检查IP阵列与Bash

我想写一个bash脚本，将采取谁的输出，解析它与awk拉IP地址，然后检查它与IP地址阵列......听起来很简单，但他们脚本我放在一起似乎没有工作，我已经在OSX和Ubuntu上测试它，任何人都可以找到原因吗？！检查IP阵列与Bash

ip_address_is_safe() { 
local address_to_test=$1; 
for safe_ip in "${safe_ips[@]}" 
do 
    if [[ $safe_ip == $address_to_test ]]; 
    then 
    return 0; 
    fi 
done; 
return 1; 
} 

who="root pts/0  2011-10-03 23:13 (99.99.999.999) 
root pts/0  2011-10-03 23:13 (12.12.121.121) 
root pts/0  2011-10-03 23:13 (14.14.141.141) 
root pts/0  2011-10-03 23:13 (127.0.0.1) 
"; 

safe_ips=("(14.14.141.141)" "(127.0.0.1)") 


old_ifs=$IFS; 
export IFS=" 
"; 
for word in $who; do 
    remote_connected_ip=`echo $word | awk '/(23)/ {print $5}'`; 
    if (ip_address_is_safe "$remote_connected_ip") 
    then 
     echo "ip was ok - $remote_connected_ip" 
    else 
     echo "ip was not ok - $remote_connected_ip" 
    fi 
done;

它使报告每个IP为 “知识产权是不正常”

干杯3

来源

2011-10-04 RailsSon

注意到和固定的错字 – RailsSon

你没有引用您的变量，所以AWK是没有看到什么你认为它看到：你需要

for word in "$who"; do 
    remote_connected_ip=`echo "$word" | awk '/(23)/ {print $5}'`

这可能是更清洁，不需要AWK：

who | while read name tty date time ip; do 
    if ip_address_is_safe $ip; then 
    echo $ip ok 
    else 
    echo $ip not ok 
    fi 
done

来源

2011-10-04 10:57:17

if [[ $safe_id == ...

应该是：

if [[ $safe_ip ==

鉴于AWK参与，你可以做到这一切awk：

who | 
    awk 'BEGIN { 
    n = split("(14.14.141.141) (127.0.0.1)", t) 
    for (i = 0; ++i <= n;) safe_ips[t[i]] 
    } 
    $NF { 
    print "ip was", ($NF in safe_ips ? x : "not"), "OK" 
    }'

来源

2011-10-04 10:37:24

检查IP阵列与Bash

回答

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