在bash中以相反的顺序读取和存储数据

Question

I have copied 2 column data to a file. Since the cluster key of my_date is set to return in descending order

     echo "copy home.admin (id,my_date) to 'myOutputFile';" > copyInputs.cql

myOutputFile -

     TEST1,2015-01-01 15:00:00+0000
     TEST1,2014-09-04 14:00:00+0000
     4.VOD,2015-08-18 04:00:00+0000
     4.VOD,2015-06-26 04:00:00+0000
     4.VOD,2015-05-13 04:00:00+0000
     000TEST8,2015-11-19 05:00:00+0000

First column is id and second is my_date. I wanted to read the data in reverse order for each id. So the output should be like this-

     TEST1,2014-09-04 14:00:00+0000
     TEST1,2015-01-01 15:00:00+0000
     4.VOD,2015-05-13 04:00:00+0000
     4.VOD,2015-06-26 04:00:00+0000
     4.VOD,2015-08-18 04:00:00+0000
     000TEST8,2015-11-19 05:00:00+0000

After getting this output am preparing an update statement to populate one new column my_rev.my_rev will start from 100 for eaach id and increment until i find a new id.

    update home.admin my_rev =100 where id = 'TEST1' and my_date = '2014-09-04 14:00:00+0000';
    update home.admin my_rev =101 where id = 'TEST1' and my_date = '2015-01-01 15:00:00+0000';
    update home.admin my_rev =100 where id = '4.VOD' and my_date = '2015-05-13 04:00:00+0000';
    update home.admin my_rev =101 where id = '4.VOD' and my_date = '2015-06-26 04:00:00+0000';
    update home.admin my_rev =102 where id = '4.VOD' and my_date = '2015-08-18 04:00:00+0000';

Any suggestion?

Answer 1

I wanted to read the data in reverse order for each id

This prints each id in reverse order:

$ awk -F, '$1==prev {s=$0 "\n" s; next} { printf "%s",s; s=$0 "\n"; prev=$1} END{printf "%s",s}' infile
TEST1,2014-09-04 14:00:00+0000
TEST1,2015-01-01 15:00:00+0000
4.VOD,2015-05-13 04:00:00+0000
4.VOD,2015-06-26 04:00:00+0000
4.VOD,2015-08-18 04:00:00+0000
000TEST8,2015-11-19 05:00:00+0000

How it works:

This script uses two variables. prev contains the ID for the previous line. s contains the lines for the most recent ID in reverse order.

• -F,

  This tells awk to use a comma as the field separator.

• $1==prev {s=$0 "\\n" s; next}

  For lines that have the same ID (field 1, denoted $1 ), this adds the new line to the beginning of variable s . The rest of the commands are skipped and awk jumps to the next line.

• printf "%s",s; s=$0 "\\n"; prev=$1

  If we get here, we are starting a new ID. In this case, we print the lines saved in s from the previous ID. We update s with the current line and we set prev to the current ID .

• END{printf "%s",s}

  After we reach the end of the file, we print s for the last ID.

Alternative

If you want to do a more complex re-ordering, this invokes sort , with all of its flexibility, for each id , keeping each id in its original order:

$ awk -F, -v s=sort '$1==prev {print | s; next} {close(s); print | s; prev=$1}' infile
TEST1,2014-09-04 14:00:00+0000
TEST1,2015-01-01 15:00:00+0000
4.VOD,2015-05-13 04:00:00+0000
4.VOD,2015-06-26 04:00:00+0000
4.VOD,2015-08-18 04:00:00+0000
000TEST8,2015-11-19 05:00:00+0000

Re-formatting

If outfile contains the output of the sorting command above, then run:

$ awk -F, '{if ($1==prev)n++; else n=100; prev=$1; printf "update home.admin my_rev =%i where id = '\''%s'\'' and my_date = '\''%s'\'';\n",n,$1,$2}' outfile
update home.admin my_rev =100 where id = 'TEST1' and my_date = '2014-09-04 14:00:00+0000';
update home.admin my_rev =101 where id = 'TEST1' and my_date = '2015-01-01 15:00:00+0000';
update home.admin my_rev =100 where id = '4.VOD' and my_date = '2015-05-13 04:00:00+0000';
update home.admin my_rev =101 where id = '4.VOD' and my_date = '2015-06-26 04:00:00+0000';
update home.admin my_rev =102 where id = '4.VOD' and my_date = '2015-08-18 04:00:00+0000';
update home.admin my_rev =100 where id = '000TEST8' and my_date = '2015-11-19 05:00:00+0000';

Answer 2

sort should do the trick

sort -r -t, -k1,2 infile

In general, the only option you need is -r .