Remove last comma within a pattern

Question

If I have the following text within a file (somefile.txt):

CREATE TABLE "table_name" (
   "id" int(11) unsigned NOT NULL,
   "some_field" varchar(10),
);

CREATE TABLE "another_table" (
   "id" int(11) unsigned NOT NULL,
   "some_other_field" varchar(10),
);

I want to remove the last trailing comma in each statement, so that it becomes:

CREATE TABLE "table_name" (
   "id" int(11) unsigned NOT NULL,
   "some_field" varchar(10)
);

CREATE TABLE "another_table" (
   "id" int(11) unsigned NOT NULL,
   "some_other_field" varchar(10)
);

I've used the regular expression \,$\n\) but I can't seem to get this to work with sed, which throws:

sed: -e expression #1, char 23: Unmatched ) or \)

when I use:

sed -i -e 's/\,$\n\)/)/g' somefile.txt

jofel · Accepted Answer · 2015-09-07 10:23:26Z

If the syntax of the file is everywhere like in the examples, you can use

sed -i -n -e '1h;1!H;${g;s/\,\n);/\n);/g;p}' somefile.txt

Explanation:

1h           # copy first line the hold buffer
1!H          # not first line -> append to the hold buffer
${           # execute at the end
   g          # copy hold buffer back to pattern buffer
   s/ ... /   # multiline replacement in pattern buffer
   p          # print pattern buffer
}

(see also http://austinmatzko.com/2008/04/26/sed-multi-line-search-and-replace/)

This way, the whole file is read, kept and modified in memory, if the file is too large for this, a different way need to be chosen.

Costas · Accepted Answer · 2015-09-07 12:53:22Z

1

sed 'N;s/,\n)/\n)/;P;D' file

or for GNU sed

sed -z 's/,\n)/\n)/g' file

or awk

awk '
    f{
        if(!/);/)
            print ","
        else 
            print ""
        f=0
    }
    /,$/{
        sub(",$", "")
        printf $0
        f=1
        next
    }
    1' file

edited Sep 7, 2015 at 12:53

answered Sep 7, 2015 at 12:32

Costas

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