Bash版本4.4.20
Ubuntu 16.04
我需要比较我拥有的特定项目的时间和扩展名。下面是与我尝试执行的操作类似的操作，但错误相同。我不确定错误在哪里，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 $filename"
fi
done<fileList.txt
错误：
Filename is one.zip
check.sh: line 8: [: too many arguments
Filename is file_1.zip
check.sh: line 8: [: too many arguments
Filename is file_2.zip
check.sh: line 8: [: too many arguments
Filename is file_3.zip
Filename is two.csv
Filename is three.sql
使用模式匹配来发挥您的优势：
while IFS= read -r filename; do
if [[ "$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
一条case语句将在任何POSIX shell中运行，而不仅仅是支持bash-like[[ ... ]]命令的shell 。
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（？），可以使用
case $value in
200[0-9]|201[0-9]|202[0-1]) echo "Year between 2000 and 2021" ;;
esac
你不能这样做，因为简单地用[[ value = ... ]]，因为|是部分case语句的语法，而不是在模式的交替操作。相反，您将需要多个匹配运算符：
if [[ $value = 200[0-9] || $value = 201[0-19] || $value = 202[0-1] ]]; then
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我来说两句