Question

我已经尝试了多次尝试在多个文件中grep ip地址的前三个八位字节，但是当第一个八位字节是一个或两个数字时，它有时会捕获误报。

shell为“SH”

（示例INPUT）

i = 5.79.78

$ data中的for i收集了需要处理的多个IP地址。为简单起见我只显示一个“i”值。

码

max=5
data="$(cat $dir* | cut -d '.' -f 1-3 | awk -v max="$max" '{a[$0]++}END{for(i in a){if(a[i] > max){print i}}}' | sed 's/.*://')"

for i in $data; do
    ii="${i}."
    list=$(grep "$ii" $dir*)
done
echo "$list"

* 输出*

/tmp/aaa.txt:5.79.78.237
/tmp/aaa.txt:5.79.78.230
/tmp/aaa.txt:5.79.78.229
/tmp/aaa.txt:5.79.78.228
/tmp/aaa.txt:5.79.78.227
/tmp/ddd.txt:5.79.78.236
/tmp/sss.txt:95.79.78.95     False
/tmp/sss.txt:95.79.78.66     False
/tmp/sss.txt:95.79.78.48     False
/tmp/sss.txt:95.79.78.216    False
/tmp/sss.txt:95.79.78.163    False

但它在我正在寻找的地址之前用数字拉取IP地址。我试图添加一个尾随“。”但它仍然无法找到所有正确的匹配。

尝试

list=$(grep -E "$i""\.[0-9]{1,3}" $dir*)
list=$(grep -e ^ "$ii" $dir*)
list=$(grep "\<$ii" $dir*)
list=$(grep "^$ii" $dir*)

(Also tried list="$grep "^$ii" $dir*)" (adding quotes)

Answer 1

供您输入

while read -r input
do
    if [ -z "$input" ]
    then
        continue
    fi
    ip3=`echo "$input." | sed 's/.*= *//'`
    regex=`echo $ip3 | sed 's/\./\\\./g'`
    grep -H "^$regex" /tmp/*.txt
done <<EOF
i = 5.79.78

i = 5.34.244
EOF

如果/tmp/a.txt包含

5.79.78.236
95.79.78.95

将打印

/tmp/a:5.79.78.236

Answer 2

#!/bin/sh

data="5.79.78 5.34.244"
octet_re='([1-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])'
n=0    
for prefix in $data; do
    n=$(expr $n + 1)
    echo "run $n"
    prefix=$( echo "$prefix" | sed 's/\./\\./g' )
    output=$( egrep "\<${prefix}\.${octet_re}\>" <<END

yes  5.79.78.123
no  95.79.78.124
no   5.34.244.256
no  15.34.244.127
yes  5.34.244.1
no   5-34.244.1
END
    )
    echo "$output"
done

run 1
yes  5.79.78.123
run 2
yes  5.34.244.1

Grep前三个八位字节的IP地址

2 个答案: