Question

我正在尝试执行以下命令，为代码列等于“code”的行设置“store”=“result”。

cursor.execute("""UPDATE pjStores SET %s=%s WHERE code=%s""", (store, result, code))

我仍然收到以下错误：

_mysql_exceptions.ProgrammingError: (1064, "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''1087'=1 WHERE code='Madness16'' at line 1")

命令中的变量是：

store=1087
result=1
code=Madness16

这是我第一次真正使用mysql，所以我是新手。我现在已经被困在这条线上约2个小时了，无法弄清楚我做错了什么。我在mysql中尝试了以下命令，它运行正常：

UPDATE pjStores SET `1087`=1 WHERE code='Madness16'

根据要求提供更多代码：

# Step through stores
cursor.execute("SHOW COLUMNS FROM pjStores")
stores = cursor.fetchall()
cursor.execute("SELECT code FROM pjStores")
codes = cursor.fetchall()

for store in stores[1:]:    # Hack to skip first entry
    pj.setStore(store[0])
    for code in codes:
        result = pj.checkCode(code[0])
        cursor.execute ("""UPDATE pjStores SET %d=%s WHERE code=%s""", (store[0],     result, code[0]))

Answer 1

你可能想尝试这样的事情，（假设这不容易受到sql注入攻击 - 意味着数据是可信的而不是用户提供的）

...
for code in code:
    result = pj.checkCode(code[0])
    query = """UPDATE pjStores SET `%s` = %%s WHERE `code` = %%s""" % store[0]
    cursor.execute(query, (result, code[0]))

Answer 2

尝试使用'代替"：

cursor.execute('''UPDATE pjStores SET %s=%s WHERE code=%s''', (store, result, code))

Python Mysql更新执行命令

2 个答案: