脚本很慢

Question

我有个问题.我需要编写一个bash脚本,它将在给定的路径中找到所有文件和目录,并显示一些有关结果的信息.允许时间:30秒.

#!/bin/bash

DIRS=0
FILES=0
OLD_FILES=0
LARGE_FILES=0
TMP_FILES=0
EXE_FILES=0
IMG_FILES=0
SYM_LINKS=0
TOTAL_BYTES=0

#YEAR_AGO=$(date -d "now - 1 year" +%s)
#SECONDS_IN_YEAR=31536000

function check_dir {
    for entry in "$1"/*
    do
        if [ -d "$entry" ]; then
            ((DIRS+=1))
            check_dir "$entry"
        else if [ -f "$entry" ]; then
                ((FILES+=1))
                #SIZE=$(stat -c%s "$entry")
                #((TOTAL_BYTES+=SIZE))
                #CREATE_DATE=$(date -r "$entry" +%s)
                #CREATE_DATE=$(stat -c%W "$entry")
                #DIFF=$((CREATE_DATE-YEAR_AGO))
                #if [ $DIFF -ge $SECONDS_IN_YEAR ]; then
                #   ((OLD_FILES+=1))
                #fi
             fi

        fi
    done
}

if [ $# -ne 2 ]; then
    echo "Usage: ./srpt path emailaddress"
    exit 1
fi

if [ ! -d $1 ]; then
    echo "Provided path is invalid"
    exit 1
fi

check_dir $1

echo "Execution time $SECONDS"
echo "Dicrecoties $DIRS"
echo "Files $FILES"
echo "Sym links $SYM_LINKS"
echo "Old files $OLD_FILES"
echo "Large files $LARGE_FILES"
echo "Graphics files $IMG_FILES"
echo "Temporary files $TMP_FILES"
echo "Executable files $EXE_FILES"
echo "Total file size $TOTAL_BYTES"

以下是使用上面的注释行执行的结果:

Execution time 1
Dicrecoties 931
Files 14515
Sym links 0
Old files 0
Large files 0
Graphics files 0
Temporary files 0
Executable files 0
Total file size 0

如果我要删除评论

SIZE=$(stat -c%s "$entry")
((TOTAL_BYTES+=SIZE))

我有:

Execution time 31
Dicrecoties 931
Files 14515
Sym links 0
Old files 0
Large files 0
Graphics files 0
Temporary files 0
Executable files 0
Total file size 447297022

31秒 我怎样才能加速我的脚本？另外+30秒可以找到日期创建更多一年的文件

Answer 1

通常情况下,在shell中使用循环表明你正在采取错误的方法.

shell之前是运行其他工具的工具.

虽然它可以进行计数,但它awk是一个更好的工具.

虽然它可以列出和查找文件,但它find更好.

最好的shell脚本是那些设法有一些工具可以完成任务的脚本,而不是那些按顺序启动数百万工具并且所有工作都由shell完成的脚本.

在这里,通常更好的方法是find找到文件并收集所需的所有数据,并将其awk删除并返回统计信息.这里使用GNU find和GNU awk(for RS='\0')和GNU date(for -d):

find . -printf '%y.%s.%Ts%p\0' |
  awk -v RS='\0' -F'[.]' -v yearago="$(date -d '1 year ago' +%s)" '
    {
      type[$1]++; 
      if ($1 == "f") {
        total_size+=$2
        if ($3 < yearago) old++
        if (!index($NF, "/")) ext[tolower($NF)]++
      }
    }
    END {
      printf("%20s: %d\n", "Directories", type["d"])
      printf("%20s: %d\n", "Total size", total_size)
      printf("%20s: %d\n", "old", old)
      printf("%20s: %d\n", "jpeg", ext["jpg"]+ext["jpeg"])
      printf("%20s: %d\n", "and so on...", 0)
    }'