bash - 如何在 bash 脚本中正确使用 elif 与 && 和 -o

Question

Bash 版本 4.4.20
Ubuntu 16.04

我需要比较我所拥有的特定项目的时间和延期。下面的内容与我尝试做的类似，但错误是相同的。我不确定错误到底在哪里，因为 shellcheck 没有产生错误。

#!/bin/bash
#

while read -r filename; do
  extension="${filename##*.}"
  if [ "$extension" == "zip" ] && [ "$filename" == "one.zip" ]; then
    echo "Filename is $filename"
  elif [ "$extension" == "zip" ] && [ "$filename" == "file_1.zip" ] -o [ "$filename" == "file_2.zip" ] -o [ "$filename" == "file_3.zip" ]; then
    echo "Filename is $filename"
  elif [ "$extension" == "csv" ] && [ "$filename" == "two.csv" ]; then
    echo "Filename is $filename"
  else
    echo "Filename is $filename"
  fi
done<fileList.txt  

错误:

Filename is one.zip
check.sh: line 8: [: too many arguments
Filename is file_1.zip
check.sh: line 8: [: too many arguments
Filename is file_2.zip
check.sh: line 8: [: too many arguments
Filename is file_3.zip
Filename is two.csv
Filename is three.sql

Answer 1

使用模式匹配来发挥你的优势:

while IFS= read -r filename; do
  if [[ "$filename" = one.zip ]; then
    echo "Filename is $filename"
  elif [[ "$filename" = file_[123].zip ]; then
    echo "Filename is $filename"
  elif [[ "$filename" = two.csv ]; then
    echo "Filename is $filename"
  else
    echo "Filename is $filename"
  fi
done < fileList.txt

一个case语句将在任何 POSIX shell 中工作，而不仅仅是支持 bash 的 shell -像[[ ... ]]命令。

while IFS= read -r filename; do
  case $filename in
    one.zip) echo "Filename is $filename" ;;
    file_[123].zip) echo "..." ;;
    two.csv) echo "..." ;;
    *) echo "..." ;;
  esac
done

要匹配年份范围(？)，您可以使用

case $value in
  200[0-9]|201[0-9]|202[0-1]) echo "Year between 2000 and 2021" ;;
esac

你不能像 [[ value = ... ]] 那样简单地做到这一点，自 |是 case 的一部分语句的语法，而不是模式中的交替运算符。相反，您需要多个匹配运算符:

if [[ $value = 200[0-9] || $value = 201[0-19] || $value = 202[0-1] ]]; then