Bash + sed/awk/cut to delete nth character - bash

Question

I trying to delete 6,7 and 8th character for each line.
Below is the file containing text format.
Actual output..
#cat test
18:40:12,172.16.70.217,UP
18:42:15,172.16.70.218,DOWN
Expecting below, after formatting.
#cat test
18:40,172.16.70.217,UP
18:42,172.16.70.218,DOWN
Even I tried with below , no luck
#awk -F ":" '{print $1":"$2","$3}' test
18:40,12,172.16.70.217,UP
#sed 's/^\(.\{7\}\).\(.*\)/\1\2/' test { Here I can remove only one character }
18:40:1,172.16.70.217,UP
Even with cut also failed
#cut -d ":" -f1,2,3 test
18:40:12,172.16.70.217,UP
Need to delete character in each line like 6th , 7th , 8th
Suggestion please

Answer 1

With GNU cut you can use the --complement switch to remove characters 6 to 8:
cut --complement -c6-8 file
Otherwise, you can just select the rest of the characters yourself:
cut -c1-5,9- file
i.e. characters 1 to 5, then 9 to the end of each line.
With awk you could use substrings:
awk '{ print substr($0, 1, 5) substr($0, 9) }' file
Or you could write a regular expression, but the result will be more complex.
For example, to remove the last three characters from the first comma-separated field:
awk -F, -v OFS=, '{ sub(/...$/, "", $1) } 1' file
Or, using sed with a capture group:
sed -E 's/(.{5}).{3}/\1/' file
Capture the first 5 characters and use them in the replacement, dropping the next 3.

Answer 2

it's a structured text, why count the chars if you can describe them?
$ awk '{sub(":..,",",")}1' file
18:40,172.16.70.217,UP
18:42,172.16.70.218,DOWN
remove the seconds.

Answer 3

The solutions below are generic and assume no knowledge of any format. They just delete character 6,7 and 8 of any line.
sed:
sed 's/.//8;s/.//7;s/.//6' <file> # from high to low
sed 's/.//6;s/.//6;s/.//6' <file> # from low to high (subtract 1)
sed 's/\(.....\).../\1/' <file>
sed 's/\(.{5}\).../\1/' <file>
s/BRE/replacement/n :: substitute nth occurrence of BRE with replacement
awk:
awk 'BEGIN{OFS=FS=""}{$6=$7=$8="";print $0}' <file>
awk -F "" '{OFS=$6=$7=$8="";print}' <file>
awk -F "" '{OFS=$6=$7=$8=""}1' <file>
This is 3 times the same, removing the field separator FS let awk assume a field to be a character. We empty field 6,7 and 8, and reprint the line with an output field separator OFS which is empty.
cut:
cut -c -5,9- <file>
cut --complement -c 6-8 <file>

Answer 4

Just for fun, perl, where you can assign to a substring
perl -pe 'substr($_,5,3)=""' file

Answer 5

With awk :
echo "18:40:12,172.16.70.217,UP" | awk '{ $0 = ( substr($0,1,5) substr($0,9) ) ; print $0}'
Regards!

Answer 6

If you are running on bash, you can use the string manipulation functionality of it instead of having to call awk, sed, cut or whatever binary:
while read STRING
do
echo ${STRING:0:5}${STRING:9}
done < myfile.txt
${STRING:0:5} represents the first five characters of your string, ${STRING:9} represents the 9th character and all remaining characters until the end of the line. This way you cut out characters 6,7 and 8 ...