How to show certain column of a record using sed

linux, sed

Solution

Update:

echo "5678:robert dylan :d.g.m. :marketing :04/19/43 85000" | 
sed 's/:/\n/g' | sed '1d;3d;$d'| sed 'N;s/\n/ - /'

yields

robert dylan - marketing

Explanation:

Works by splitting the line into several lines based on `:`, deleting the first, third and last lines, and then joining them up again.

NOTE: In the last expression, you can specify what separates the name from the designation by putting something else between the final set of `/ /` in `'N;s/\n/ /'`

Previous AWK solution:

Not `sed`, but `awk` is quite a natural tool for this if it is acceptable:

$ echo "5678:robert dylan :d.g.m. :marketing :04/19/43 85000" | awk -F":" '{print $2, $4}'

yields

robert dylan  marketing

Alternatively, if your data was stored in a file named `data.txt`:

awk -F":" '{print $2, $4}' data.txt

would produce the same output.

awk is really well suited for these sort of tasks.

Explanation:

awk -F":" '{print $2, $4}'

`-F` sets the field separator to `:`, `print $2, $4` print resulting fields 2 and 4 respectively. You can use `printf` to format the output as specific as you need.

Problem

Original source