编写Shell脚本复制SAC文件并按规则截取重命名

Here's a straightforward bash script that does exactly what you need—copying your SAC files while renaming them to keep only the first timestamp part:

#!/bin/bash

# Loop through all SAC files in the current directory
for sac_file in *.sac; do
    # Skip if no SAC files exist (avoids errors when wildcard doesn't match)
    [[ -e "$sac_file" ]] || continue

    # Step 1: Remove the .sac extension from the filename
    without_ext="${sac_file%.sac}"

    # Step 2: Remove everything after the last underscore (separates the two timestamps)
    new_filename="${without_ext%_*}"

    # Copy the original file to the new name (verbose mode for feedback)
    cp -v "$sac_file" "$new_filename"
done

How it works: #

Let’s break down the logic using your first filename as an example:

1. Original filename: AU.GIRL..BHZ_2016-08-01T07_54_45.882_2016-08-01T08_18_05.882.sac
2. After stripping .sac: AU.GIRL..BHZ_2016-08-01T07_54_45.882_2016-08-01T08_18_05.882 (using bash parameter expansion ${sac_file%.sac} to remove the shortest matching suffix)
3. After trimming the second timestamp: AU.GIRL..BHZ_2016-08-01T07_54_45.882 (using ${without_ext%_*} to strip everything after the last underscore)

Quick Notes: #

• The -v flag in cp prints each copy operation to the terminal—remove it if you don’t need visual feedback.
• If your SAC files are in a different folder, replace *.sac with the full path (e.g., /home/your/data/*.sac).
• The check [[ -e "$sac_file" ]] ensures the script doesn’t throw errors if there are no .sac files present.

内容的提问来源于stack exchange，提问作者rehman