Perl正则替换YAML文件版本号未达预期问题求助

Hey Stefan, let's walk through why your current regex and Perl command aren't working, and get that version number updated correctly.

What's Going Wrong? #

You've hit two key issues with your approach:

• Shell variables aren't being passed to Perl
  When you wrap your Perl script in single quotes ('s/$search/$replace/m'), the shell doesn't expand the $search and $replace variables—Perl sees literal $search and $replace instead of their values. Since those variables aren't defined in Perl, your regex never matches anything.

• Your replacement breaks YAML structure
  Your regex version:\s*([^\s]+) matches the entire string version: 1.2. If it did work, replacing that with just 1.3 would turn the line into 1.3, losing the version: key entirely—definitely not what you want for a valid YAML file.

Working Solutions #

Here are two solid ways to fix this:

Option 1: Hardcode the pattern directly in Perl (simplest & most reliable) #

Use a capture group to preserve the version: part, then replace just the version number:

perl -pie 's/(version:\s*)\S+/\11.3/' ./test.yml

• (version:\s*) captures the key and any following whitespace
• \S+ matches the existing version number (all non-whitespace characters)
• \11.3 inserts the captured key/whitespace followed by your new version

Option 2: Use shell variables with proper escaping #

If you want to keep using variables for flexibility, wrap your Perl script in double quotes so the shell expands the variables. Just remember to escape backslashes so Perl sees them correctly:

SEARCH_PATTERN="(version:\\s*)\\S+"
NEW_VERSION="\\11.3"
perl -pie "s/$SEARCH_PATTERN/$NEW_VERSION/" ./test.yml

• We escape \ as \\ because double quotes let the shell parse escape characters
• \1 becomes \\1 to prevent the shell from treating it as a special sequence

Result #

After running either command, your test.yml will look exactly like you want:

app:
  version: 1.3
  name: abc

内容的提问来源于stack exchange，提问作者Stefan