获取Python SQL错误-必须声明标量变量"@p1offset"；使用Tkinter时

我正在尝试运行2个SQL查询，并将它们分别输出到Tkinter中的2个标签：

    self.output = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output.grid(column=0, row=6, columnspan=5, padx=0, pady=0)
    self.output2 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output2.grid(column=0, row=7, columnspan=5, padx=0, pady=0)

我得到了一个标量变量错误，我不确定是什么原因造成的。 在Tkinter中，我不能将多行输出到一个标签，因此我必须创建额外的标签并放置

            OFFSET 1 ROWS
            FETCH NEXT 1 ROWS ONLY

在第二个SQL查询中，以便在第二个标签上显示第二行

以下是我正在尝试的：

def calculate(self):
    firstname = str(self.first_entry.get())
    lastname = str(self.last_entry.get())     
    license = str(self.lic_entry.get())
    if (firstname and not lastname and not license):  # "You entered first name."

        try:
            connection = pypyodbc.connect('Driver={SQL Server};Server=myServer;Database=myDatabase;Trusted_Connection=yes;')
        except pypyodbc.Error as ex:
            sqlstate = ex.args[0]
            if sqlstate == '28000':
                self.answer_label['text'] = "You do not have access." 
        cursor = connection.cursor() 
        SQLCommand = ("SELECT LASTNAME, FIRSTNAME, L_LICNUMBER "      
            "FROM dbo.my_table "   # table name
            "with (nolock)"
            "WHERE FIRSTNAME = ?")
        Values = [firstname]
        cursor.execute(SQLCommand,Values)
        results = cursor.fetchall()
        if results:
            self.output['text'] = (results[0]) # display  results

        else:
            self.output['text'] = "That name does not exist."

        cursor = connection.cursor() 
        SQLCommand = ("SELECT LASTNAME, FIRSTNAME, L_LICNUMBER "      
            "FROM dbo.my_table "   # table name
            "with (nolock)"
            "WHERE FIRSTNAME = ?"
            "order by FIRSTNAME"
            "offset 1 rows"
            "fetch next 1 rows only")
        Values = [firstname]
        cursor.execute(SQLCommand,Values)
        results = cursor.fetchall()
        if results:
            self.output2['text'] = (results[0]) # display  results
            connection.close()

---

我找了一整天才找到解决办法。我在谷歌上找不到任何东西。 偏移行和获取不起作用

这是：

def calculate(self):
    firstname = str(self.first_entry.get())
    lastname = str(self.last_entry.get())     
    license = str(self.lic_entry.get())
    if (firstname and not lastname and not license):  # "You entered first name."

        try:
            connection = pypyodbc.connect('Driver={SQL Server};Server=myServerName;Database=myDatabBaseName;Trusted_Connection=yes;')
        except pypyodbc.Error as ex:
            sqlstate = ex.args[0]
            if sqlstate == '28000':
                self.answer_label['text'] = "You do not have access." 
        cursor = connection.cursor() 
        SQLCommand = ("SELECT LASTNAME, FIRSTNAME, L_LICNUMBER "      
            "FROM dbo.my_table "   # table name
            "with (nolock)"
            "WHERE FIRSTNAME = ?")
        Values = [firstname]
        cursor.execute(SQLCommand,Values)
        results = cursor.fetchmany(10)
        if results:
            self.output0['text'] = (results[0]) # display  results
            self.output1['text'] = (results[1])
            self.output2['text'] = (results[2])
            self.output3['text'] = (results[3])
            self.output4['text'] = (results[4])
            self.output5['text'] = (results[5])
            self.output6['text'] = (results[6])
            self.output7['text'] = (results[7])
            self.output8['text'] = (results[8])
            self.output9['text'] = (results[9])
            connection.close()

        else:
            self.output['text'] = "That name does not exist."

