awk field separator not working for first line - awk

Question

echo 'NODE_1_length_317516_cov_18.568_ID_4005' | awk 'FS="_length" {print $1}'
Obtained output:
NODE_1_length_317516_cov_18.568_ID_4005
Expected output:
NODE_1
How is that possible? I'm missing something.

Answer 1

When you are going through lines using Awk, the field separator is interpreted before processing the record. Awk reads the record according the current values of FS and RS and goes ahead performing the operations you ask it for.
This means that if you set the value of FS while reading a record, this won't have effect for that specific record. Instead, the FS will have effect when reading the next one and so on.
So if you have a file like this:
$ cat file
1,2 3,4
5,6 7,8
And you set the field separator while reading one record, it takes effect from the next line:
$ awk '{FS=","} {print $1}' file
1,2 # FS is still the space!
5
So what you want to do is to set the FS before starting to read the file. That is, set it in the BEGIN block or via parameter:
$ awk 'BEGIN{FS=","} {print $1}' file
1,2 # now, FS is the comma
5
$ awk -F, '{print $1}' file
1
5
There is also another way: make Awk recompute the full record with {$0=$0}. With this, Awk will take into account the current FS and act accordingly:
$ awk '{FS=","} {$0=$0;print $1}' file
1
5

Answer 2

awk Statement used incorrectly
Correct way is
awk 'BEGIN { FS = "#{delimiter}" } ; { print $1 }'
In your case you can use
awk 'BEGIN { FS = "_length" } ; { print $1 }'

Answer 3

Inbuilt variables like FS, ORS etc must be set within a context i.e in 1 of the following blocks: BEGIN, condition blocks or END.
$ echo 'NODE_1_length_317516_cov_18.568_ID_4005' | awk 'BEGIN{FS="_length"} {print $1}'
NODE_1
$
You can also pass the delimiter using -F switch like this:
$ echo 'NODE_1_length_317516_cov_18.568_ID_4005' | awk -F "_length" '{print $1}'
NODE_1
$