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python - pypyodbc - 使用存储过程并将数据存储到数据框中

问题描述

所以我有一个存储过程，我想连接并在 python 中执行，这样我就可以将数据导出为一个花哨的表。

这是我使用的代码，注释是我尝试过的代码。

import plotly as py
import plotly.figure_factory as ff
import pandas as pd
import pypyodbc
import pyodbc
import datetime
import seaborn as sns
import matplotlib.pyplot as plt


#force the ODBC driver to use case-sensitive column names
pypyodbc.lowercase = True  

startdate = '2018-10-23 08:00'
enddate = '2018-10-24 12:00'

params = startdate,enddate

##Establishing a connection to SQL database

connection = pypyodbc.connect('Driver={SQL Server};'
                                'Server=LOLOL;'
                                'Database=LOL_Collection;'
                                'trusted_connection=yes;')
crsr = connection.cursor()

SQL = """

SET NOCOUNT ON
declare @start = '2018-10-23 08:00'
declare @end = '2018-10-24 12:00'
exec [dbo].[AM_SIGNAL_Histogram] @start, @end

"""


#SQL = """
#    
#exec [dbo].[AM_SIGNAL_Histogram] @start, @end
#
#"""


#result = crsr.execute(SQL)
#         
#print(result.fetchall())


df = pd.read_sql(SQL,connection)
connection.close()

这是我不断得到的错误：

ProgrammingError: ('42000', "[42000] [Microsoft][ODBC SQL Server Driver][SQL Server]'=' 附近的语法不正确。")

感谢您的帮助，肯尼斯

标签： pythonsql-serverpypyodbc

解决方案

问题不在于您的 Python 代码，而在于您的 T-SQL。直接在 SQL Server Management Studio (SSMS) 中运行它，

SET NOCOUNT ON
declare @start = '2018-10-23 08:00'
declare @end = '2018-10-24 12:00'
exec [dbo].[AM_SIGNAL_Histogram] @start, @end

生产

Msg 102, Level 15, State 1, Line 2
Incorrect syntax near '='.
Msg 137, Level 15, State 2, Line 4
Must declare the scalar variable "@start".

您的declare语句缺少数据类型。以下修改对我有用：

SET NOCOUNT ON;
declare @start varchar(50) = '2018-10-23 08:00';
declare @end varchar(50) = '2018-10-24 12:00';
exec [dbo].[AM_SIGNAL_Histogram] @start, @end;

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