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好吧，我目前正在编写一个脚本，它将获取一个带有“~”分隔符的文件并将其拆分。但是它们是数组中的一项，需要取出该元素的最后 3 个单词的 [5] 元素，将它们切掉并将它们分配给数组中的不同值，然后重新调整元素 [5] 以删除这些项目。我尝试过 sed、cut 和其他命令，但我迷路了。我正在为这个脚本使用 bash，下面是我的问题的简短形式。

#!#!/bin/bash 
STR="FAILED~LOSS~Positive~MULTICOUNT~1~LOSS SUMMARY - Log: One vs TWO DAD MAR  DE~5~489646.22~469646.22~5" 
IFS="~" read -ra STR_ARRAY <<< "$STR"

for x in "${STR_ARRAY[@]}"
    do
            echo "> [$x]"
    done

当前打印：

[0] = FAILED
[1] = LOSS
[2] = Positive
[3] = MULTICOUNT
[4] = 1
[5] = LOSS SUMMARY - Log: One vs TWO DAD MAR DE
[6] = 5
[7] = 489646.22
[8] = 469646.22
[9] = 5

想要打印：

[0] = FAILED
[1] = LOSS
[2] = Positive
[3] = MULTICOUNT
[4] = 1
[5] = LOSS SUMMARY - Log: One vs TWO 
[6] = DAD 
[7] = MAR 
[8] = DE
[9] = 5
[10] = 489646.22
[11] = 469646.22
[12] = 5
4

2 回答 2

2

你可以这样做：

#!#!/bin/bash 
STR="FAILED~LOSS~Positive~MULTICOUNT~1~LOSS SUMMARY - Log: One vs TWO DAD MAR  DE~5~489646.22~469646.22~5" 
IFS="~" read -ra STR_ARRAY <<< "$STR"
IFS=" " read -ra T <<< "${STR_ARRAY[5]}"
STR_ARRAY2=("${STR_ARRAY[@]:0:5}" "${T[*]:0:${#T[@]} - 3}" "${T[@]:(-3)}" "${STR_ARRAY[@]:6}")
printf '%s\n' "${STR_ARRAY2[@]}"

输出：

FAILED
LOSS
Positive
MULTICOUNT
1
LOSS SUMMARY - Log: One vs TWO
DAD
MAR
DE
5
489646.22
469646.22
5
于 2014-07-18T15:53:40.823 回答
1

将您的 STR 调整为 NEW_STR 并将其与 IFS=~ 一起使用（小 2 行更改）

$ cat 1.sh
#!/bin/bash
STR="FAILED~LOSS~Positive~MULTICOUNT~1~LOSS SUMMARY - Log: One vs TWO DAD MAR  DE~5~489646.22~469646.22~5"

NEW_STR="$(echo $STR | sed "s/TWO DAD MAR DE/TWO~DAD~MAR~DE/")"

IFS="~" read -ra STR_ARRAY <<< "$NEW_STR"
for x in "${STR_ARRAY[@]}"
    do
            echo "> [$x]"
done

输出：

$ ./1.sh
> [FAILED]
> [LOSS]
> [Positive]
> [MULTICOUNT]
> [1]
> [LOSS SUMMARY - Log: One vs TWO]
> [DAD]
> [MAR]
> [DE]
> [5]
> [489646.22]
> [469646.22]
> [5]
于 2014-07-18T16:03:24.220 回答