regex - 在 bash 中查找 2 个字符串中的指定字符串

Question

你好， 我有这样的文件:

today#123 
2934
9236
monday 

today#12341
4246
58234
monday

today#456
7768
32347
monday

但在我的文件中大约有 200k+ 行，但它是由以“今天”开始并以“星期一”结束的部分组成的

我可以轻松地将一个或所有部分分开:

awk '/today/ {show=1} show; /monday/ {show=0}' file.txt 

但我不知道如何找到带有特殊字符串的部分(在本例中为 7768).谁能帮我 ？

1.) 每个部分的行数都是随机的

2.) 文件不断变化(每天一次或两次)

结果应该是这样的:

    today#456
    7768
    32347
    monday

谢谢。

Answer 1

遵循 awk 也可能对您有所帮助。我正在设置一个名为 value 的变量，您可以在其中提供您想要查找的任何值，并且除了名为 value 的变量的值之外，也无需更改代码中的任何内容。

awk -v value="7768" '
/monday/ && flag{
  print;
  flag=val=""
}
/today/{
  val=$0;
  next
}
$0 ~ value{
  flag=1;
  print val RS $0;
  next
}
flag && val
'    Input_file

输出如下。

today#456
7768
32347
Monday

说明:现在也为上述代码添加说明。

awk -v value="7768" '  ##Creating a variable named value where OP could define its variable value which OP wants to search in any line.
/monday/ && flag{      ##Searching for a string monday in any line and variable flag is NOT NULL then do following:
  print;               ##printing the current line then.
  flag=val=""          ##Nullifying the values of variable flag and val here.
}
/today/{               ##Searching for a string today here if it is found on any line then do following.
  val=$0;              ##Assigning current line value to variable val here.
  next                 ##next is out of the box keyword of awk it will skip all further statements from here.
}
$0 ~ value{            ##Checking condition here if any line value is equal to variable value then do following:
  flag=1;              ##Making variable flag value to 1 or in other words making flag value to TRUE here.
  print val RS $0;     ##Printing the value of variable val with RS(record separator, whose default value is a new line) and current line then.
  next                 ##Mentioning next will skip all further statements now.
}
flag && val            ##checking condition here if variable flag and val is NOT NULL then do following.
'  Input_file          ##mentioning Input_file name here.