Question

尝试开发一些代码，以便在按下按钮时我可以返回information_Schema的内容并允许自己从生成的列表中选择一个数据库，创建按钮并请求登录不是问题而且有效，然而，返回结果是失败的，我到目前为止创建的代码是：

def mysqlConnect():
    import pymysql
    import subprocess
    sqlUsr = MysqlUsr.get()
    sqlpwd = Mysqlpwd.get()
    conn = pymysql.connect(host='192.168.0.27', user= sqlUsr, passwd=sqlpwd, db='information_schema')
    cursor = conn.cursor()
    conn.query("SELECT SCHEMA_NAME FROM SCHEMATA")
    data = cursor.fetchall()
    print (data)

行data = conn.fetchall()似乎给出了与括号()相关的错误，因为代码必须首先处理它，但我不明白为什么，我看到的所有示例都有这种语法？我想我需要schema_name中的行进入元组，所以我可以将这些信息用作“下拉”选择框？有没有人做过类似的事情？在我可以返回行之前，我无法创建下拉列表，此时我可以返回的是fetchall()命令失败时的行数。

Exception in Tkinter callback Traceback (most recent call last):
File "C:\Python33\lib\tkinter_init_.py", line 1475, in
    call return self.func(*args)
File "S:\python\jon\wrt_toolkit_v5\wrt_toolkit_v6.py", line 97, in
    mysqlConnect data = cursor.fetchall()
File "C:\Python33\lib\site-packages\pymysql3-0.4-py3.3.egg\pymysql\cursors.py", line 194,
    in fetchall self._check_executed()
File "C:\Python33\lib\site-packages\pymysql3-0.4-py3.3.egg\pymysql\cursors.py", line 64,
    in _check_executed self.errorhandler(self, ProgrammingError, "execute() first")
File "C:\Python33\lib\site-packages\pymysql3-0.4-py3.3.egg\pymysql\connections.py", line 184,
    in defaulterrorhandler raise errorclass(errorvalue)
pymysql.err.ProgrammingError: execute() first

Answer 1

您需要在致电cursor.execute(query, args=None)

之前致电fetch_all

python查询mysql添加结果下拉选择？

1 个答案: