Question

我想从IP池文件中读取ip并从中获取whois，然后将其报告给CSV电子表格文件。我写了这个脚本来执行：

#!/bin/bash
echo "ip,netname,org-name,remarks,descr,country,person,address,phone,origin" > csv
while read -r ip
do
whois $ip > whoisip
netname= cat whoisip | grep "netname"
orgname= cat whoisip | grep "org-name"
remarks= cat whoisip | grep "remarks"
descr= cat whoisip | grep "descr"
country= cat whoisip | grep "Country"
person= cat whoisip | grep "person"
address= cat whoisip | grep "address"
phone= cat whoisip | grep "phone"
origin= cat whoisip | grep "origin"
echo $ip,$netname,$orgname,$remarks,$descr,$country,$person,$address,$phone,$origin >> csv
done <pool

但是，我的脚本生成了这个CSV文件：

ip,netname,org-name,remarks,descr,country,person,address,phone,origin
x.x.x.x,,,,,,,,
y.y.y.y,,,,,,,,
z.z.z.z,,,,,,,,
...

为什么第二个值为空？

Answer 1

我试图修复你的脚本：

#!/bin/bash
echo "ip,netname,org-name,remarks,descr,country,person,address,phone,origin" > csv
while read -r ip
do
    whois $ip > whoisip
    netname=`cat whoisip | grep -i "netname"`
    orgname=`cat whoisip | grep -i "org-name"`
    remarks=`cat whoisip | grep -i "remarks"`
    descr=`cat whoisip | grep -i "descr"`
    country=`cat whoisip | grep -i "Country"`
    person=`cat whoisip | grep -i "person"`
    address=`cat whoisip | grep -i "address"`
    phone=`cat whoisip | grep -i "phone"`
    origin=`cat whoisip | grep -i "origin"`
    echo $ip,$netname,$orgname,$remarks,$descr,$country,$person,$address,$phone,$origin >> csv
done <pool

``执行随附的commando并返回输出

-i使grep不区分大小写

在变量名称，等号和变量声明和/或赋值中，不允许空格：

var=value  #Correct -> var has the value value
var= value #Incorrect -> var is empty

报告whois ip到csv文件

1 个答案: