我认为上次人们不愿意帮助我,因为我不想发布专栏,但我很想解决这个问题,所以我放弃了我更好的判断并把它暴露出来。
好的,我有两个相似的表,我使用 MySQL 和 PHP 来生成表。到目前为止,我有 3 张桌子:
CPE:
fastethernet00,subnet00,fastethernet01,subnet01,hsrp,vip,bgp,bgp_neighbor,remote_as,bgp_customer_net,next_hop,loopback,cpe,hostname,cpe_cust_index,int_next_hop_1,int_next_hop_2,voice,solution
Liverpool:
apn1,apn2,apn3,apn4,apn5,customer,vpi_vci,vlan1,cust_index,RADIUS,nexthop,atinterface1,atinterface2,feinterface1,feinterface2,spinterface1,spinterface2,ip_address_range1,ip_address_range2,handset_address_range1,handset_address_range2,handset_address_range3,handset_address_range4,handset_address_range5,handset_address_range6,handset_address_range7,handset_address_range8,handset_address_range9,handset_address_range10,handset_address_range11,handset_address_range12,handset_address_range13,handset_address_range14,handset_address_range15,handset_address_range16,DNS_Server1,DNS_Server2,OLVDMVPN,live
Greenwich:
apn_1,apn_2,apn_3,apn_4,apn_5,customer_,vpi_vci_,vlan_1,cust_index_,RADIUS_,nexthop_,atinterface_1,atinterface_2,feinterface_1,feinterface_2,spinterface_1,spinterface_2,ip_address_range_1,ip_address_range_2,handset_address_range_1,handset_address_range_2,handset_address_range_3,handset_address_range_4,handset_address_range_5,handset_address_range_6,handset_address_range_7,handset_address_range_8,handset_address_range_9,handset_address_range_10,handset_address_range_11,handset_address_range_12,handset_address_range_13,handset_address_range_14,handset_address_range_15,handset_address_range_16,DNS_Server_1,DNS_Server_2,OLVDMVPN_,live_
我需要完成的第一步是通过 apn_1/apn1 将格林威治连接到利物浦。相同的记录不一定在两者上,甚至可能存在重复。我用来创建这个表的 php 在 UNION ALL 周围抛出了一个语法错误:
mysql_query ("CREATE TABLE Both (
SELECT * FROM `Greenwich` left outer join Liverpool on Greenwich.apn_1 = Liverpool.apn1
UNION ALL
SELECT * FROM Greenwich
RIGHT OUTER JOIN Liverpool on Greenwich.apn_1 = Liverpool.apn1 where Greenwich.apn_1 is null)")
谁能看出这有什么问题?
之后,我需要通过 cust_index/cust_index_ 将结果表链接到 cpe_cust_index,所以理想情况下我认为我需要合并这些单元格。如果这些细胞存在于格林威治和利物浦,它们将永远是相同的。
提前致谢!