我必须将数百万行更新为MySQL。我目前正在使用for循环来执行查询。为了使更新更快,我想使用Python MySQL Connector的executemany()
,这样我就可以使用单个查询批量更新每个批次。
答案 0 :(得分:6)
我不认为mysqldb有一种方法可以同时处理多个UPDATE查询。
但是你可以在最后使用ON DUPLICATE KEY UPDATE条件的INSERT查询。
为了便于使用和提高可读性,我编写了以下示例。
import MySQLdb
def update_many(data_list=None, mysql_table=None):
"""
Updates a mysql table with the data provided. If the key is not unique, the
data will be inserted into the table.
The dictionaries must have all the same keys due to how the query is built.
Param:
data_list (List):
A list of dictionaries where the keys are the mysql table
column names, and the values are the update values
mysql_table (String):
The mysql table to be updated.
"""
# Connection and Cursor
conn = MySQLdb.connect('localhost', 'jeff', 'atwood', 'stackoverflow')
cur = conn.cursor()
query = ""
values = []
for data_dict in data_list:
if not query:
columns = ', '.join('`{0}`'.format(k) for k in data_dict)
duplicates = ', '.join('{0}=VALUES({0})'.format(k) for k in data_dict)
place_holders = ', '.join('%s'.format(k) for k in data_dict)
query = "INSERT INTO {0} ({1}) VALUES ({2})".format(mysql_table, columns, place_holders)
query = "{0} ON DUPLICATE KEY UPDATE {1}".format(query, duplicates)
v = data_dict.values()
values.append(v)
try:
cur.executemany(query, values)
except MySQLdb.Error, e:
try:
print"MySQL Error [%d]: %s" % (e.args[0], e.args[1])
except IndexError:
print "MySQL Error: %s" % str(e)
conn.rollback()
return False
conn.commit()
cur.close()
conn.close()
一个衬垫的说明
columns = ', '.join('`{}`'.format(k) for k in data_dict)
与
相同column_list = []
for k in data_dict:
column_list.append(k)
columns = ", ".join(columns)
以下是使用示例
test_data_list = []
test_data_list.append( {'id' : 1, 'name' : 'Marco', 'articles' : 1 } )
test_data_list.append( {'id' : 2, 'name' : 'Keshaw', 'articles' : 8 } )
test_data_list.append( {'id' : 3, 'name' : 'Wes', 'articles' : 0 } )
update_many(data_list=test_data_list, mysql_table='writers')
查询输出
INSERT INTO writers (`articles`, `id`, `name`) VALUES (%s, %s, %s) ON DUPLICATE KEY UPDATE articles=VALUES(articles), id=VALUES(id), name=VALUES(name)
值输出
[[1, 1, 'Marco'], [8, 2, 'Keshaw'], [0, 3, 'Wes']]
答案 1 :(得分:2)