如何让 grep 在 bash 中“读取行时”迭代工作功能？

Question

我有一个文本文件 (capital_names.txt)，其中包含如下行：

Warsaw_  
London_  
Oslo_  
...

在另一个文件 (capital_info.txt) 中，我有以下几行：

London_1_  
London_2  
cityLondon_3  
capitalWarsaw_1  
Warsaw_2  
...

我想编写一个 shell 脚本仅当大写名称采用以下格式“Name_”时才会对其进行 grep。
所需的输出是这样的多个文件：

$ cat Warsaw_output.txt  
Warsaw_2

$ cat London_output.txt   
London_1  
London_2  

这是脚本的关键部分：

$outp=$"output"  
while read line; do  
grep ^$line capital_info.txt > $line$outp  
done < capital_names.txt

但是，输出文件是空的（0 字节）并且具有以下名称：

'Warsaw_$'\r''output'  
'London_'$'\r''output'

当我运行单个命令时 (grep -f ^"London_" Capital_info.txt）一切正常，但我无法手动对 Capital_names.txt 中的 50000 个条目执行此操作。我该如何解决这个问题？

Answer 1

根据您显示的示例和尝试，请尝试以下 awk 代码。在 GNU awk 中编写和测试，应该可以在任何 awk 中工作。

awk '
BEGIN  { FS=OFS="_" }
FNR==NR{
  arr[$1]
  next
}
($1 in arr) && $2~/^[0-9]+$/{
  outFile=($1"_output.txt")
  if(prev!=outFile){ close(prev) }
  print ( $1,$2 ) > (outFile)
  prev=outFile
}
' capital_names.txt capital_info.txt

说明：为上述内容添加详细说明。

awk '                                  ##Starting awk program from here.
BEGIN  { FS=OFS="_" }                  ##In BEGIN section of awk setting FS and OFS as _ here.
FNR==NR{                               ##Checking condition FNR==NR then do following.
  arr[$1]                              ##Creating array arr with index of $1.
  next                                 ##next will skip all further statements from here.
}
($1 in arr) && $2~/^[0-9]+$/{          ##Checking if $1 is in arr AND 2nd field is digits.
  outFile=($1"_output.txt")            ##Creating outFile which has output file name in it.
  if(prev!=outFile){ close(prev) }     ##Checking if previous output file name is NOT same as current output file name then close previous one, to avoid too many open files error.
  print ( $1,$2 ) > (outFile)          ##printing 1st and 2nd field to outFile here.
  prev=outFile                         ##Setting prev to outFile value here.
}
' capital_names.txt capital_info.txt   ##Mentioning Input_file names here.