Question

我有400个文件夹里面有几个文件，我感兴趣的是：

计算每个文件夹中扩展名为.solution的文件数量，以及
然后仅输出那些少于440个元素的文件夹

点1）很容易得到命令：

for folder in $(ls -d */ | grep "sol_cv_");
do
    a=$(ls -1 "$folder"/*.solution | wc -l); 
    echo $folder has "${a}" files;
done

但有没有简单的方法来过滤少于440个元素的文件？

Answer 1

这个简单的脚本可以为您服务： -

 #!/bin/bash

 MAX=440

 for folder in sol_cv_*; do
     COUNT=$(find "$folder" -type f -name "*.solution" | wc -l)
     ((COUNT < MAX)) && echo "$folder"
 done

Answer 2

以下脚本

counterfun(){
count=$(find "$1" -maxdepth 1 -type f -iname "*.solution" | wc -l)
(( count < 440 )) && echo "$1"
}
export -f counterfun
find /YOUR/BASE/FOLDER/ -maxdepth 1 -type d -iname "sol_cv_*" -exec bash -c 'counterfun "$1"' _ {} \;
#maxdepth 1 in both find above as you've confirmed no sub-folders

应该这样做

Answer 3

避免解析ls命令并使用printf '%q\n来计算文件：

for folder in *sol_cv_*/; do
    # if there are less than 440 elements then skip
    (( $(printf '%q\n' "$folder"/* | wc -l) < 440 )) && continue
    # otherwise print the count using safer printf '%q\n'
    echo "$folder has $(printf '%q\n' "$folder"*.solution | wc -l) files"
done

使用少于n的文件来回显ls命令的输出

3 个答案: