How to print first column of next line in current line

awktext processing

I have some file like this :

abc 123    
abc 789  
bcd 456  
acb 135

I would like to print first column of next line in current line.

Desired output:

abc  123 abc  
abc 789 bcd  
bcd 456 acb  
acb 135

I prefer to use awk.

Best Answer

Memorise the previous line:

awk 'NR > 1 { print prev, $1 } { prev = $0 } END { print prev }'

This processes the input as follows:

if the current line is the second or greater, print the previous line (stored in prev, see the next step) and the first field of the current line, separated by the output field separator (the space character by default);
in all cases, store the current line in the prev variable;
at the end of the file, print the previous line.

Related Solutions

Print only lines where the first column is unique

awk normally reads each line of the input and invokes the script on it. The cases where you would use getline are few and far between. When your script is run with six lines of input, this is an overview of what happens:

Read line 1 normally

Set variables
Call getline, which reads line 2
Compare variables

Read line 3 normally

Set variables
Call getline, which reads line 4
Compare variables

Read line 5 normally

Set variables
Call getline, which reads line 6
Compare variables

Obviously this isn’t going to work.
Secondly, you made a common mistake in your awk code. In awk, fields from the input are referenced as $number and variables are referenced as variable_name. This is different from shell scripts, where command line arguments are referenced as $number and variables are referenced as $variable_name. Your test
```
if ($1 != $path)
```
should be
```
if ($1 != path)
```
Your overall approach is flawed. You can’t identify strings that occur only once in the file by looking at two lines at a time. I believe that you can do it by looking at three lines at a time (i.e., by keeping the two previous lines in variables), but things like that get complicated and messy. It’s probably simpler to count occurrences. Here’s a minimal modification on your script to do that.
```
awk '{
  if ($1 != path) {
    if (count == 1) {
      print prev
    }
    count=1
  }
  else count++
  prev=$0; path=$1
}
END {
    if (count == 1) {
      print prev
    }
}'
```
I deleted type, since you never used it.

Disclosure: This is essentially the same as the last part of glenn’s answer.

How to print a specific column condition using awk

Just replace print with printf to print the value on the same line, and insert new line after finishing the iteration on the the line:

awk '{{for(i=1;i<=NF;i++)if($i == "name:") printf $(i+1)" "$(i+2)" "} print ""; }' yourfile

The output:

1 AAA 
2 AAA 3 BBB 
1 AAA 2 BBB 5 CCC

Best Answer

Related Solutions

Print only lines where the first column is unique

How to print a specific column condition using awk

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