51 lines
2.0 KiB
Python
51 lines
2.0 KiB
Python
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# The Mia! Accounting Flask Project.
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# Author: imacat@mail.imacat.idv.tw (imacat), 2023/1/30
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# Copyright (c) 2023 imacat.
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#
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# Licensed under the Apache License, Version 2.0 (the "License");
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# you may not use this file except in compliance with the License.
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# You may obtain a copy of the License at
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#
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# http://www.apache.org/licenses/LICENSE-2.0
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#
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# Unless required by applicable law or agreed to in writing, software
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# distributed under the License is distributed on an "AS IS" BASIS,
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# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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# See the License for the specific language governing permissions and
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# limitations under the License.
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"""The account query.
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"""
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import sqlalchemy as sa
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from flask import request
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from accounting.utils.query import parse_query_keywords
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from .models import Account, AccountL10n
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def get_account_query() -> list[Account]:
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"""Returns the accounts, optionally filtered by the query.
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:return: The accounts.
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"""
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keywords: list[str] = parse_query_keywords(request.args.get("q"))
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if len(keywords) == 0:
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return Account.query.order_by(Account.base_code, Account.no).all()
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code: sa.BinaryExpression = Account.base_code + "-" \
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+ sa.func.substr("000" + sa.cast(Account.no, sa.String),
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sa.func.char_length(sa.cast(Account.no,
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sa.String)) + 1)
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conditions: list[sa.BinaryExpression] = []
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for k in keywords:
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l10n: list[AccountL10n] = AccountL10n.query\
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.filter(AccountL10n.title.contains(k)).all()
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l10n_matches: set[str] = {x.account_id for x in l10n}
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conditions.append(sa.or_(Account.base_code.contains(k),
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Account.title_l10n.contains(k),
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code.contains(k),
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Account.id.in_(l10n_matches)))
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return Account.query.filter(*conditions)\
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.order_by(Account.base_code, Account.no).all()
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