3

我有一系列目录

079/af3
100/af4
120/af3
  . 
  .
  .

每个???/af?目录都包含一个很长的文件results.stdout。接近这个文件的结尾，可以找到字符串

 Normal termination: iterations complete!

如果 af3 (resp. af4) 中的计算成功，则在文件中写入一条或多条错误消息。为了避免手动检查每个文件，我正在编写一个生成摘要文件的脚本：

 Massflow        af3      af4 
      079    Success  Failure
      100    Failure  Success
      120    Success  Success
        .      .       .
        .      .       .

到目前为止，我已经能够烹饪以下内容：

#!/bin/bash

strlen="9" # want to keep the format flexible, instead than hardcode it
format0="%"$strlen"s %"$strlen"s %"$strlen"s\n"
# write the header of file summary
awk -v format="$format0" ' BEGIN { printf format, "Massflow", "af3", "af4"
                             } ' >> summary


for dir in ??? # loop on all the directories
do
    for j in 3 4 # loop on the two subdirs
    do
    result[$j]=$(tac $dir/af$j/results.stdout | awk '
    /TACOMA:- Normal termination: iterations complete!/ {success = 1; exit}
    END { if (success == 1)
              print "Success"
          else
              print "Failure"
        }')
    done
done
exit

但是，我不知道如何编写摘要文件...我想将result数组传递给另一个 awk 程序，但 awk 不接受数组变量。有什么建议么？如果您认为我的编程风格、工具选择或两者都很糟糕，请随意更改方法甚至工具：)

4

4 回答 4

2

我只是printf循环时的结果：

printf 'Massflow        af3      af4\n'
for dir in $(find -maxdepth 1 -type d) # loop on all the directories
do
    printf '     %d  ' "$(printf '%s' "$dir" | sed -e 's/[^0-9]//g')"
    for j in 3 4 # loop on the two subdirs
    do
    result[$j]=$(tac $dir/af$j/tacoma.stdout | awk '
    /TACOMA:- Normal termination: iterations complete!/ {success = 1; exit}
    END { if (success == 1)
              print "Success"
          else
              print "Failure"
        }')
        printf '  %s' "$result[j]"
    done
    printf '\n'
done
于 2012-10-22T13:12:32.997 回答
2

首先，不要使用 tac，因为反转整个文件没有任何好处。只需将文件交给 awk。

您可以省略第二个 for 循环并保存两个结果并在之后打印它们：

for dir in ??? # loop on all the directories
do
    for j in 3 4; do
        af[$j]=$(awk '/TACOMA:- Normal termination: iterations complete!/ {success = 1; exit}
                   END { if (success == 1)
                             print "Success"
                         else
                             print "Failure"
                   }'  $dir/af$j/results.stdout)
     done

     awk -v format="$format0" "BEGIN { printf format, \"$dir\", \"${af[3]}\", \"${af[4]}\"; } " >> summary
done

来自@EdMorton，bash只有没有awk：

for dir in ??? # loop on all the directories
do
    for j in 3 4; do
        if grep -q "TACOMA:- Normal termination: iterations complete!" "$dir/af$j/results.stdout"; then
            af[$j]="Success"
        else
            af[$j]="Failure"
        fi
     done

     printf "$format0" "$dir" "${af[3]}" "${af[4]}" >> summary
done
于 2012-10-22T13:22:31.070 回答
2

这是查看问题的另一种方法：使用grep而不是awk，并使用column来格式化输出。

isSuccess() {
    if tac "$1" | grep -q 'Normal termination: iterations complete'; then
        echo Success
    else
        echo Failure
    fi
}

{
    echo Massflow af3 af4
    for dir in ???; do
        echo "$dir" $(isSuccess "$dir/af3/results.stdout") $(isSuccess "$dir/af4/results.stdout")
    done
} | column -t
于 2012-10-22T21:32:47.410 回答
1

首先使用 find 或循环或任何您喜欢的方式构建输出文件列表，然后将整个列表传递给 awk，例如

for dirName in ???
do
   for subName in af3 af4
   do
      files="$files $dirName/$subName/results.stdout"
   done
done

awk '
FNR == 1 {
   split(FILENAME,dfA,"/")
   dirName = dfA[1]
   subName = dfA[2]
   dirNames[dirName]
   subNames[subName]
}

/Normal termination: iterations complete!/ {
   succ[dirName,subName]
}

END {
   printf "Massflow"
   for (subName in subNames) {
      printf "\t%s",subName
   }
   print ""

   for (dirName in dirNames) {
      printf "%s", dirName
      for (subName in subNames) {
         printf "\t%s", ( (dirName,subName) in succ ? "Success" : "Failure" )
      }
      print ""
   }
}
' $files

请注意，我没有在末尾引用 $files ，因此它会为您的示例正确扩展。我刚刚编辑了我的答案，因为我看到目录结构是 dir/subdir/results.stdout 而不是我最初想到的 dir/file。

根据@DeltaIV 的回复评论版本

for dirName in ???
do
   for subName in af3 af4
   do
      files="$files $dirName/$subName/results.stdout"
   done
done

awk '
# FNR == 1 is true at the first line of each input file
FNR == 1 {

   split(FILENAME,dfA,"/")
   dirName = dfA[1]
   subName = dfA[2]

   # Use array dirNames as the set of all top level directory names
   # and array subNames as the set of all sub-directory names so later
   # we can loop through them all to produce output.
   dirNames[dirName]
   subNames[subName]
}

# Check if the current line of the current input file contains the
# success indication text.
/Normal termination: iterations complete!/ {

   # The success indication text was found in the current file so
   # updated array succ which is the set of all dirName/SubName
   # pairs that had the success indication in their results file.
   succ[dirName,subName]
}

# "END" is true after all input files have been processed.
END {

   # Print the header line consisting of Massflow followed by the
   # sub-directory names
   printf "Massflow"
   for (subName in subNames) {
      printf "\t%s",subName
   }
   print ""

   # Loop through the set of dirNames so you get one per row
   # and for each dirName process all sub-directory names
   for (dirName in dirNames) {
      printf "%s", dirName
      # Loop through the set of subNames and process each one
      # as a new tab-separated column of output
      for (subName in subNames) {
         # If the current dirName/subName combination is in the succ
         # set then print "Success", otherwise print "Failure".
         printf "\t%s", ( (dirName,subName) in succ ? "Success" : "Failure" )
      }
      # Move down to the next row of output.
      print ""
   }
}
' $files
于 2012-10-22T16:51:33.713 回答