Question

我正在尝试学习sed / awk，我打算在以下任务中使用它。我有一个命令打印出一个文件列表（每行可能超过一个），如下所示：

--- /section/1 ---
appname1/detail1/something appname1/detail2/somethingelse another/app/2.0
sillyapp/details/here  bug/2.5
--- /section2/details/here ---
apname2/3.2.5  apname2/3.2.6 apname3/something.0.4/here

我想做两件事：

（1）使用sed仅 文件的第一部分（从＆＃39;＆＃39;到＆＃39; /＆＃39;）以便我们拥有

--- /section/1 --- appname1 appname1 another sillyapp bug --- /section2/details/here --- apname2 apname2 apname3

（2）使用awk（我认为？）找出每个应用程序列出的次数，以便我们可以

appname1: 2 another: 1 sillyapp: 1 bug: 1 apname2: 2 apname3: 1

可以使用sed / awk吗？如果是这样，有人可以详细说明如何完成每个（任何原因）吗？

Answer 1

我将grep与-o一起使用以仅提取匹配项，并使用-P来获取与Perl兼容的正则表达式：

grep -Po '(^|\s)\K\w+(?=/)' file | sort | uniq -c

  1 another
  2 apname2
  1 apname3
  2 appname1
  1 bug
  1 sillyapp

正则表达式是：

(^|\s)  # either the beginning of the line, or a space
\K      # forget about what came before (i.e. don't remember the space)
\w+     # some word characters
(?=/)   # the next character is a slash (look-ahead)

sed：我不是大师，但我想出了这个：

sed -nr '/^---/d; s/(^| +)([^/]+)[^ ]+/\2 /g; H; ${x;s/\n//g;s/ $//; s/ /\n/g;p}' file

appname1
appname1
another
sillyapp
bug
apname2
apname2
apname3

那是

sed -nr '          # -n suppress printing; -r enable extended regular expressions
    /^---/d                      # delete "header" lines
    s/(^| +)([^/]+)[^ ]+/\2 /g   # extract the words you want, add a trailing space
    H                            # append this transformed line to the hold space
    ${                           # on the last line of input:
        g                        # bring the hold space contents into the pattern space
        s/\n//g                  # remove newlines
        s/ $//                   # remove a trailing space
        s/ /\n/g                 # change spaces into newlines
        p                        # and, finally, print the results
    }
' file

在此之后，按上述方式添加| sort | uniq -c

sed / awk采取＆＃39;单词的第一部分＆＃39;

1 个答案: