使用`executemany`更新现有SQLite3数据库中的条目(使用Python sqlite3)

时间:2015-11-19 01:58:28

标签: python sqlite

我知道executemany可以用来方便地将新条目添加到数据库中;与for循环中的单个execute相比,减少Python方法调用开销很有用。但是,我想知道这是否适用于SQLite的UPDATE

更具体地说,请考虑以下设置:

cnx = sqlite3.connect(DATABASE)
c = cnx.cursor()
for path in paths:
    for data in some_computation(path):
        c.execute("UPDATE TABLENAME SET cont=? WHERE id=?", (data[1], data[0]))
cnx.commit()
cnx.close()

我甚至不确定下面的方法是否会更快(必须对它进行基准测试),但问题是它不起作用,因为我做错了我假设。有关在下面的代码段中使用executemany来完成我上面发布的任务的提示吗?

cnx = sqlite3.connect(DATABASE)
c = cnx.cursor()

for path in paths:
    data_ids, data_conts = [], []
    for data in some_computation(path):
        if len(data_ids) >= CHUNKSIZE:
            c.executemany("UPDATE TABLENAME SET cont=? WHERE id=?", (data_conts, data_ids))
            cnx.commit()
            data_ids, data_conts = [], []
        data_ids.append(data[0])
        data_conts.append(data[1])
    c.executemany("UPDATE TABLENAME SET cont=? WHERE id=?", (data_conts, data_ids))      
    cnx.commit()

cnx.commit()
cnx.close()

非常感谢您的提示和见解!

编辑1:

底部示例的问题:

ProgrammingError: Incorrect number of bindings supplied. The current statement uses 2, and there are 50000 supplied.

(其中CHUNKSIZE = 50000)

编辑2:

发生同样的错误

cnx = sqlite3.connect(DATABASE)
c = cnx.cursor()

for path in paths:
    data_conts = []
    for data in some_computation(path):
        if len(data_ids) >= CHUNKSIZE:
            c.executemany("UPDATE TABLENAME SET cont=? WHERE id=?", (data_conts,))
            cnx.commit()
            data_conts = []

        data_conts.append([data[1], data[0]])
    c.executemany("UPDATE TABLENAME SET cont=? WHERE id=?", (data_conts,))   
    cnx.commit()

cnx.commit()
cnx.close()

但是感谢@falsetru我注意到了我的错误,它应该是

... WHERE id=?", data_conts)

而不是

... WHERE id=?", (data_conts,))

2 个答案:

答案 0 :(得分:9)

您需要传递一系列序列([[cont,id], [cont,id], [cont,id], ...],而不是[cont, cont, cont, ...], [id, id, id, ..]):

for path in paths:
    params = []
    for data in some_computation(path):
        if len(data_ids) >= CHUNKSIZE:
            c.executemany("UPDATE TABLENAME SET cont=? WHERE id=?", params)
            cnx.commit()
            params = []
        params.append([data[1], data[0]])
    if params:
        c.executemany("UPDATE TABLENAME SET cont=? WHERE id=?", params)
    cnx.commit()

答案 1 :(得分:0)

你所拥有的是完美的,除了你应该使用zip(conts,id),其中conts和id是列表。这会自动为您重新排列。