bash 更新具有完整路径的文件名的文件列表

Question

我有一个如下所示的日志文件：

 Jan 1 06:09:23 somefile.txt
 Jan 2 12:18:27 somefile1.txt
 Jan 3 04:16:00 somefile2.txt

我想找到每个文件并将每个文件的完整路径插入到这个文件中。我认为 find、awk 和 sed 的某种组合可以实现这一点，但到目前为止，我还没有提出一个可行的解决方案来更新文件，使其如下所示。

Jan 1 06:09:23 /path/to/file/somefile.txt
Jan 2 12:18:27 /path/to/file1/somefile1.txt
Jan 3 04:16:00 /path/to/file2/somefile2.txt

我已经能够删除文件名并毫无问题地找到文件，但到目前为止我想出的内容会写出一个新文件并丢失原始文件内容。我曾希望保持原始文件完好无损。

#!/bin/bash
#functions
getup(){

for i in `cat /home/work/uploadtmp`
do
     find /home/uploads/*$i 2> /dev/null >> /home/work/upfile
done
}

listfile(){
while read line; do ls -lt $line; done < /home/work/upfile

}

#run functions
getup
listfile | awk '{print $1 " " $2 " " $3 " " $4}' | sort -k1M -k2 -k3 > /home/log/newfile

Answer 1

# create a temporary output file, so we only overwrite the destination when complete
tempfile=$(mktemp /home/log/newfile.XXXXXX)

# ...and tell the shell to delete that temporary file if it's still around when we exit
# ...won't work for SIGKILL or power failures, but better than nothing.
trap 'rm -f -- "$tempfile"' EXIT

# iterate over lines in the input file...
while read -r mon day time filename; do
  # ...quoting each name to only match itself...
  filename_pat=$(sed -e 's@[]*?[]@\\&@g'  <<<"$filename")
  # ...using find to locate the first file with the given name for each...
  fullname=$(find /home/uploads -name "$filename_pat" -print -quit)
  # ...and printing that new name on our stdout
  printf '%s\n' "$mon $day $time $fullname"
done </home/work/uploadtmp >"$tempfile" # ...redirecting the whole loop to our tempfile...

# ...then performing a single atomic rename to overwrite the final destination
mv "$tempfile" /home/log/newfile

Answer 2

在awk中，外部使用find收集文件路径：

$ cat program.awk
NR==FNR {                 # read in the files file records
    a[$NF]=$0; next }         # hash them to a and tskip o the next record
{                         # find produced list processing
    n=split($0,b,"/");        # basename functionality, filename part in b[n]
    if(sub(b[n],$0,a[b[n]]))  # replace filename in a with full path version
        print a[b[n]]         # and print
}
$ awk -f program.awk files <(find .)
Jan 3 04:16:00 ./file2/somefile2.txt
Jan 1 06:09:23 ./file/somefile.txt
Jan 2 12:18:27 ./file1/somefile1.txt

此解决方案（或旧解决方案）不允许文件名中有空格。不过，通过放弃 $NF 用法，在第一个块中很容易解决：

f=$0                      # current record to var f
sub(/^([^ ]+ ){3}/,"",f)  # remove timestamp
a[f]=$0                   # hash to a on f
next                      # ...

旧版本 @CharlesDuffy 在 cmets 中批评（++ 表示）。出于教育目的而留在这里：

$ awk -v path=".." '{ s="find " path " -name " $NF; s | getline $NF } 1' file
Jan 1 06:09:23 ../test/file/somefile.txt
Jan 2 12:18:27 ../test/file1/somefile1.txt
Jan 3 04:16:00 ../test/file2/somefile2.txt

• find 命令字符串被收集到 var s
• 执行并输出写回最后一个字段（$NF）