使用find和sed将文件名添加到文件的开头 [英] add filename to beginning of file using find and sed

问题描述

使用以下我将文件名称添加到每行的前面，并将输出发送到单个文件。

  ls |同时读取文件;做sed -es / ^ / $ file / g$ file>出;完成
  

我想执行相同的 sed 替换，但使用 find 和 exec 或 xargs 命令 -

  find。 -type f -exec seds / ^ / {} / g{}> out + 
  

但我得到一个错误 -

< blockquote>

find：只有一个{}实例支持-exec ... +

输入文件是这样的 -

fileA.txt

  A1 
 A2 
  

fileB.txt

  B1 
 B2 
  

所需输出

  fileA.txt A1 
 fileA.txt A2 
 fileB.txt B1 
 fileB.txt B2 
  

我知道如何用awk做这个，但是我想用sed来做， find和exec或xargs。

解决方案

  find。 -type f | xargs awk'$ 0 = FILENAME $ 0'>出
  

当我回答这个问题时，你的不awk行还没有。无论如何，请看下面我更新的答案：

根据评论更新

所以你想用find，exec / xargs和sed来做。我的脚本需要 GNU Sed ，我希望你能拥有它。

首先看一个班轮：（好，> ; out 被省略了，你可以把它添加到行尾。）

  find。 -type f | xargs -i echo {} | sed -r's＃（。\ /）(.*)# cat& \ | seds：^：file \ 2：g#ge'
 
 
 
现在让我们进行一个测试，见下：
 
 
  kent $ head * .txt 
 ==> a.txt< == 
 A1 
 A2 
 
 ==> b.txt< == 
 B1 
 B2 
 
 kent $ find。 -type f | xargs -i echo {} | sed -r's＃（。\ /）(.*)# cat& \ | seds：^：file \ 2：g#ge'
文件b.txt B1 
文件b.txt B2 
文件a.txt A1 
文件a.txt A2 
  
是你期望的结果吗？  
 
 
 简短的解释 
 
 
 
 
  find .... | xargs -i echo {} 没有什么需要解释的，只需要在每行打印
文件名（前导./ code>）
 
 
然后将文件名传递给一个sed行，如 sed -r's＃（。\ / /（（*）） 
 #ge' 
 
 
请记住，在上面的行中，我们有两个组 \ 1：./ 和
  \ 2a.txt（文件名） 
 
 
在$ sed行结尾的 e ， MAGIC 部分将作为shell命令执行
（需要GNU sed）
 > 
 
  MAGIC ： cat& \ | seds：^：file \ 2：g  cat& amp ;只是输出文件
内容，并管道到另一个sed。做替换（ s：..：..：g ）
 
 
最后，MAGIC的执行结果将会是
的外部替代。
 
 
 
 
 
关键是Gnu sed的e / p> 
using the following I add the file name to the front of each line and send the output to a single file. 
ls | while read file; do sed -e "s/^/$file/g" $file > out; done
I want to perform the same sed replacement but using a find and exec or xargs  command - 
find . -type f -exec sed "s/^/{}/g" {} > out +
but I get an error - 

  
find: Only one instance of {} is supported with -exec ... +
Input files are like this -

fileA.txt
A1
A2
fileB.txt
B1
B2
desired output 
fileA.txt A1
fileA.txt A2
fileB.txt B1
fileB.txt B2
I know how to do this with awk, but I'd like to do it with sed, find and exec or xargs.
解决方案
 find . -type f |xargs awk '$0=FILENAME$0' > out
as I answered this, your "no awk" line not yet there. anyway, take a look my updated answer below:

updated based on comment

so you want to use find, exec/xargs, and sed to do it. My script needs GNU Sed, i hope you have it.

see the one liner first: (well, > out is omitted. You could add it to the end of the line. )
find . -type f | xargs -i echo {}|sed -r 's#(.\/)(.*)#cat &\|sed  "s:^:file \2 :g"#ge'
now let's take a test, see below:
kent$  head *.txt
==> a.txt <==
A1
A2

==> b.txt <==
B1
B2

kent$  find . -type f | xargs -i echo {}|sed -r 's#(.\/)(.*)#cat &\|sed  "s:^:file \2 :g"#ge'
file b.txt B1
file b.txt B2
file a.txt A1
file a.txt A2
is the result your expectation? 

Short explanation 


find ....|xargs -i echo {} nothing to explain, just print the
filename per line (with leading "./")
then pass the filename to a sed line like sed -r 's#(.\/)(.*)# MAGIC
#ge'
remember that in the above line, we have two groups \1: "./" and
\2 "a.txt"(filename)
since we have e at the end of sed line, the MAGIC part would be
executed as shell command.(GNU sed needed)
MAGIC: cat &\|sed  "s:^:file \2 :g  cat & is just output the file
content, and pipe to another sed. do the replace (s:..:..:g)
finally, the execution result of MAGIC would be the Replacement of
the outer sed.


the key is the 'e' of Gnu sed.

                        
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