Question

试图找到一种方法来对100个文件的所有名称进行grep。 grepping每个文件中所有可用的名称必须出现在同一行。

FILE1

"company":"COMPANY1","companyDisplayName":"CM1","company":"COMPANY2","companyDisplayName":"CM2","company":"COMPANY3","companyDisplayName":"CM3",

FILE2

"company":"COMPANY99","companyDisplayName":"CM99"

我真正想要的输出是（包括文件名作为前缀。）

FILE1:COMPANY1,COMPANY2,COMPANY3    
FILE2:COMPANY99

我尝试了grep -oP '(?<="company":")[^"]*' *，但得到了这样的结果：

FILE1:COMPANY1
FILE1:COMPANY2
FILE1:COMPANY3
FILE2:COMPANY99

Answer 1

请您尝试以下。

awk -F'[,:]' '
BEGIN{
  OFS=","
}
{
  for(i=1;i<=NF;i++){
    if($i=="\"company\""){
      val=(val?val OFS:"")$(i+1)
    }
  }
  gsub(/\"/,"",val)
  print FILENAME":"val
  val=""
}
'   Input_file1  Input_file2

说明： 添加上述代码的说明。

awk -F'[,:]' '                          ##Starting awk program here and setting field separator as colon OR comma here for all lines of Input_file(s).
BEGIN{                                  ##Starting BEGIN section of awk here.
  OFS=","                               ##Setting OFS as comma here.
}                                       ##Closing BEGIN BLOCK here.
{                                       ##Starting main BLOCK here.
  for(i=1;i<=NF;i++){                   ##Starting a for loop which starts from i=1 to till value of NF.
    if($i=="\"company\""){              ##Checking condition if field value is equal to "company" then do following.
      val=(val?val OFS:"")$(i+1)        ##Creating a variable named val and concatenating its own value to it each time cursor comes here.
    }                                   ##Closing BLOCK for if condition here.
  }                                     ##Closing BLOCK for, for loop here.
  gsub(/\"/,"",val)                     ##Using gsub to gklobally substitute all " in variable val here.
  print FILENAME":"val                  ##Printing filename colon and variable val here.
  val=""                                ##Nullifying variable val here.
}                                       ##Closing main BLOCK here.
'  Input_file1  Input_file2             ##Mentioning Input_file names here.

输出如下。

Input_file1:COMPANY1,COMPANY2,COMPANY3
Input_file2:COMPANY99

编辑： ：在OP需要使用grep并希望从其输出中获得最终输出的情况下添加解决方案（尽管我建议使用{{ 1}}解决方案本身，因为我们没有使用多个命令或子外壳。

awk

Answer 2

有两种工具可以获取grep命令的输出，并以所需的方式对其进行重新格式化。第一个工具是GNU datamash。第二个是eBay tsv-summarize软件包中的tsv-utils（免责声明：我是作者）。两种工具都可以通过类似的方式解决此问题：

$ # The grep output
$ echo $'FILE1:COMPANY1\nFILE1:COMPANY2\nFILE1:COMPANY3\nFILE2:COMPANY99' > grep-output.txt
$ cat grep-output.txt
FILE1:COMPANY1
FILE1:COMPANY2
FILE1:COMPANY3
FILE2:COMPANY99

$ # Using GNU datamash
$ cat grep-output.txt | datamash -field-separator : --group 1 unique 2
FILE1:COMPANY1,COMPANY2,COMPANY3
FILE2:COMPANY99

$ # Using tsv-summarize
$ cat grep-output.txt | tsv-summarize --delimiter : --group-by 1 --unique-values 2 --values-delimiter ,
FILE1:COMPANY1,COMPANY2,COMPANY3
FILE2:COMPANY99

从多个文件中收集同一行上的所有字符串

2 个答案: