如何使用sqlite为一组行创建唯一的ID

时间:2014-05-23 18:55:30

标签: python sqlite flask

我有以下数据库架构:

CREATE TABLE orders (
    transaction_id integer primary key autoincrement,
    total_price integer not null
);
CREATE TABLE order_items (
    transaction_id integer REFERENCES orders(transaction_id),
    SKU integer not null,
    product_name text not null,
    unit_price integer not null,
    quantity integer not null
);

通过此烧录代码将数据发送到数据库:

@app.route('/load_ajax', methods=["GET", "POST"])
def load_ajax():
    if request.method == "POST":
        data = request.get_json()
        for group in groupby(data, itemgetter('name')):
            id, data_list = group
            for d in data_list:
                print d['subtotal']
                db = get_db()
                db.execute('insert into order_items (SKU, product_name, unit_price, quantity) values (?, ?, ?, ?)',
                           [d['sku'], d['name'], d['price'], d['quantity']])
                db.execute('insert into orders (total_price) values (?)',
                           [d['subtotal']])
                db.commit()
        return jsonify(location=url_for('thankyou'))

读取订单表会给出:

1|1187.86
2|1187.86
3|1187.86
4|1187.86
5|1187.86
6|1187.86
7|102.92
8|102.92
9|102.92
10|102.92

并且读取order_items表给出了:

|ASD|Hot Sauce|10.99|1
|JKL|Chilli Peppers|8.99|1
|UIO|Sip 'n' Sizzle T-Shirt|10.5|1
|MEE|Energy Drinks|10.99|1
|FUE|Literally, Anything|1|1
|POL|Burger Ryan|1000|1
|ASD|Hot Sauce|10.99|1
|JKL|Chilli Peppers|8.99|1
|UIO|Sip 'n' Sizzle T-Shirt|10.5|1
|MEE|Energy Drinks|10.99|1

我想要完成的是生成一个唯一的交易ID,该ID将按特定顺序分配给每个项目。

首先,我不确定为什么order_items表中的transaction_id列没有出现。其次,我怎样才能得到一个看起来像这样的表:

订单:

1|1187.86
1|1187.86
1|1187.86
1|1187.86
1|1187.86
1|1187.86
2|102.92
2|102.92
2|102.92
2|102.92

ORDER_ITEMS:

1|ASD|Hot Sauce|10.99|1
1|JKL|Chilli Peppers|8.99|1
1|UIO|Sip 'n' Sizzle T-Shirt|10.5|1
1|MEE|Energy Drinks|10.99|1
1|FUE|Literally, Anything|1|1
1|POL|Burger Ryan|1000|1
2|ASD|Hot Sauce|10.99|1
2|JKL|Chilli Peppers|8.99|1
2|UIO|Sip 'n' Sizzle T-Shirt|10.5|1
2|MEE|Energy Drinks|10.99|1

这是我可以做sqlite的事吗?

编辑: 这就是我改变CL。的答案以适应我的代码:

@app.route('/load_ajax', methods=["GET", "POST"])
def load_ajax():
    if request.method == "POST":
        data = request.get_json()
        db = get_db()
        c = db.cursor()
        c.execute('insert into orders (total_price) values (?)', [data[0]['subtotal']])
        transaction_id = c.lastrowid
        for group in groupby(data, itemgetter('name')):
            id, data_list = group
            for d in data_list:
                db.execute('insert into order_items (transaction_id, SKU, product_name, unit_price, quantity) values (?, ?, ?, ?, ?)',
                           [transaction_id, d['sku'], d['name'], d['price'], d['quantity']])
        db.commit()
    return jsonify(location=url_for('thankyou'))

2 个答案:

答案 0 :(得分:1)

transaction_id表中的order_items列未显示,因为您没有在其中插入任何内容。

您不希望orders表中有重复的行。 仅插入一次订单(我假设单个POST是单个订单):

    data = request.get_json()
    subtotal = data......
    c = db.cursor()
    c.execute('insert into orders (total_price) values (?)', [subtotal])
    transaction_id = c.lastrowid
    for group in groupby(data, itemgetter('name')):
        id, data_list = group
        for d in data_list:
            db.execute('insert into order_items (transaction_id, SKU, product_name, unit_price, quantity) values (?, ?, ?, ?)',
                       [transaction_id, d['sku'], d['name'], d['price'], d['quantity']])
    db.commit()

答案 1 :(得分:0)

您可以在订单和order_item之间使用映射表

CREATE TABLE orders (
    transaction_id integer primary key autoincrement,
);

CREATE TABLE order_mapping (
    id integer primary key autoincrement,
    transaction_id foreign key references orders(transaction_id),
    item_id foreign key references order_items(item_id)
);


CREATE TABLE order_items (
    item_id integer primary key autoincrement,
    SKU integer not null,
    product_name text not null,
    unit_price integer not null,
    quantity integer not null
);
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