MySQL-将查询结果与Python中的字符串进行比较

Question

I would like to check whether the email returned by a query is empty. I am using the following code:

query = "select email from u_code where code = '"+code+"'"
cursor.execute(query)
result_set = cursor.fetchall()

length = cursor.rowcount
if (length==1):
   print ' result: ' + str(result_set[0]) + ' OK'
   print ' length of result: ' + str(len(result_set[0]))

   if (result_set[0] == ''):
       print("empty email")
       result = 1;
   else:
       print("email taken")
       result = 0


print "result: " + str(result)

The output is wrong. It should return 1 since the email field is empty

 result: (u'',) OK
 length of result: 1
 email taken
 result: 0

Any advice?

Answer 1

result_set is an array, which you access the first tuple: result_set[0] => (u'',) which indeed has a length of 1.

Accessing the first element (such as via result_set[0][0] ) would give you the empty string you're actually looking for.

Answer 2

.fetchall() returns a list of tuples. So your condition will never be met.

You can either do:

if (result_set[0][0] == ''):
    print("empty email")
    result = 1;
else:
    print("email taken")
    result = 0

Or more simply use .fetchone() :

result_set = cursor.fetchone()

if (result_set[0] == ''):
    print("empty email")
    result = 1;
else:
    print("email taken")
    result = 0

Also you should query your database like this to avoid the chance of SQL injection attacks:

query = "select email from u_code where code = %s"
cursor.execute(query,(code,))

Answer 3

For checking an existing value in table, i would recommend COUNT query

query = "SELECT COUNT(email) FROM u_code WHERE code = %s"
cursor.execute(query, [code,])

# COUNT query always return value, if no matching data 
# it simply return integer value 0, so we can safely take data
# from result[0]
result_check = cursor.fetchone()

if not result_check[0]:
    print("empty email")
    result = 1;
else:
    print("email taken")
    result = 0

print "result: " + str(result)