Question

我有一个要基于功能更新的sql表到目前为止，这是我的代码：

def read(conn):
    cursor = conn.cursor()
    #cursor2 = conn.cursor()
    cursor.execute("SELECT All [id],[keyword],[result],[status],[createddate] FROM [Table1].[dbo].[aa]")
    index = 1
    cursor.execute("set nocount on")
    for row in cursor:
        s = row[1]
        s = re.sub(r'[^\w\s]', '', s)
        a=do_func(s)
        if a:
            cursor.execute("update [Table1].[dbo].[aa] set status = 'completed', result = 'True' where id ={}".format(index))
        else:
            cursor.execute("update [Table1].[dbo].[aa] set status = 'completed', result = 'False' where id ={}".format(index))
        if index == 10:
            break
        index += 1

我得到pyodbc.ProgrammingError: No results. Previous SQL was not a query. 我添加了“ set nocount on”但没有解决，我也尝试制作第二个光标但也没有解决问题

Answer 1

好的，请参见代码：您将需要拆分光标以进行选择和光标进行更新，您不能同时使用两者。更新后，您将需要提交。让我知道它是否有效。

def read(conn):
    selectcursor = conn.cursor()
    updatecursor = conn.cursor()
    selectcursor.execute("SELECT [id],[keyword],[result],[status],[createddate] FROM [Table1].[dbo].[aa]")
    index = 1
    result = selectcursor.fetchall()
    for row in result:
        s = row[1]
        s = re.sub(r'[^\w\s]', '', s)
        a=do_func(s)
        if a:
            updatecursor.execute("update [Table1].[dbo].[aa] set status = 'completed', result = 'True' where id ={}".format(index))
            updatecursor.commit()
        else:
            updatecursor.execute("update [Table1].[dbo].[aa] set status = 'completed', result = 'False' where id ={}".format(index))
            updatecursor.commit()
        if index == 10:
            break
        index += 1
        selectcursor.close()
        updatecursor.close()

我无法从python更新数据库

1 个答案: