Question

我按如下方式创建一个简单的SQLite表：

import sqlite3
sqlite_file = '/Users/Dom/Desktop/Test.sqlite'
conn = sqlite3.connect(sqlite_file)
c = conn.cursor()

c.execute('''CREATE TABLE Results(Col1 text, Col2 text)''')

c.execute('''CREATE TABLE ListIDTable(Int numeric, ID text)''')

values_to_insert = [(1,"1a"), (2,"1b"), (3,"2a"), (4,"2b"), (5,"3a"), 
(6,"3b"), (7,"4a"), (8,"4b"), (9,"5a"), (10,"5b"), (11,"6a"), (12,"6b"),
(13,"7a"), (14,"7b") ]

c.executemany("INSERT INTO ListIdTable (Int, ID) values (?,?)",
values_to_insert)

conn.commit()
conn.close()

一切看起来都不错。然后我按如下方式遍历上面创建的表：

import sqlite3

sqlite_file = '/Users/Dom/Desktop/Test.sqlite'
conn = sqlite3.connect(sqlite_file)
conn.row_factory = sqlite3.Row
c = conn.cursor()
c2 = conn.cursor()

c.execute('select * from ListIDTable ORDER BY Int ASC')
for r in c:
    TblInt = r['Int']
    print (TblInt)

    c2.execute("INSERT INTO Results (Col1 , Col2) values (?,?)", ("XXX",
    "YYY"))
    conn.commit()

我希望输出：＆＃34; 1,2,3,4,5,6 ......＆＃34;等

然而，我得到：＆＃34; 1,2,1,2,3,4,5,6 ......＆＃34;等

当我删除最后conn.commit（）语句的缩进时，我得到了预期的输出。有人可以帮助我理解为什么只是＆＃34; 1＆amp; 2＆＃34;重复，但一切都正常进行？

由于

Answer 1

以下页面确定并解释了该问题： Using multiple cursors in a nested loop in sqlite3 from python-2.7

为了解决上述问题，我基本上完成了整个查找表并消除了对多个游标的需求：

import sqlite3
sqlite_file = '/Users/Dom/Desktop/Test.sqlite'
conn = sqlite3.connect(sqlite_file)
conn.row_factory = sqlite3.Row
c = conn.cursor()

c.execute('select * from ListIDTable ORDER BY Int ASC')
rows = c.fetchall()

for r in rows:
    TblInt = r['Int']
    print (TblInt)

    c.execute("INSERT INTO Results (Col1 , Col2) values (?,?)", ("XXX", 
    "YYY"))
    conn.commit()

Python SQLite3：为什么缩进conn.commit（）语句会产生这种影响？

1 个答案: