How to print empty spaces in first column using awk or sed

Question

Here is my sample input file.

ip.txt:

john    math
        science
paul    math
        science
rosy    math
jill    science
rob     math
        science
hary    math

Desired output:

john

paul

rosy
jill
rob

hary

When I use:

awk '{print $1}' ip.txt

My output is:

john
science
paul
science
rosy
jill
rob
science
hary

I don not want the second column values printed, I want the blank spaces to be printed out to a file.

How can I achieve this? I am using Solaris 5.10 with ksh.

glenn jackman · Accepted Answer · 2015-07-16 18:41:51Z

4

awk:

awk '{print (NF>1) ? $1 : ""}' file

If the number of fields is more than 1, print the first field, otherwise print an empty line.

A couple of extra thoughts:

If your data is tab-separated, then
```
awk -F '\t' '{print $1}' file
```
If you want to extract the first 8 characters
```
awk '{print substr($0,1,8)}' file
```

answered Jul 16, 2015 at 18:31

glenn jackman

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cuonglm · Accepted Answer · 2015-07-16 18:29:30Z

1

With POSIX sed:

sed -e 's/^\([^[:blank:]]*\)[[:blank:]].*/\1/;t' -ed <file

answered Jul 16, 2015 at 18:29

cuonglm

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Digital Trauma · Accepted Answer · 2015-07-16 18:30:39Z

1

If you want all the extra spaces before the subject field, you can do:

grep -o "^[^ ]* \+" ip.txt

If you don't want the extra spaces:

sed 's/ \+.*//' ip.txt

Both of these will preserve lines with empty name fields.

answered Jul 16, 2015 at 18:30

Digital Trauma

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3 Answers 3