Escaping Regex in SED

Question

I need to use Sed to do a search and replace. I'm replacing /**# for define('WP_POST_REVISIONS', 3);\n\n/**#.

But I can't figure out the proper escaping. Even after escaping the (obviously needed) single quotes, I still get a bash: syntax error near unexpected token ')'

What is the proper escaping in this case?

Sounded link you wanted to stop using regex in sed. You know... escape it? — Almo
– Almo, Commented Jun 6, 2013 at 15:23

Olivier Dulac · Accepted Answer · 2013-06-06 16:04:18Z

2

try to replace your:

find /start/path -name *.html -exec sed -ie 's|/**#|define(\'WP_POST_REVISIONS\', 3);|g' '{}' \;

with:

find /start/path -name '*.html' -print0  \
  | xargs -0 -n 1 sed -ie 's|/\*\*#|define('\''WP_POST_REVISIONS'\'', 3);\n/\*\*#|g'

and tell us what it gives you

(I tried to guess you were looking for the actual string "/**#" in your file(s) ... please give us examples of what you are really looking for, if it isn't that actual string)

edited Jun 6, 2013 at 16:04

answered Jun 6, 2013 at 15:57

Olivier Dulac

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5 Comments

Gaia Over a year ago

this is what I actually need, without any escaping, for clarity sed -ie "s|/**#|define('WP_POST_REVISIONS', 3);\n/**#|g"

Olivier Dulac Over a year ago

@Gaia: your /**# regexp is probably invalid: what kind of string do you need to match: "////#" or "/#" or "#" ? (ie: /*#) , or "/.......#" (...=anything) ? (ie: /.*#)

Olivier Dulac Over a year ago

I assumed you were trying to match the actual string : /**# so I replaced your regexp with the proper thing : /\*\*# (in sed, an un-escaped * means "the preceding character 0, 1 or more times")

shellter Over a year ago

@Gaia: Welcome to stackoverflow. Please consider updating your question, rather than maintaining multiple conversations (essentially about the same issue) that any new readers will have to read and digest. Good luck.

Olivier Dulac Over a year ago

@Gaia: you may want to read about regular expressions (for example your colleagues or local library may have this book : shop.oreilly.com/product/9780596528126.do)

Luke H · Accepted Answer · 2018-09-17 19:57:45Z

2

It is not sed escaping, but bash escaping.

Escaping does not work within single-quotes (')

You can use double-quotes ("), if you have no special characters like "$\ in the parameter (or escape them there if necessary):

find /start/path -name *.html -exec sed -ie "s/abc/define('WP_POST_REVISIONS', 3);/g" '{}' \;

Or quote using $', which supports escaping:

find /start/path -name *.html -exec sed -ie $'s/abc/define(\'WP_POST_REVISIONS\', 3);/g' '{}' \;

edited Sep 17, 2018 at 19:57

Luke H

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answered Jun 6, 2013 at 15:50

BeniBela

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4 Comments

Gaia Over a year ago

How do I add a CRLF with double quotes?

Gaia Over a year ago

this is what I actually need, without any escaping, for clarity sed -ie "s|/**#|define('WP_POST_REVISIONS', 3);\n/**#|g"

BeniBela Over a year ago

Type the quote, press enter, if you need a line break. And use /[*][*] to match /**

Gaia Over a year ago

Thank you, but Olivier nailed it first.

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