Question

I am trying to execute the following command, setting “store”=”result”, for the row where the code column equals “code”.

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

I keep receiving the following error though:

_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")

Where the variables from the command were:

store=1087
result=1
code=Madness16

This is my first time really using mysql, so I am new to this. I’ve been stuck on this line for ~2 hours now, and cannot figure out what I am doing wrong. I tried the following command in mysql, and it ran correctly:

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

More code as requested:

# 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

You may want to try something like this instead, (assuming this is not prone to sql injection attacks – meaning the data is trusted and not user provided)

...
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]))