如何将文件中旧格式日期转换为$(date +%Y-%m-%d %H:%M)格式？

问题描述 #

我有一个包含约200行内容的大文件，部分内容示例如下：

started at Wed Jun  5 08:45:01 PM +0330 2024 -- ended at Wed Jun  5 10:35:34 PM +0330 2024.
started at Thu Jun  6 01:30:01 AM +0330 2024 -- ended at Thu Jun  6 03:17:18 AM +0330 2024.
started at Thu Jun  6 07:30:01 AM +0330 2024 -- ended at Thu Jun  6 09:19:19 AM +0330 2024.
started at Thu Jun  6 01:30:01 PM +0330 2024 -- ended at Thu Jun  6 03:19:16 PM +0330 2024.

希望将文件中$(date)输出格式的日期，转换为$(date +%Y-%m-%d %H:%M:%S)格式，预期输出如下：

started at 2024-06-05 20:45:01 -- ended at 2024-06-05 22:35:34.
started at 2024-06-06 01:30:01 -- ended at 2024-06-06 03:17:18.
started at 2024-06-06 07:30:01 -- ended at 2024-06-06 09:19:19.
started at 2024-06-06 13:30:01 -- ended at 2024-06-06 15:19:16.

请问该如何实现这一转换？是否可行？

实现方案 #

完全可行，以下是两种高效的命令行处理方法，适合200行规模的文件：

方法一：awk结合date命令逐行处理 #

利用awk提取日期字段，调用date命令完成格式转换：

awk '{
    # 拼接起始日期的原始字符串
    start_raw = $3 " " $4 " " $5 " " $6 " " $7 " " $8;
    # 调用date转换格式
    cmd = "date -d \"" start_raw "\" +\"%Y-%m-%d %H:%M:%S\"";
    cmd | getline start_fmt;
    close(cmd);
    # 拼接结束日期的原始字符串
    end_raw = $12 " " $13 " " $14 " " $15 " " $16 " " $17;
    cmd = "date -d \"" end_raw "\" +\"%Y-%m-%d %H:%M:%S\"";
    cmd | getline end_fmt;
    close(cmd);
    # 输出格式化后的行
    print "started at " start_fmt " -- ended at " end_fmt ".";
}' input.txt > output.txt

• 替换input.txt为你的源文件路径，output.txt为结果文件路径
• 依赖GNU date（Linux默认支持），macOS用户需安装coreutils后用gdate替换date

方法二：sed提取日期+shell循环转换 #

先用sed正则提取原始日期字符串，再通过shell循环调用date转换：

while read -r line; do
    # 提取起始日期原始字符串
    start_raw=$(echo "$line" | sed -E 's/started at (.*) -- ended at.*/\1/')
    start_fmt=$(date -d "$start_raw" +"%Y-%m-%d %H:%M:%S")
    # 提取结束日期原始字符串
    end_raw=$(echo "$line" | sed -E 's/.*-- ended at (.*)\./\1/')
    end_fmt=$(date -d "$end_raw" +"%Y-%m-%d %H:%M:%S")
    # 输出结果
    echo "started at $start_fmt -- ended at $end_fmt."
done < input.txt > output.txt

• 逻辑更直观，适合对正则熟悉的用户
• 同样需要GNU date，macOS替换为gdate

内容的提问来源于stack exchange，提问作者Saeed