Command to remove all but select columns for each file in unix directory - bash

Question

I have a directory with many files in it and want to edit each file to only contain a select few columns.
I have the following code which will only print the first column
for i in /directory_path/*.txt; do awk -F "\t" '{ print $1 }' "$i"; done
but if I try to edit each file by adding >'$I' as below then I lose all the information in my files
for i in /directory_path/*.txt; do awk -F "\t" '{ print $1 }' "$i" > "$i"; done
However I want to be able to remove all but a select few columns in each file for example 1 and 3.

Answer 1

Given:
cat file
1 2 3
4 5 6
You can do in place editing with sed:
sed -i.bak -E 's/^([^[:space:]]*).*/\1/' file
cat file
1
4
If you want freedom to work with multiple columns and have in place editing, use GNU awk that also supports in place editing:
gawk -i inplace '{print $1, $3}' file
cat file
1 3
4 6
If you only have POSIX awk or wanted to use cut you generally do this:
Modify the file with awk, cut, sed, etc
Redirect the output to a temp file
Rename the temp file back to the original file name.
Like so:
awk '{print $1, $3}' file >tmp_file; mv tmp_file file
Or with cut:
cut -d ' ' -f 1,3 file >tmp_file; mv tmp_file file
To do a loop on files in a directory, you would do:
for fn in /directory_path/*.txt; do
awk -F '\t' '{ print $1 }' "$fn" >tmp_file
mv tmp_file "$fn"
done

Answer 2

Just to add a little more to #dawg's perfectly well working answer according to my use case.
I was dealing with CSVs, and standard CSV can have , in some values as long as it's in double quotes like for example, the below-mentioned row will be a valid CSV row.
col1,col2,col2
1,abc,"abc, inc"
But the command above was treating the , between the double quotes as delimiter too.
Also, the output file delimiter wasn't specified in the command.
These are the modifications I had to make for it handle the above two problems:
for fn in /home/ubuntu/dir/*.csv; do
awk -F ',' '{ FPAT = "([^,]*)|(\"[^\"]+\")"; OFS=","; print $1,$2 }' "$fn" >tmp_file
mv tmp_file "$fn"
done
The OSF delimiter will be the diameter of the output/result file.
The FPAT handles the case of , between quotation mark.
The regex and the information for that is mentioned ins awk's official documentation in section 4.7 Defining Fields by Content.
I was led to that solution through this answer.