在Python SQLite中执行INSERT后,表中没有结果
我使用Python和SQLite包创建表并将数据插入表中。但是,在我启动执行后,表中没有任何内容。有人能帮我弄清楚吗?谢谢在Python SQLite中执行INSERT后,表中没有结果,python,sqlite,Python,Sqlite,我使用Python和SQLite包创建表并将数据插入表中。但是,在我启动执行后,表中没有任何内容。有人能帮我弄清楚吗?谢谢 def conSqlite(): conn = sqlite3.connect('C:\\Users\jet.cai\Documents\Logsitic.db') json_path = r'C:\Users\jet.cai\PycharmProjects\VJSF\txtToJson.json' try: create_table = ('''
def conSqlite():
conn = sqlite3.connect('C:\\Users\jet.cai\Documents\Logsitic.db')
json_path = r'C:\Users\jet.cai\PycharmProjects\VJSF\txtToJson.json'
try:
create_table = ('''
CREATE TABLE IF NOT EXISTS CODE2
(Delivery TEXT,
Customer_Name NCHAR(50),
Shipment_Priority TEXT
)''')
conn.execute(create_table)
except:
print("Table Failed")
return False
with open(json_path, 'r') as jsonf:
lines = json.load(jsonf)
for line in lines:
sql = "insert into CODE2(Delivery,Customer_Name,Shipment_Priority) values('%s','%s','%s')"%(line['Delivery'],line['Customer Name'],line['Shipment Priority'])
conn.execute(sql)
# No results can be selected out
df = pd.read_sql("select Delivery from CODE2", conn)
print(df)
commit
是处理数据库时要记住的一个非常重要的概念。伙计们,我发现了,我没有提交insert操作,如果发生了,很抱歉打扰您Hi Alan,没错,谢谢您的评论!我们同时看到了它。一切都很好。Stack应该允许我们用拇指仰视:)