Question

特殊参数＆＃34; $ @＆＃34;包含以下字符串变量

回声$ @ 输出：a.bdf，b.bdf，c.nas，d.nas

我想用扩展名＆＃39; bdf＆＃39;提取字符串变量。并将其保存在另一个数组中。是否有可能在bash中这样做？

Answer 1

使用for循环迭代它：

for ARG in "$@";do
    if [[ "$ARG" == *.bdf ]];then
        BDF_ARRAY+=("$ARG")    #you don't need to initialize this array before the loop in bash
    else                       #optional block, if you want to split $@ in 2 arrays
        OTHER_ARRAY+=("$ARG")  
    fi
done

echo ${BDF_ARRAY[@]}
echo ${OTHER_ARRAY[@]}

Answer 2

使用for循环

for i in "$@";do
    if [[ "${i##*.}" == bdf ]]; then
        ARRAY2+=("$i")
    fi
done

Answer 3

在通用案例中：

#!/bin/bash

for i in "$@"; do
    case "$i" in
        *.bdf)  BDF_ARRAY+=("$i")
                ;;
        *.nas)  NAS_ARRAY+=("$i")
                ;;
    esac
done

for i in "${BDF_ARRAY[@]}"; do echo "BDF: $i"; done
for i in "${NAS_ARRAY[@]}"; do echo "NAS: $i"; done

提取基于模式的数组中定义的字符串变量

3 个答案: