呼應一個ls命令的輸出與小於n文件

Question

我有400個文件夾裏面的幾個文件，我感興趣的是：

1. 計數與擴展.solution許多文件是如何每個文件夾中，並
2. 然後輸出只有那些文件夾具有小於440層的元件

的點1）很容易獲得與命令：

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

但是，有沒有簡單的方法來過濾只有少於440個元素的文件？

Answer 1

這個簡單的腳本可以爲你工作： -

#!/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

下面

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 

腳本應該做

Answer 3

避免解析ls命令，並使用printf '%q\n計數文件：

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