BASH-在数组中查找正则表达式，打印找到的数组项

Question

我已经在这里测试过正则表达式： http : //regexr.com/3bchs

我无法获取仅打印正则表达式搜索项的数组。

files=(`ls $BACKUPDIR`)
daterange='(2015\-06\-)[0-9]+\s?'

for i in "${files[@]}"
do
        if [[ "$files[$i]" =~ $daterange ]];
         then
                 echo $i
         fi
done

输入： 2015-06-06 2015-06-13 2015-06-20 2015-06-27 2015-07-04 2015-07-11

输出：

2015-06-06 
2015-06-13 
2015-06-20 
2015-06-27 
2015-07-04 
2015-07-11

Answer 1

通过运行bash -vx <script>我发现它正在编译的文件是错误的。 我需要将$files[$i]更改为$i 。

$files[$i] = 2015-06-06 [2015-06-06]

感谢Etan Reisner的评论，我的答案得到了进一步的改善。 通过不解析ls输出。

参考： https : //stackoverflow.com/a/11584156/3371795

#!/bin/bash

# Enable special handling to prevent expansion to a
# literal '/example/backups/*' when no matches are found. 
shopt -s nullglob

        # Variables
        YEAR=`date +%Y`
        MONTH=`date +%m`

        DIR=(/home/user/backups/backup.weekly/*)

        # REGEX - Get curent month
        DATE_RANGE='('$YEAR'\-'$MONTH'\-)[0-9]+\s?'


# Loop through dir
for i in "${DIR[@]}";
do
        # Compare dir name with date range
        if [[ "$i" =~ $DATE_RANGE ]];
        then
                # I have no idea what this is, works fine without it.
                [[ -d "$i" ]] 

                # Echo found dirs
                echo "$i"
        fi
done