2017-07-15 50 views
-1

我想从IP Pool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。 我写这个剧本来执行:举报whois ip到csv文件

#!/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 

,但我的脚本生成此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,,,,,,,, 
... 

为什么第二个值是空的?

+0

“var =”后面不能有空格。 'var = cat whoisip | grep“netname”'会尝试*设置var等于一个字符串,而不是grep的输出。例如,使用NETNAME =“$(grep netname whoisip)”。请注意,变量名应该大写,grep的输出应该被引用。请参阅https://www.shellcheck.net/中的其他几个指针。 –

+0

我替换了NETNAME =“$(grep netname whoisip)”,但仍然是第二个值为空! – mbzadegan

+0

'netname'!='NETNAME' – Cyrus

回答

0

我试着把你的脚本:

#!/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 

``执行封闭突击队,并返回输出

-i使得grep的不区分大小写

变量名,等号和值变量之间声明和/或分配不允许空间:

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