python mysql 2014 Commands out of sync; you can't run this command now

这个问题出现再 mysql和c 的api。

简单的解决方法是不使用api直接把整个连接和命令传过去。

例如，cmd = 'mysql -h 192.168.32.210 -P 3316 -u bfdroot -pqianfendian -D DMP_GDMP_Cbehe -e "%s"' % update_sql2

问题原因：

Mysql文档：Commands out of sync
If you get Commands out of sync; you can't run this command now in
your client code, you are calling client functions in the wrong order.
This can happen, for example, if you are using mysql_use_result() and
try to execute a new query before you have called mysql_free_result().
It can also happen if you try to execute two queries that return data without calling mysql_use_result() or mysql_store_result() in
between.

第一种 存储结果未释放，然后又查询
第二种两次查询之间没有存储结果

解决方案：

do

{

result = mysql_store_result( mysql );

mysql_free_result(result);

}while( !mysql_next_result( mysql ) );

例子 http://www.linuxidc.com/Linux/2013-04/82619.htm

: Update OK!sql: insert into t_alarm_record_file (recordPath,recordName,hostIp,startTime,endTime,deviceId,programNumber,deviceType,interfaceNo,alarmType,alarmTime) values ('/figure/data/AlarmRecord/StreamTS/1-码流_魅力音 乐主路/2011-11-07/20111107150116.ts','','10.0.60.2','2011-11-07
15:01:16','1970-01-01 08:00:00',37486602,3905,'0',1,12,'2011-11-07 15:01:20');update t_alarm set fileId = LAST_INSERT_ID() where deviceId = 37486602 and programNumber = 3905 and alarmType = 12 and alarmDate = '2011-11-07 15:01:20' and fileId is null

2011-11-07 15:01:35,331: ERROR : Update failed!sql: insert into t_alarm_record_file (recordPath,recordName,hostIp,startTime,endTime,deviceId,programNumber,deviceType,interfaceNo,alarmType,alarmTime) values ('/figure/data/AlarmRecord/StreamTS/1-码流_魅力音 乐主路/2011-11-07/20111107150116.ts','','10.0.60.2','2011-11-07
15:01:16','1970-01-01 08:00:00',37486602,3905,'0',1,13,'2011-11-07 15:01:27');update t_alarm set fileId = LAST_INSERT_ID() where deviceId = 37486602 and programNumber = 3905 and alarmType = 13 and alarmDate = '2011-11-07 15:01:27' and fileId is null, ERROR:Commands
out of sync; you can't run this command now

在第一次调用Update()执行多条sql语句时成功，但以后的所有调用都失败了。



经过查找，发现问题在于：在Update()中执行多条sql语句时，

如果仅仅是插入等不需要返回值的SQL语句，也一样得读完整个resault集并释放，最小化的写法：



do

{

result = mysql_store_result( mysql );

mysql_free_result(result);

}while( !mysql_next_result( mysql ) );



经过这么一处理，问题终于解决了。