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The following one-liner Perl statement didn't work when using the "*" globbing:

>>> perl -i'.bak' -pe 's#__SUB__CWD__#'$(pwd)'#g' *.sql
Can't open *.sql: Invalid argument.

The following statement works but only when you give a single file as an argument:

>>> perl -p -i'.bak' -e 's#__SUB__CWD__#'$(pwd)'#g' A_SINGLE_FILE.sql

Any idea what could make perl to realize "*" must be read in a REGEX way?

I'm trying to replace all the occurence of "SUB__CWD" in tons of files.

C.f. https://github.com/oracle/db-sample-schemas/blob/master/README.md

tmangin
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2 Answers2

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That's not Perl's fault, it's Windows' fault. Linux shells expand the wildcards for you, cmd.exe does not. I don't know about PowerShell, but it seems it doesn't either.

Also, you don't want to interpret the asterisk the regex way, but the glob pattern way. In a regex, * means "the previous thing should be repeated zero or more times".

You can use Perl's glob function:

perl -p -i'.bak' -e 'BEGIN {@ARGV = glob shift} s#__SUB__CWD__#'$(pwd)'#g' *.sql

Note that you need to backslash backslashes in the path to prevent Perl from interpreting them, i.e. use $($pwd -replace '\\', '\\') instead of just $(pwd). You also need to backslash the # characters in the path as they would otherwise be understood as the delimiters, so use $($pwd -replace '[\\#]', '\$&').

Therefore, it would be even cleaner not to use $(pwd), but let Perl know what the replacement is:

perl -p -i'.bak' -e 'BEGIN{ ($pwd, @ARGV) = (shift, map glob, @ARGV) }
                   s/__SUB__CWD__/$pwd/g' -- $(pwd) *.sql

The first argument is assigned to $pwd via shift, the remaining attributes are mapped to glob, which means you can specify more than one wildcard pattern.

choroba
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  • Thanks very much for the detailed answer. That worked (with -i'.bak' instead of -i.bak ) . It's indeed a Powershell issue, which is the reason why I mentionned it in the Top Question so that other people find your answer. The command worked totally fine on Mac OSX. – tmangin Apr 03 '20 at 10:53
  • Just another question: when do we use of the hashtag # in "s#string#replacement_string#g" vs. the slash "/" in "s/string/replacement_string/g" ? – tmangin Apr 03 '20 at 10:59
  • Thanks very much also for the precision on the distinction between the REGEX wildcard vs. GLOB wildcard. That was a very helpful! – tmangin Apr 03 '20 at 11:00
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    @tmangin: You can use any delimiter you like. The slash is the standard one, a programmer usually switches to a different one if they need to use literal slashes in the regex or replacement, so they don't need to backslash them. Using `s{}{}` in such a case is recommended by the Perl Best Practices book. – choroba Apr 03 '20 at 12:44
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choroba's answer provides good pointers and solutions focused more on the perl side of things.

A direct adaptation of your command to PowerShell on Windows is (both Windows PowerShell and PowerShell [Core]):

perl -i'.bak' -pe "s#__SUB__CWD__#$($pwd -replace '[\\#]', '\$&')#g" (Get-ChildItem *.sql).Name

As choroba points out, PowerShell on Windows does not do globbing (expansion of wildcard patterns to matching filenames) for you; however, it does on Unix-like platforms - see below.

  • (Get-ChildItem *.sql).Name uses the Get-ChildItem cmdlet to perform the globbing manually.

  • $pwd -replace '[\\#]', '\$&' uses PowerShell's regex-based -replace operator to ensure that both the \ path separators in the current-directory path as well as any # characters are \-escaped, to ensure that \ isn't interpreted as the start of an escape sequence by Perl s function, and that a # isn't misinterpreted as a delimiter character.

For PowerShell [Core] on Unix-like platforms (macOS, Linux):

Essentially, your command would work as-is - as long as the current directory path doesn't contain spaces (which would affect your command in POSIX-like shells such as bash too).

The following variant addresses this problem:

# PowerShell [Core] on macOS and Linux
# If the current-directory path contains '#' chars.,
# use the -replace operation shown above.
perl -i'.bak' -pe "s#__SUB__CWD__#$(pwd)#g" *.sql

Note: $($pwd) would be slightly more efficient, because it uses the automatic $PWD variable rather than the pwd command, an alias for the Get-Location cmdlet.

mklement0
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