如何使用命令输出中带有空格的元素初始化数组

如何使用命令输出中带有空格的元素初始化数组

以下脚本:

DYN_HOSTS_START_ARRAY=($(grep -E "STARTING HOST" sample.log | cut -d' ' -f 1,2))
for ((i=0; i< ${#DYN_HOSTS_START_ARRAY[@]}; i++))
do
    echo "$i:  start: "${DYN_HOSTS_START_ARRAY[$i]}""
done

使用以下示例.log 文件:

2019-11-11 19:05:55,823 DEBUG  STARTING HOST 46
2019-11-11 19:05:55,831 DEBUG  STARTING HOST 703
2019-11-11 19:05:55,837 DEBUG  STARTING HOST 505
2019-11-11 19:05:55,858 DEBUG  STARTING HOST 93
2019-11-11 19:05:55,859 DEBUG  STARTING HOST 486
2019-11-11 19:05:55,861 DEBUG  STARTING HOST 72
2019-11-11 19:05:55,879 DEBUG STARTING HOST 855
2019-11-11 19:05:55,913 DEBUG  STARTING HOST 560
2019-11-11 19:05:56,067 DEBUG STARTING HOST 199

产生以下不需要的输出:

0:  start: 2019-11-11
1:  start: 19:05:55,823
2:  start: 2019-11-11
3:  start: 19:05:55,831
4:  start: 2019-11-11
5:  start: 19:05:55,837
6:  start: 2019-11-11
7:  start: 19:05:55,858
8:  start: 2019-11-11
9:  start: 19:05:55,859
10:  start: 2019-11-11
11:  start: 19:05:55,861
12:  start: 2019-11-11
13:  start: 19:05:55,879
14:  start: 2019-11-11
15:  start: 19:05:55,913
16:  start: 2019-11-11
17:  start: 19:05:56,067

所需输出应仅包含 9 个元素(而不是 18 个),每个元素包含日期和时间,并由原始空格分隔。

如何修复我的脚本,同时保留数组初始化,仅 9 个元素,以实现这一点?

答案1

您可以使用mapfile -t并读取流程替换中的数据,如下所示@Kusalananda指出。

mapfile -t dyn_hosts_start_array < <(grep 'STARTING HOST' sample.log | cut -d' ' -f 1,2)    
for i in "${!dyn_hosts_start_array[@]}"; do
  printf '%s:  start: %s\n' "$i" "${dyn_hosts_start_array[i]}"
done

或者您可以将IFS用于分词的变量更改为换行符(默认值为空格符、制表符和换行符),然后将其更改回原始值。

oldifs=$IFS
IFS=$'\n'
dyn_hosts_start_array=( $(grep 'STARTING HOST' sample.log | cut -d' ' -f 1,2) )
IFS=$oldifs    
for i in "${!dyn_hosts_start_array[@]}"; do
  printf '%s:  start: %s\n' "$i" "${dyn_hosts_start_array[i]}"
done

输出(两种变体):

0:  start: 2019-11-11 19:05:55,823
1:  start: 2019-11-11 19:05:55,831
2:  start: 2019-11-11 19:05:55,837
3:  start: 2019-11-11 19:05:55,858
4:  start: 2019-11-11 19:05:55,859
5:  start: 2019-11-11 19:05:55,861
6:  start: 2019-11-11 19:05:55,879
7:  start: 2019-11-11 19:05:55,913
8:  start: 2019-11-11 19:05:56,067

有关的:

答案2

awk '{print NR-1 ": ", "start:", $1, $2;}' sample.log

使用问题打印的输入

0:  start: 2019-11-11 19:05:55,823
1:  start: 2019-11-11 19:05:55,831
2:  start: 2019-11-11 19:05:55,837
3:  start: 2019-11-11 19:05:55,858
4:  start: 2019-11-11 19:05:55,859
5:  start: 2019-11-11 19:05:55,861
6:  start: 2019-11-11 19:05:55,879
7:  start: 2019-11-11 19:05:55,913
8:  start: 2019-11-11 19:05:56,067

要更好地控制输出格式,您也可以printf这样使用

awk '{printf "%d:  start: %s %s\n", NR-1, $1, $2;}' sample.log

命令替换在空格或特殊字符方面存在一些缺陷。显然,输入中的两个字段保存在单独的数组元素中。该脚本根据您的原始脚本将它们组合起来。

DYN_HOSTS_START_ARRAY=($(grep -E "STARTING HOST" sample.log | cut -d' ' -f 1,2))
for ((i=0; i< ((${#DYN_HOSTS_START_ARRAY[@]} / 2)); i++))
do
    echo "$i:  start: ${DYN_HOSTS_START_ARRAY[((2 * $i))]} ${DYN_HOSTS_START_ARRAY[((2 * $i + 1))]}"
done

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