您必须创建一组LabelFrame、列表框或消息来显示结果

    self.output0 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output0.grid(column=0, row=6, columnspan=5, padx=0, pady=0)
    self.output1 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output1.grid(column=0, row=7, columnspan=5, padx=0, pady=0)
    self.output2 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output2.grid(column=0, row=8, columnspan=5, padx=0, pady=0)
    self.output3 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output3.grid(column=0, row=9, columnspan=5, padx=0, pady=0)
    self.output4 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output4.grid(column=0, row=10, columnspan=5, padx=0, pady=0)
    self.output5 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output5.grid(column=0, row=11, columnspan=5, padx=0, pady=0)
    self.output6 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output6.grid(column=0, row=12, columnspan=5, padx=0, pady=0)
    self.output7 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output7.grid(column=0, row=13, columnspan=5, padx=0, pady=0)
    self.output8 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output8.grid(column=0, row=14, columnspan=5, padx=0, pady=0)
    self.output9 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output9.grid(column=0, row=15, columnspan=5, padx=0, pady=0)  

---

我找了一整天才找到解决办法。我在谷歌上找不到任何东西。 偏移行和获取不起作用

这是：

def calculate(self):
    firstname = str(self.first_entry.get())
    lastname = str(self.last_entry.get())     
    license = str(self.lic_entry.get())
    if (firstname and not lastname and not license):  # "You entered first name."

        try:
            connection = pypyodbc.connect('Driver={SQL Server};Server=myServerName;Database=myDatabBaseName;Trusted_Connection=yes;')
        except pypyodbc.Error as ex:
            sqlstate = ex.args[0]
            if sqlstate == '28000':
                self.answer_label['text'] = "You do not have access." 
        cursor = connection.cursor() 
        SQLCommand = ("SELECT LASTNAME, FIRSTNAME, L_LICNUMBER "      
            "FROM dbo.my_table "   # table name
            "with (nolock)"
            "WHERE FIRSTNAME = ?")
        Values = [firstname]
        cursor.execute(SQLCommand,Values)
        results = cursor.fetchmany(10)
        if results:
            self.output0['text'] = (results[0]) # display  results
            self.output1['text'] = (results[1])
            self.output2['text'] = (results[2])
            self.output3['text'] = (results[3])
            self.output4['text'] = (results[4])
            self.output5['text'] = (results[5])
            self.output6['text'] = (results[6])
            self.output7['text'] = (results[7])
            self.output8['text'] = (results[8])
            self.output9['text'] = (results[9])
            connection.close()

        else:
            self.output['text'] = "That name does not exist."

您必须创建一组LabelFrame、列表框或消息来显示结果

    self.output0 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output0.grid(column=0, row=6, columnspan=5, padx=0, pady=0)
    self.output1 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output1.grid(column=0, row=7, columnspan=5, padx=0, pady=0)
    self.output2 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output2.grid(column=0, row=8, columnspan=5, padx=0, pady=0)
    self.output3 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output3.grid(column=0, row=9, columnspan=5, padx=0, pady=0)
    self.output4 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output4.grid(column=0, row=10, columnspan=5, padx=0, pady=0)
    self.output5 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output5.grid(column=0, row=11, columnspan=5, padx=0, pady=0)
    self.output6 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output6.grid(column=0, row=12, columnspan=5, padx=0, pady=0)
    self.output7 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output7.grid(column=0, row=13, columnspan=5, padx=0, pady=0)
    self.output8 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output8.grid(column=0, row=14, columnspan=5, padx=0, pady=0)
    self.output9 = tkinter.LabelFrame(self, height=20, width=830, bg='#F7F7F7', text=' ', bd=0, labelanchor='n')
    self.output9.grid(column=0, row=15, columnspan=5, padx=0, pady=0)  

---

我不认为这是问题所在，但如果第二个查询（在

连接之前.close（）

）是有意的，那么在最后一个查询之后您似乎没有

else

？是的。我把它放在那里尝试它，我仍然得到标量变量错误。我认为这可能与用户输入被用作变量有关。它只允许我在第一次查询中使用它一次。我不认为这是问题所在，但如果第二次查询（在

连接之前.close（）

）是有意的，那么在最后一次查询之后您似乎没有

else

。是的。我把它放在那里尝试它，我仍然得到标量变量错误。我认为这可能与用户输入被用作变量有关。它只允许我在第一次查询中使用它一次。