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有一個數據:perl的選擇特定的列並存儲到不同的陣列
set interfaces ge-1/0/2 unit 101 family inet address 10.187.132.3/27 vrrp-group 1 virtual-address 10.187.132.1
set interfaces ge-1/0/2 unit 102 family inet address 10.187.132.35/27 vrrp-group 2 virtual-address 10.187.132.33
set interfaces ge-1/0/2 unit 103 family inet address 10.187.132.67/27 vrrp-group 3 virtual-address 10.187.132.65
set interfaces ge-1/0/2 unit 104 family inet address 10.187.132.99/27 vrrp-group 4 virtual-address 10.187.132.97
set interfaces ge-1/0/2 unit 105 family inet address 10.187.132.131/27 vrrp-group 5 virtual-address 10.187.132.129
set interfaces ge-1/0/2 unit 106 family inet address 10.187.132.163/27 vrrp-group 6 virtual-address 10.187.132.161
set interfaces ge-1/0/2 unit 107 family inet address 10.187.132.195/27 vrrp-group 7 virtual-address 10.187.132.193
set interfaces ge-1/0/2 unit 108 family inet address 10.187.132.227/27 vrrp-group 8 virtual-address 10.187.132.225
我們需要提取一些列,並將其存儲到不同的變量。數據來自過濾文件後,因此它將是$ _。有什麼方法將它存儲到不同的陣列?
$address[0] = "10.187.132.3/27"
$vrrp-group[0] = "1"
$virtual-address[0] = "10.187.132.1"
$address[1] = "10.187.132.35/27"
$vrrp-group[1] = "2"
$virtual-address[1] = "10.187.132.33"
我採用分體式嘗試,但我不知道如何選擇上是很容易使用awk做perl的spesific柱(AWK {「打印$ 8」})。
@address = split(/\s+/, $_);
但它失敗了。
預期結果:
@address:
$VAR1 = '10.187.132.3/27'
$VAR2 = '10.187.132.35/27'
$VAR3 = '10.187.132.67/27'
$VAR4 = '10.187.132.99/27'
$VAR5 = '10.187.132.131/27'
$VAR6 = '10.187.132.163/27'
$VAR7 = '10.187.132.195/27'
$VAR8 = '10.187.132.227/27'
@ VRRP備份組:
$VAR1 = '1'
$VAR2 = '2'
$VAR3 = '3'
$VAR4 = '4'
$VAR5 = '5'
$VAR6 = '6'
$VAR7 = '7'
$VAR8 = '8'
@虛擬地址:
$VAR1 = '10.187.132.1'
$VAR2 = '10.187.132.33'
$VAR3 = '10.187.132.65'
$VAR4 = '10.187.132.97'
$VAR5 = '10.187.132.129'
$VAR6 = '10.187.132.161'
$VAR7 = '10.187.132.193'
$VAR8 = '10.187.132.225'
它如何在腳本里面實現? – rabka
@rabka:哪個腳本?這是一個單線。在'END'部分中,您可以使用哈希數組來完成數據所需的任何操作。 – Birei
是否可以將其插入到腳本中?我們如何做到這一點? – rabka