Fiddle 的架构
结果是:我想获取 msr_name 为“PRE”或“POST”或两者的记录
id | msr_name1(only pre) | msr_data1(only pre) | msr_name2(only post) | msr_data2(only post)
16703 PRE pre_data POST post_data
16711 PRE pre_data NULL NULL
16715 NULL NULL POST post_data