Shell – Sort an output by year month and date

Awk

You can adapt the linked pipe in a straight forward way:

$ ./script | awk '{ print $NF,$0 }' | sort -k1,1 -n | cut -f2- -d' '

In awk the expression $x references the x-th column of the current line (starting with 1) - and the predefined variable NF stores the number of columns of the current line, thus print $NF,$0 prints for each line the last column and the complete line (because $0 denotes the complete line). The cut command then outputs the 2nd to the last column of each line.

The -k1,1 part of sort means that only the first column is used as sort key - this only makes a difference when more than one line have the same value in the first column. Without -k1,1 the following columns will influence the relative order (as secondary and so on sorting key) in that case. With -k1,1 only the first column is used as sorting key - and the relative order of lines with the same key is not changed (i.e. a stable sort is performed).

sed

Alternatively you can solve it via sort and sed:

$ ./script | sed 's/^\(.\+[ \t]\+\)\([0-9]\+ *\)$/\2 \1/' | \
     sort -k1,1 -n | sed 's/^\([0-9]\+\) \(.\+\)$/\2 \1/'

Where the newline at the end of the first line is escaped via \ - you can remove that \ and enter the pipe as one line.

The idea is to first move the last column to the front, sort by the first column and then put it to the back again.

It is assumed that the last column is separated via white space, i.e. [ \t]\+ (spaces or tabs).

The sed expressions do the swapping via group references (e.g. \2 \1) - the groups are marked in the pattern via escaped parentheses: \(...\)

You're using the system() function in the wrong way. Or rather, that's not what to use here as it doesn't return the output from the command, only the exit status (the output goes to the terminal).

Assuming date is GNU date, this is an awk script that will perform the date reformatting:

BEGIN { OFS = FS = "," }

$1 {
    cmd = sprintf("date -d '%s' '+%%d-%%m-%%Y' 2>/dev/null", $1);
    cmd | getline $1;
    print;
    close(cmd);
}

Running it:

$ awk -f script.awk data.csv
Bogført,Tekst,Beløb,Saldo
03-02-2017, random text,-425,-611524.54

The script will discard empty input lines. It creates a command string cmd that does the actual date conversion using GNU date. Errors from date will be thrown away (and $1 will remain unchanged).

To do it with cvssql (from csvkit):

$ sed '1,2d' data.csv | csvsql -H --query 'SELECT strftime("%d-%m-%Y", a), b, c, d FROM stdin' | sed '1d' >new_data.csv

The data in new_data.csv will be without the header row. To add it back:

$ cat <( head -n 1 data.csv ) new_data.csv >even_newer_data.csv