CSV文件正则匹配需求：从'B08-1506'起匹配至下一同类模式并合并多行

Got it, let's work through this problem step by step. Your goal is to split your CSV into blocks starting with identifiers like B08-1506 (following the BXX-XXXX pattern, where X is a digit), capture each block up to the next identifier, and merge any multi-line content in each block into a single line. Here's how to do it with regex:

Regex Pattern (With DOTALL Mode Enabled) #

First, we need a regex that can span lines and stop at the next identifier. Use this pattern (make sure your regex engine supports DOTALL mode, which lets . match newlines):

(B\d{2}-\d{4})(.*?)(?=B\d{2}-\d{4}|$)

Breakdown of the Pattern: #

• (B\d{2}-\d{4}): Captures the starting identifier (e.g., B08-1506) as Group 1. This matches exactly two digits after B, a hyphen, then four digits.
• (.*?): Non-greedy match of all characters (including newlines, thanks to DOTALL) for the rest of the block. The non-greedy ? ensures we stop at the next identifier instead of matching all the way to the end of the file.
• (?=B\d{2}-\d{4}|$): Positive lookahead that tells the regex to stop when it hits either the next BXX-XXXX identifier or the end of the file.

Merging Multi-Line Content into Single Lines #

Once you capture each block, you'll need to clean up line breaks and extra whitespace. Here's how to implement this in Python as an example:

import re

# Your raw CSV content (with line breaks)
raw_csv = """B08-1506,324873, st, $0.0, ljkflka,,,,,
1 of 37 jksdfhjfhjk jkdsfh,,,,,,,
B08-1606,324873, st, $0.0, ljkflka,,,,,
1 of 37 jksdfhjfhjk jkdsfh,,,,,,,
B09-0680,324873, st, $0.0, ljkflka,,,,,
1 of 37 jksdfhjfhjk jkdsfh,,,,,,,
B09-1234,..."""

# Compile regex with DOTALL mode
pattern = re.compile(r'(B\d{2}-\d{4})(.*?)(?=B\d{2}-\d{4}|$)', re.DOTALL)

# Process each matched block
cleaned_lines = []
for match in pattern.finditer(raw_csv):
    id_segment = match.group(1)
    content_segment = match.group(2)
    # Replace all newlines and multiple spaces with a single space, then trim edges
    cleaned_content = re.sub(r'\s+', ' ', content_segment).strip()
    # Combine into a single line and add to results
    cleaned_lines.append(f"{id_segment}{cleaned_content}")

# Print or save the cleaned lines
for line in cleaned_lines:
    print(line)

For Text Editors (Like VS Code) #

If you're using a text editor to do this manually:

1. Enable regex mode and the "Match newlines with .*" option (DOTALL equivalent)
2. Find: (B\d{2}-\d{4})([\s\S]*?)(?=B\d{2}-\d{4}|$)
3. Replace: $1${2//\n/ } (adjust syntax if needed—some editors use \1 instead of $1, and you may need to replace newlines with spaces in a separate step if the editor doesn't support inline replacements)

Key Notes: #

• Always use non-greedy matching (.*?) here—without it, the regex will match from the first identifier all the way to the last one, which isn't what you want.
• DOTALL mode is essential to capture content across line breaks in each block.
• Cleaning whitespace ensures your final CSV lines don't have messy line breaks or redundant spaces.

内容的提问来源于stack exchange，提问作者veilupearl