property.


def dict_path_get(dicttree, path, default=None):
    """Use a /-separated path to query a structure of nested dicts.

    Note: If the path does not exist, this will return None instead of
    raising an exception.

    >>> nested = {
    ...     'a': {
    ...         'b': {
    ...             'c': 'foo'
    ...         }
    ...     }
    ... }
    ...
    >>> dict_path_get(nested, 'a/b/c')
    'foo'

    """
    d = dicttree
    try:
        for k in path.split('/'):
            d = d[k]
    except KeyError:
        d = default
    return d