Question

我有一个日志文件如下。每行记录一些字符串和线程ID。每个主题属于进程，进程可以包含N个主题。

基于以下示例，我想提取（使用bash工具，grep，sed等等）属于给定进程的所有线程的所有行。请注意，该过程仅在线程序列的顶部提及一次：

line1 thread= 150 process= 200
line2 thread= 152 whatever
line3 thread= 150 whatever
line4 thread= 150 whatever
line5 thread= 130 whatever
line6 thread= 130 process= 200
line7 thread= 150 process= 201
line8 thread= 130 whatever
line9 thread= 130 whatever

对于此示例，请提供进程200，输出应为：

line1 thread= 150 process= 200
line3 thread= 150 whatever
line4 thread= 150 whatever
line6 thread= 130 process= 200
line8 thread= 130 whatever
line9 thread= 130 whatever

Answer 1

awk 解决方案：

filter_threads.awk 脚本：

#!/bin/awk -f
function get_thread(s){           # extracts thread number from the string
    t = substr(s,index(s,"=")+1); # considering `=` as separator (e.g. `thread=150`) 
    return t; 
} 
BEGIN { 
    pat = "process="p   # regex pattern to match the line with specified process
}
$3~pat {    # on encountering "process" line
    thread = get_thread($2); print; next   # getting base thread number 
}
{ 
    t = get_thread($2); 
    if (t==thread) print  # comparing current thread numbers with base thread number
}

用法：

awk -f filter_threads.awk -v p=200 yourfile

- 其中p是进程号

输出：

line1 thread=150 process=200
line3 thread=150 whatever
line4 thread=150 whatever
line6 thread=130 process=200
line8 thread=130 whatever
line9 thread=130 whatever

<强> 更新：

当您更改初始输入时，新解决方案如下：

awk -v p=200 '$4~/process=/ && $5==p{ thread=$3; print; next }$3==thread{ print }' yourfile

bash命令根据起始键

1 个答案: