Is it possible to pass arguments to a sed script

That was a hard one. Assuming you have a file like this:

$ cat file
word
line with a word and words and wording wordy words.

Where:

• Line 1: is the search pattern that should be held in the hold space and quoted to `word`.
• Line 2: is the line to seach and replace globally.

The sed command:

sed -n '1h; 2{x;G;:l;s/^\([^\n]\+\)\n\(.*[^`]\)\1\([^`]\)/\1\n\2`\1`\3/;tl;p}' file

Explanation:

• 1h; save the first line to the hold space (this is wait we want to search for).
  • hold space contains: word
• 2{...} applies to the second line.
• x; exchange the pattern space and the hold space.
• G; append the hold space to the pattern space. In the pattern space we have now:

word # I will call this line the "pattern line" from now on
line with a word and words and wording wordy words.

• :l; set a label called l as point for later.
• s/// do the actual search/replace in the pattern space mentioned above:
  • ^\([^\n]\+\)\n search in the "pattern line" for all characters (from the beginning of the line ^) which are not a newline [^\n] (one or more times \+), until a newline \n. This is now stored in the back-reference \1. It contains the "pattern line".
  • (.*[^`]) search for any character .* followed by a character, which is not a backtick [^`]. This is stored in \2. \2 contains now: line with a word and words and wording wordy, until the last occurence of word, because...
  • \1 is the next search term (the back-reference \1, word), hence what the "pattern line" contains.
  • ([^`]) this is followed by another character which is not a backtick; saved to reference \3. If we don't do this (and the part in \2 from above), we would end of in an endless loop quoting the same word, again and again -> ````word````, because s/// would always be successful and tl; jumps back to :l (see tl; further down).
  • \1\n\2\1\3 all of the above is replaced by the back-references. The second \1 is the one we should quote (note the first reference is the "pattern line").
• tl; if the s/// was successful (we replaced something) jump to the label called l and start again until there is nothing more to search and replace. This is the case, when all occurences of word are replaced/quoted.
• p; when all is done, print the altered line (pattern space).

The output:

$ sed -n '1h; 2{x;G;:l;s/^\([^\n]\+\)\n\(.*[^`]\)\1\([^`]\)/\1\n\2`\1`\3/;tl;p}' file
word
line with a `word` and `word`s and `word`ing `word`y `word`s.

The complication here is that if you replace all ' with " and then all " with ', you will be left with only '. So, you will first need to replace ' with something else—for example, the NULL character (\0) which you can safely assume won't be in your input file—then " with ' and then replace that something else with " again. For example, in perl:

$ perl -pe "if(/KEYWORD){s/'/\0/g; s/\"/'/g; s/\0/\"/g}" file
KEYWORD_1 table name column = 'string' AND column = 'string'
Additional text
      .
      .
      .
KEYWORD_2 text "text in quote" etc. 

Explanation

• -pe : print each input line after applying the script given by -e.
• if(/KEYWORD/){something} : do something only if this line matches KEYWORD.
• s/foo/bar/g : replace all occurrences of foo in the line with bar. The g means "global". Without it, only the 1st occurrence on each line would be replaced.

Note that since the script itself is inside double quotes, the double quotes inside the script need to be escaped (\").

As pointed out in the comments, there's a more direct way I should have thought of in the first place:

perl -pe "tr/'\"/\"'/ if /^KEYWORD/" file

The tr operator transliterates lists of characters. The general format is tr/searchlist/replacementlist/. So this will replace all ' with " and all " with ' only on lines matching KEYWORD.