使用awk或sed添加空白列

Question

I have a file with the following structure (comma delimited)

116,1,89458180,17,FFFF,0403254F98

I want to add a blank column on the 4th field such that it becomes

116,1,89458180,,17,FFFF,0403254F98

Any inputs as to how to do this using awk or sed if possible ?

thank you

Answer 1

Assuming that none of the fields contain embedded commas, you can restate the task as replacing the third comma with two commas. This is just:

sed 's/,/,,/3'

With the example line from the file:

$ echo "116,1,89458180,17,FFFF,0403254F98" | sed 's/,/,,/3'
116,1,89458180,,17,FFFF,0403254F98

Answer 2

You can use this awk ,

awk -F, '$4="," $4' OFS=, yourfile

(OR)

awk -F, '$4=FS$4' OFS=, yourfile

If you want to add 6th and 8th field,

awk -F, '{$4=FS$4; $1=FS$1; $6=FS$6}1' OFS=, yourfile

Answer 3

Through awk

$ echo '116,1,89458180,17,FFFF,0403254F98' | awk -F, -v OFS="," '{print $1,$2,$3,","$4,$5,$6}'
116,1,89458180,,17,FFFF,0403254F98

• It prints a , after third field(delimited) by ,

Through GNU sed

$ echo 116,1,89458180,17,FFFF,0403254F98| sed -r 's/^([^,]*,[^,]*,[^,]*)(.*)$/\1,\2/'
116,1,89458180,,17,FFFF,0403254F98

• It captures all the characters upto the third command and stored it into a group. Characters including the third , upto the last are stored into another group. In the replacement part, we just add an , between these two captured groups. Through Basic sed,

Through Basic sed

$ echo 116,1,89458180,17,FFFF,0403254F98| sed 's/^\([^,]*,[^,]*,[^,]*\)\(.*\)$/\1,\2/'
116,1,89458180,,17,FFFF,0403254F98

Answer 4

echo 116,1,89458180,17,FFFF,0403254F98|awk -F',' '{print $1","$2","$3",,"$4","$5","$6}'

Answer 5

Non-awk

t="116,1,89458180,17,FFFF,0403254F98"
echo $(echo $t|cut -d, -f1-3),,$(echo $t|cut -d, -f4-)

Answer 6

You can use bellow awk command to achieve that.Replace the $3 with what ever the column that you want to make it blank.

awk -F, '{$3="" FS $3;}1' OFS=, filename

Answer 7

sed -e 's/\([^,]*,\)\{4\}/&,/' YourFile

自行替换4 [内容（非逗号），而不是逗号）的序列，然后用逗号替换