[SOLVED] bash substitution on apostrophes help

otkaz · 10-04-2011, 09:19 PM

Hi I'm having a problem with part of a script where I clean a variable of certain chars by adding an escape \ to the beginning of them.
Example

Code:

FILENAME="${FILENAME//[/\[}"
FILENAME="${FILENAME//]/\]}"
FILENAME="${FILENAME//(/\(}"
FILENAME="${FILENAME//)/\)}"
FILENAME="${FILENAME//&/\&}"
FILENAME="${FILENAME// /\ }"

got stuck on ' apostrophes
this does not work

Code:

FILENAME="${FILENAME//'/\'}"

recognises them as syntax
./testscript.sh: eval: line 6: unexpected EOF while looking for matching `''
./testscript.sh: eval: line 7: syntax error: unexpected end of file
Sorry I'm sure this is pretty basic stuff I should be able to get on my own, but I'm really stuck any help is greatly appreciated.

Tinkster · 10-04-2011, 10:01 PM

Quote:

Originally Posted by otkaz

Hi I'm having a problem with part of a script where I clean a variable of certain chars by adding an escape \ to the beginning of them.
Example

Code:

FILENAME="${FILENAME//[/\[}"
FILENAME="${FILENAME//]/\]}"
FILENAME="${FILENAME//(/\(}"
FILENAME="${FILENAME//)/\)}"
FILENAME="${FILENAME//&/\&}"
FILENAME="${FILENAME// /\ }"

got stuck on ' apostrophes
this does not work

Code:

FILENAME="${FILENAME//'/\'}"

recognises them as syntax
./testscript.sh: eval: line 6: unexpected EOF while looking for matching `''
./testscript.sh: eval: line 7: syntax error: unexpected end of file
Sorry I'm sure this is pretty basic stuff I should be able to get on my own, but I'm really stuck any help is greatly appreciated.

Code:

echo $FILENAME 
huh'hah
FILENAME="${FILENAME//\'/\'}"
echo $FILENAME 
huh\'hah

That what you're after?

David the H. · 10-04-2011, 10:22 PM

The field inside the parameter substitution is considered a separate quoting environment from the main line, so quotes, the backslash itself, and a couple of other characters need to be individually escaped.

Code:

FILENAME="${FILENAME//\'/\'}"
FILENAME="${FILENAME//\"/\\\"}"

It may seem odd that for the single quote, the replacement string needs to be kept literal, but for the double quote the backslash needs to be escaped as well.

I believe this is due to the whole line being parsed after the substitution is made, where the outer double-quotes protect the single quotes in the variable string , but not the doubles. \" and \\ are parsed inside double quotes, but \' is not.

See, and carefully read, the QUOTING section of the bash man page for more details on this.

otkaz · 10-04-2011, 11:51 PM

using FILENAME="${FILENAME//\'/\'}" instead of FILENAME="${FILENAME//'/\'}" did the trick. I would have probably never figured this out on my own thanks guys!

grail · 10-05-2011, 04:41 AM

My question would be around what is the point of replacing an apostrophe with itself?

David the H. · 10-05-2011, 08:22 AM

Quote:

Originally Posted by grail

My question would be around what is the point of replacing an apostrophe with itself?

It's not. It's replacing the apostrophe with a backslashed apostrophe. The replacement string outputs literally. Try it out.

It appears to work similarly to sed substitutions. The LHS needs to strictly parse escapes, but the RHS does not. The only backslash interpretation that occurs on the substituted string comes after the parameter has been expanded, as I pointed out before.

grail · 10-06-2011, 12:38 AM

I must be missing something then:

Code:

$ x="one'two"
$ echo $x
one'two
$ echo ${x//\'/\'}
one'two

So I would consider this tried out and not what has been advertised

Code:

$ bash --version
GNU bash, version 4.2.10(1)-release (x86_64-unknown-linux-gnu)
Copyright (C) 2011 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

David the H. · 10-06-2011, 01:14 AM

You didn't quote the variable, so the backslash was subsequently removed again after the expansion.

The behavior of backslashes is different inside double quotes than when unquoted. Whereas all escapes are processed in unquoted strings, soft quotes only reduce \$, \`, \", \\, and \<newline> into their literal equivalents, along with certain escape sequences like \n. All other combinations are retained as-is.

grail · 10-06-2011, 02:26 AM

@David - thank you for the clarity ... my bad on forgetting to quote

So what about replacing say one of the letters in the line with an apostrophe and still having the variable quoted?

Code:

$ x="one'two"
$ echo $x
one'two
$ echo "${x//e/\'}"
on\''two
$ echo "${x//e/'}"
>
# desired output
on''two

David the H. · 10-06-2011, 10:44 AM

Wow, that is a tricky one. I honestly can't see any direct way to accomplish it. It looks like you may have discovered a genuine weakness in the syntax.

Quoting and backslashing doesn't work, obviously, and I even tried this, with no success:

Code:

$ echo "${x//e/$'\047'}"
bash: bad substitution: no closing `}' in "${x//e/'}"

But this does work, probably because the ascii conversion takes place after the substitution:

Code:

$ x="one'two"
$ echo -e "${x//e/\047}"
on''two

This works too:

Code:

$ y="'"
$ echo "${x//e/$y}"
on''two