Question

我想解析一个Apache日志文件，例如：

1.1.1.1 - - [12/Dec/2019:18:25:11 +0100] "GET /endpoint1/ HTTP/1.1" 200 4263 "-" "Mozilla/5.0 (Windows NT 6.0; rv:34.0) Gecko/20100101 Firefox/34.0" "-"
1.1.1.1 - - [13/Dec/2019:18:25:11 +0100] "GET /endpoint1/ HTTP/1.1" 200 4263 "-" "Mozilla/5.0 (Windows NT 6.0; rv:34.0) Gecko/20100101 Firefox/34.0" "-"
2.2.2.2 - - [13/Dec/2019:18:27:11 +0100] "GET /endpoint1/ HTTP/1.1" 200 4263 "-" "Mozilla/5.0 (Windows NT 6.0; rv:34.0) Gecko/20100101 Firefox/34.0" "-"
2.2.2.2 - - [13/Jan/2020:17:15:13 +0100] "GET /endpoint2/ HTTP/1.1" 200 4263 "-" "Mozilla/5.0 (Windows NT 6.0; rv:34.0) Gecko/20100101 Firefox/34.0" "-"
3.3.3.3 - - [13/Jan/2020:17:15:13 +0100] "GET /endpoint2/ HTTP/1.1" 200 4263 "-" "Mozilla/5.0 (Windows NT 6.0; rv:34.0) Gecko/20100101 Firefox/34.0" "-"
1.1.1.1 - - [13/Feb/2020:17:15:13 +0100] "GET /endpoint2/ HTTP/1.1" 200 4263 "-" "Mozilla/5.0 (Windows NT 6.0; rv:34.0) Gecko/20100101 Firefox/34.0" "-"
4.4.4.4 - - [13/Feb/2020:17:15:13 +0100] "GET /endpoint2/ HTTP/1.1" 200 4263 "-" "Mozilla/5.0 (Windows NT 6.0; rv:34.0) Gecko/20100101 Firefox/34.0" "-"
4.4.4.4 - - [13/Feb/2020:17:15:13 +0100] "GET /endpoint2/ HTTP/1.1" 200 4263 "-" "Mozilla/5.0 (Windows NT 6.0; rv:34.0) Gecko/20100101 Firefox/34.0" "-"
4.4.4.4 - - [13/Feb/2020:17:15:13 +0100] "GET /endpoint2/ HTTP/1.1" 200 4263 "-" "Mozilla/5.0 (Windows NT 6.0; rv:34.0) Gecko/20100101 Firefox/34.0" "-"

我需要获取每月访问的客户端IP列表。我有这样的东西

awk '{print $1,$4}' access.log | grep Dec | cut -d" " -f1 | uniq -c

但这是错误的，因为它计算每天的访问IP。

预期结果应该像（缩进无关紧要）：

Dec 2019
1.1.1.1 2
2.2.2.2 1
Jan 2020
2.2.2.2 1
3.3.3.3 1
Feb 2020
4.4.4.4 3
1.1.1.1 1

其中2是截至2019年12月1.1.1.1 ip的总访问量。

您能建议我一种方法吗？

Answer 1

一个用于GNU awk的设备，它按照输入数据的顺序输出（即，按日志顺序输出时间顺序数据，例如日志记录）：

$ gawk '                     # using GNU awk
BEGIN {
    a[""][""]                # initialize a 2D array
}
{
    split($4,t,/[/:]/)       # split datetime 
    my=t[2] OFS t[3]         # my=month year
    if(!(my in mye)) {       # if current my unseen
        mye[my]=++myi        # update month year exists array with new index
        mya[myi]=my          # chronology is made
    }
    a[mye[my]][$1]++         # update record to a hash
}
END {                        # in the end
    # PROCINFO["sorted_in"]="@val_num_desc"  # this may work for ordering visits
    for(i=1;i<=myi;i++) {    # in fed order 
        print mya[i]         # print month year
        for(j in a[i])       # then related ips in no particular order
            print j,a[i][j]  # output ip and count
    }
}' file

输出：

Dec 2019
1.1.1.1 2
2.2.2.2 1
Jan 2020
2.2.2.2 1
3.3.3.3 1
Feb 2020
1.1.1.1 1
4.4.4.4 3

Answer 2

尽管您的示例预期输出看起来与所显示的示例不匹配，但根据您显示的示例输出和描述，您可以尝试以下方法。另外，由于这是一个日志文件，因此我将使用awk的字段分隔符方法，因为日志将采用固定模式。

awk -F':| |-|/+|]' '
{
  ind[$7 OFS $8 OFS $1]++
  value[$7 OFS $8 OFS $1]=$1
}
END{
  for(i in value){
    split(i,arr," ")
    print arr[1],arr[2] ORS value[i],ind[i]
  }
}' Input_file

说明： 添加以上详细说明。

awk -F':| |-|/+|]' '                             ##Starting awk program from here and setting field separators as : space - / ] here.
{
  ind[$7 OFS $8 OFS $1]++                        ##Creating ind array whose index is 7th 8th and 1st field and keep increasing value with 1 here.
  value[$7 OFS $8 OFS $1]=$1                     ##Creating value with index of 7th, 8th and 1st field and its value is 1st field.
}
END{                                             ##Starting END block of this program from here.
  for(i in value){                               ##Traversing through value elements here.
    split(i,arr," ")                             ##Splitting i into array arr with delimiter as space here.
    print arr[1],arr[2] ORS value[i],ind[i]      ##Printing 1st and 2nd element of arr with ORS(new line) and array value and ind value here.
  }
}' Input_file                                    ##Mentioning Input_file name here.

Answer 3

尝试一下。

外壳：

#!/usr/bin/env bash
LOG_FILE=$1

#regex to find mmm/yyyy
dateUniq=`grep -oP '(Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)\/\d{4}' $LOG_FILE | sort | uniq`


for i in $dateUniq
do  
    #output mmm yyyy
    echo $i | sed 's/\// /g'
    
    #regex to find ip
    ipUniq=`grep $i $LOG_FILE | grep -oP '(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)'  | sort | uniq`
    
    for x in $ipUniq
    do  
        count=`grep $i $LOG_FILE |grep -c $x`
        #output count ip
        echo $count $x
    done
    echo
done

输出：

Dec 2019
2 1.1.1.1
1 2.2.2.2

Feb 2020
1 1.1.1.1
3 4.4.4.4

Jan 2020
1 2.2.2.2
1 3.3.3.3

用bash解析Apache日志

3 个答案: