Regex 在bash中的2个字符串中查找指定字符串

你好,， 我有这样的文件：

但在我的文件中，大约有200000多行，但它是从“今天”开始到“星期一”结束的

我可以使用以下工具轻松分离一个或所有部分：

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

但我无法找到如何查找带有特殊字符串的节（在本例中为7768） . 有人能帮我吗

1.）每个部分都有随机数量的行

2.）文件不断变化（每天一到两次）

结果应该是这样的：

    today#456
    7768
    32347
    monday

多谢各位

使用awk：

awk 'show && c{
        if(show=$1==7768)print c;
        c=""
     }
     show;
     /monday/{
         show=0
     }
    /today/{
         show=1;
         c=$0
    }
    ' infile

输出：

$ awk 'show && c{if(show=$1==7768)print c;c=""}show;/monday/{show=0}/today/{show=1;c=$0}' infile
today#456
7768
32347
monday

输入：

$ cat infile
today#123 
2934
9236
monday 

today#12341
4246
58234
monday

today#456
7768
32347
monday

---

您可以为此编写bash脚本，比如

recordfinder.sh

。它可以如下所示：

# cat recordfinder.sh
#!/bin/bash
exitfn(){
echo "Usage : recordfinder.sh <filename> <searchstring>"
[ "$1" -eq 1 ] && echo "Couldn't open file" && exit 1
[ "$1" -eq 2 ] && echo "No search string provided" && exit 2
}
[ -f "$1" ] || exitfn 1
[ -z "$2" ] && exitfn 2 
awk -v str="$2" -v RS="" '$0 ~ str'  "$1"

# ./recordfinder.sh filename 7768
today#456
7768
32347
monday

#cat recordfinder.sh
#!/bin/bash
exitfn（）{
echo“用法：recordfinder.sh”
[“$1”-eq 1]&&echo“无法打开文件”&&exit 1
[“$1”-等式2]&&echo“未提供搜索字符串”&&exit 2
}
[-f“$1”]| |退出1
[-z“$2”]&退出2
awk-v str=“$2”-v RS=“”$0~str”“$1”
#./recordfinder.sh文件名7768
今天#456
7768
32347
星期一

---

希望它能给你一些灵活性：-）

下面的

awk

也可以帮助你。我正在设置一个名为

value

的变量，在这个变量中，除了名为value的变量的值之外，您可以提供您想要查找的任何值，并且不需要在代码内部进行任何更改

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

解释：现在也为上述代码添加解释

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.

测试：

---

$sed-n'/today/{:a；/monday/{\n4246\n/p；b}；n；ba}'
'/today/{show=1}（show^^$0~/srchTarget/）？祝你好运。它不起作用了：/它显示语法错误，awk不知道^^^，我尝试只使用一个，但没有匹配：（也很抱歉打扰您，但您能解释一下^^^^应该怎么做吗？^抱歉，可能是个愚蠢的问题，但我是一个新来的，在第1行中找不到关于2^^^^^的任何信息。对不起，输入错误，
^^
应该是
&
。您好，Akshay，这看起来很酷，但它只有在字符串位于第一行时才有效。）在“今天”之后加上e。字符串可以在“今天”和“星期一”之间的任意位置，你也可以解释它为什么有效（仅适用于第一行）？对不起，我试图理解，但对初学者来说并不容易：/谢谢！！！这对我的目的来说非常有效！！但是，你也能解释为什么吗？有一个想法，awk显示了以“今天”开头的部分第二行是带“value”的行（忽略“today”和“value”之间的所有行），然后正常继续。但为了我的目的，我需要知道第一行以“today”开头的完整字符串
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.

                    sed -n '/today/{:a;/monday/{/\n4246\n/p;b};N;ba}'
                         ^     ^     ^    ^          ^    ^ ^  ^  ^
                         |     |     |    |          |    | |  |  |
dont print all lines-----+     |     |    |          |    | |  |  |
                               |     |    |          |    | |  |  |
if found start of block (today)+     |    |          |    | |  |  |
  then start loop with label (a)-----+    |          |    | |  |  |
  if found end of block (monday)----------+          |    | |  |  |
    then check if patterm (4246) found---------------+    | |  |  |
      if found, then print this buffer--------------------+ |  |  |
    break the loop------------------------------------------+  |  |
  load another line into buffer--------------------------------+  |
  and loop (goto label (a))---------------------------------------+

$ sed -n '/today/{:a;/monday/{/\n4246\n/p;b};N;ba}' <sample.txt 
today#12341
4246
58234
monday