Question

假设我的文件data.log包含以下内容：

[12/Mar/2015] /var/lib/file1.txt
[12/Mar/2015] /var/lib/file2.txt
[12/Mar/2015] /var/lib/file3.txt

如何使用awk和bash将此文件的内容存储到一个数组中，其中[12/Mar/2015]是元素，/var/lib/fileN.txt是其关键字？

Answer 1

击：

# the data
$ cat data.log
[12/Mar/2015] /var/lib/file1.txt
[12/Mar/2015] /var/lib/file2.txt
[12/Mar/2015] /var/lib/file3.txt

# the associative array declaration
$ declare -A map

# read the data from the file into the array
$ while read -r date file; do map[$file]=$date; done < data.log

# iterate through the data
$ for key in "${!map[@]}"; do printf "%s => %s\n" "$key" "${map[$key]}"; done
/var/lib/file3.txt => [12/Mar/2015]
/var/lib/file2.txt => [12/Mar/2015]
/var/lib/file1.txt => [12/Mar/2015]

Answer 2

在awk中：

$ awk '{map[$2]=$1}END{for(key in map){printf "%s => %s\n",key,map[key]}}' infile
/var/lib/file2.txt => [12/Mar/2015]
/var/lib/file3.txt => [12/Mar/2015]
/var/lib/file1.txt => [12/Mar/2015]

或者，作为非单行代表：

# For each line, use the second field as key and the first field as value
{
    map[$2] = $1
}

END {
    # Iterate over all keys
    for (key in map) {
        printf "%s => %s\n", key, map[key]
    }
}

这会进入一个文件，例如script.awk，然后用

调用

awk -f script.awk infile

将数组中的两列存储为元素 - ＆gt;密钥使用awk

2 个答案: