如何批量重命名文件夹内文件：提取指定字符串加字符移至文件名前缀

Hey there, let's tackle these two file renaming scenarios with practical, easy-to-use solutions. I'll cover cross-platform Python scripts (perfect if you work across Windows, macOS, or Linux) and system-specific command-line tools for quick one-off tasks.

Scenario 1: Find a Specific String, Extract It + Extra Characters as Prefix #

Suppose you have files like report_Q32024_final.pdf and you want to extract the string Q3 plus the following 4 characters (2024) to make the prefix Q32024_report_final.pdf. Here's how to do it:

Python Script (Cross-Platform) #

This uses regular expressions to target your specific string and capture the extra characters you need:

import os
import re

# Configure these values to match your needs
target_dir = "/path/to/your/folder"  # Replace with your folder path
search_str = "Q3"  # The specific string to look for
extra_chars = 4  # Number of additional characters to capture after the search string

# Regex pattern: matches the search string plus the specified number of characters
pattern = re.compile(f"({re.escape(search_str)}.{{{extra_chars}}})")

for filename in os.listdir(target_dir):
    # Skip directories, only process files
    if not os.path.isfile(os.path.join(target_dir, filename)):
        continue
    
    match = pattern.search(filename)
    if match:
        captured_prefix = match.group(1)
        # Remove the captured part from the original filename (clean up leftover separators)
        new_filename = filename.replace(captured_prefix, "", 1).strip("_- ")
        # Prepend the captured prefix
        new_filename = f"{captured_prefix}_{new_filename}"
        # Rename the file
        os.rename(
            os.path.join(target_dir, filename),
            os.path.join(target_dir, new_filename)
        )
        print(f"Renamed: {filename} → {new_filename}")

• How to use: Update target_dir, search_str, and extra_chars to fit your case. The script skips folders and only renames files that match your pattern.

Scenario 2: Batch Rename Files to Move a Fixed-Format String (e.g., Date) to Prefix #

Your example mentions moving 8-digit date strings like 20171202 to the start of filenames (e.g., vacation_20171202_beach.jpg → 20171202_vacation_beach.jpg). If you need to extract the date plus 4 extra characters (e.g., 20171202abcd), just adjust the regex pattern.

Python Script (Cross-Platform) #

This targets 8-digit date strings (YYYYMMDD) but can be modified for other fixed patterns:

import os
import re

target_dir = "/path/to/your/folder"

# Regex pattern for 8-digit dates (YYYYMMDD) + optional 4 extra characters
# Remove the `.{4}` if you only want the date itself
pattern = re.compile(r"(\d{8}.{4})")

for filename in os.listdir(target_dir):
    if not os.path.isfile(os.path.join(target_dir, filename)):
        continue
    
    match = pattern.search(filename)
    if match:
        captured_prefix = match.group(1)
        # Remove the captured part and clean up any leftover underscores/hyphens
        rest_of_name = filename.replace(captured_prefix, "", 1).strip("_- ")
        new_filename = f"{captured_prefix}_{rest_of_name}"
        os.rename(
            os.path.join(target_dir, filename),
            os.path.join(target_dir, new_filename)
        )
        print(f"Renamed: {filename} → {new_filename}")

• Modify for other patterns: If your fixed string is something else (e.g., ID- followed by 5 digits), change the regex to (ID-\d{5}).

Command-Line Options #

Windows (PowerShell)

For the date example (move 8-digit date to prefix):

$folderPath = "C:\path\to\your\folder"
Get-ChildItem -Path $folderPath -File | ForEach-Object {
    if ($_.Name -match '(\d{8})') {
        $datePrefix = $matches[1]
        $newName = "$datePrefix_" + $_.Name.Replace($datePrefix, "").Trim('_ -')
        Rename-Item -Path $_.FullName -NewName $newName
        Write-Host "Renamed: $($_.Name) → $newName"
    }
}

macOS/Linux (Bash)

Using rename (note: some systems use prename for Perl-based rename):

cd /path/to/your/folder
# For moving 8-digit date to prefix
rename 's/^(.*?)(\d{8})(.*)$/$2_$1$3/' *

• To add 4 extra characters: Adjust the regex to s/^(.*?)(\d{8}.{4})(.*)$/$2_$1$3/

Pro Tips:

• Always test scripts on a copy of your files first to avoid accidental data loss!
• If you need to handle file extensions carefully (e.g., don't modify the extension), split the filename into name and extension before processing.

内容的提问来源于stack exchange，提问作者rykman