awk比較7個文件，打印匹配和區別根據提起：

我需要comapre 7個文件Ref.txt和Jan.txt到Jun.txt並獲得匹配，並且不匹配，對於這種情況我期待檢查Ref.txt的第二個字段，將Jan.txt的所有第一個字段設置爲Jun.txt，如果是，則打印Ref.txt（主轉儲）的所有文件，，然後將Jan.txt的整行打印到Jun。文本。並且在Jan.txt和Jun.txt之間找不到匹配來聲明「NotFound」。awk比較7個文件，打印匹配和區別根據提起：

Ref.txt

abc 10 xxyyzz 
bdc 20 xxyyzz 
edf 30 xxyyzz 
ghi 40 xxyyzz 
ofg 50 xxyyzz 
mgf 60 xxyyzz

Jan.txt

10 Jan 100 
30 Jan 300 
50 Jan 500

Feb.txt

Mar.txt

20 Mar 600 
50 Mar 1500

Apr.txt

10 Apr 100 
30 Apr 300 
50 Apr 500

May.txt

10 May 200 
20 May 400 
40 May 800 
60 May 1200

Jun.txt

20 Jun 600 
50 Jun 1500

所需的輸出：

Ref.txt Ref.txt Ref.txt Jan.txt Jan.txt Jan.txt Feb.txt Feb.txt Feb.txt Mar.txt Mar.txt Mar.txt Apr.txt Apr.txt Apr.txt May.txt May.txt May.txt Jun.txt Jun.txt Jun.txt 
abc 10 xxyyzz 10 Jan 100 10 Feb 200 Notfound Notfound Notfound 10 Apr 100 10 May 200 Notfound Notfound Notfound 
bdc 20 xxyyzz Notfound Notfound Notfound 20 Feb 400 20 Mar 600 Notfound Notfound Notfound 20 May 400 20 Jun 600 
edf 30 xxyyzz 30 Jan 300 Notfound Notfound Notfound Notfound Notfound Notfound 30 Apr 300 Notfound Notfound Notfound Notfound Notfound Notfound 
ghi 40 xxyyzz Notfound Notfound Notfound 40 Feb 800 Notfound Notfound Notfound Notfound Notfound Notfound 40 May 800 Notfound Notfound Notfound 
ofg 50 xxyyzz 50 Jan 500 Notfound Notfound Notfound 50 Mar 1500 50 Apr 500 Notfound Notfound Notfound 50 Jun 1500 
mgf 60 xxyyzz Notfound Notfound Notfound 60 Feb 1200 Notfound Notfound Notfound Notfound Notfound Notfound 60 May 1200 Notfound Notfound Notfound

預先感謝您的回覆

來源

2014-01-27 SVR

謝謝你回想起來，爲您的文章的可讀性。 – Noctua

謝謝你的有趣問題。我們無聊到死。 – devnull

爲什麼在預期的輸出中沒有'4月10日100'？ –

這裏有一個禮物：請向東西問題，你不明白

awk ' 
    FNR == 1 { 
     printf "%s %s %s\t", FILENAME, FILENAME, FILENAME 
     if (NR > FNR) file[++num_files] = FILENAME 
    } 
    NR == FNR { 
     id[NR] = $2 
     ref[NR] = $0 
     num_ids++ 
     next 
    } 
    { value[FILENAME,$1] = $0 } 
    END { 
     print "" 
     for (row=1; row<=num_ids; row++) { 
      printf "%s\t", ref[row] 
      for (f=1; f<=num_files; f++) { 
       key = file[f] SUBSEP id[row] 
       printf "%s\t", (key in value ? value[key] : "Notfound") 
      } 
      print "" 
     } 
    } 
' {Ref,Jan,Feb,Mar,Apr,May,Jun}.txt

Ref.txt Ref.txt Ref.txt Jan.txt Jan.txt Jan.txt Feb.txt Feb.txt Feb.txt Mar.txt Mar.txt Mar.txt Apr.txt Apr.txt Apr.txt May.txt May.txt May.txt Jun.txt Jun.txt Jun.txt 
abc 10 xxyyzz 10 Jan 100 10 Feb 200 Notfound 10 Apr 100 10 May 200 Notfound  
bdc 20 xxyyzz Notfound 20 Feb 400 20 Mar 600 Notfound 20 May 400 20 Jun 600 
edf 30 xxyyzz 30 Jan 300 Notfound Notfound 30 Apr 300 Notfound Notfound  
ghi 40 xxyyzz Notfound 40 Feb 800 Notfound Notfound 40 May 800 Notfound  
ofg 50 xxyyzz 50 Jan 500 Notfound 50 Mar 1500 50 Apr 500 Notfound 50 Jun 1500  
mgf 60 xxyyzz Notfound 60 Feb 1200 Notfound Notfound 60 May 1200 Notfound

來源

2014-01-27 16:26:13

+1不錯。它仍然令人困惑，爲什麼OP的每一行預期輸出包含「1 + 6 * 3 = 19」元素（而您的解決方案包含「1 + 6 = 7」）:) –

非常感謝Glenn Jackman爲您及時提供幫助＆禮物，它工作正常.. – SVR

awk比較7個文件，打印匹配和區別根據提起：

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