Shell脚本：基于某个字符创建文件列表

Question

我需要为以下文件创建一个文件列表：

APPLE_001

APPLE_002

BBB_004

APPLE_003

我只需要为

创建file_list

APPLE_001

APPLE_002

APPLE_003

谢谢， Ipsita

Answer 1

您的规格不是那么窄，但这里的 bash 脚本符合您的要求：

#!/bin/bash

if [[ $# < 2 ]]
then
    echo "[ERROR] this script expects two arguments, input file and output file" >&2
    exit 1
fi

input_file=$1
output_file=$2

if [[ ! -f $input_file ]]
then
    echo "[ERROR] your input file '$input_file' is missing" >&2
    exit 1
fi

if [[ -f $output_file ]]
then
    echo "[ERROR] your output file '$ouput_file' already exists please move it away" >&2
    exit 1
fi

while read LINE
do
    if [[ $LINE =~ APPLE_[0-9]+ ]]
    then
    echo $LINE >> $output_file
    else
    echo "'$LINE' does not match expected pattern, skip it"
    fi
done < $input_file

if [[ -f $ouput_file ]]
then
    echo "'$output_file' generated."
else
    echo "[WARNING] no pattern found in '$input_file' no file generated"
fi

使其可执行（chmod + x ./list_starting_with.sh）

使用./list_starting_with.sh运行它file_in.txt file_out.txt