将 grep 结果折叠成一行，同时保留唯一值并删除重复值？

2024-6-6 • tag-icon

我正在使用，但也zsh愿意接受解决方案。bash假设我有以下命令：

ls **/*/assessment(.) | xargs egrep "(^  \[\./.*\]$|^    input = .*$)"

这给了我类似这样的输出：

<path-01-to>/assessment:  [./input_file_01]
<path-01-to>/assessment:    input = 'input_file_01.i'
<path-02-to>/assessment:  [./input_file_02]
<path-02-to>/assessment:    input = 'input_file_02.i'
<path-02-to>/assessment:  [./input_file_02b]
<path-02-to>/assessment:    input = 'input_file_02b.i'
<path-03-to>/assessment:  [./input_file_03]
<path-03-to>/assessment:    input = 'input_file_03.i'

我想折叠行，以便每[./title] input对 ( ) 位于同一行。因此，预期的输出将如下所示：

<path-01-to>/assessment:  [./input_file_01] input = 'input_file_01.i'
<path-02-to>/assessment:  [./input_file_02] input = 'input_file_02.i'
<path-02-to>/assessment:  [./input_file_02b] input = 'input_file_02b.i'
<path-03-to>/assessment:  [./input_file_03] input = 'input_file_03.i'

我看过 usingtr -d "\n"但它将所有内容合并到一行中。我在想awk或者不同的语言可能更适合这个。

编辑：

这是输入文件的示例assessment：

路径/01/到/评估：

[Tests]
  [./input_file_01]
    type = RunApp
    input = 'input_file_01.i'
  [../]
[]

路径/02/至/评估：

[Tests]
  [./input_file_02]
    type = RunApp
    input = 'input_file_02.i'
    cli_args = 'blah blah'
  [../]
  [./input_file_02b]
    type = CSVDiff
    input = 'input_file_02b.i'
  [../]
[]

答案1

这将在每个 UNIX 机器上的任何 shell 中使用任何 awk 来工作：

$ awk '/^ *\[\.\//{title=$1} /^ *input =/{print FILENAME ":", title, "=", $NF}' */*/assessment
01/to/assessment: [./input_file_01] = 'input_file_01.i'
02/to/assessment: [./input_file_02] = 'input_file_02.i'
02/to/assessment: [./input_file_02b] = 'input_file_02b.i'

答案1

相关内容