CODEWITHSUNDEEP
bashprintingawkoutput
user3443445
user3443445

Reputation: 1

Awk: field not printing to file, but works to stdout

awk noob here.

Fundamentally, I have an awk command, ending in {print $4}, that, when fed to an echo > file command, does not print the value of the requested field. Output is blank.

I'm writing a bash script with the goal of creating a file containing four fields: subject, structure, side, and volume, e.g.

136     Hippocampus     Left     3450

And I'm drawing from subject-specific files with a bunch of header lines and then a tab-delimited table of brain structures, volumes and other data.

12    17    3450    3450.7    Left-Hippocampus

The script iterates over these files with loops for subject, then structure and side, and for each combination uses awk to match $side-$struct with the appropriate field in the file, then print the field with the volume. Then it echoes all that data to the outfile.

set volume = `awk -v side="$side" -v struct="$struct" '$5 ~ /$side-$struct/ {print $4}' filename.stats

echo "${subj}    ${struct}    ${side}    ${volume}" >> $outfile

However, the file always shows the first three columns' appropriate values, but never the volume value; that column is blank.

When I run a sample awk command in stdout,

awk '$5 ~ /Left-Hippocampus/ {print $4}' filename.stats

It returns the number I want.

Why won't the script correctly print to $outfile the volume data as well as that for the looped variables?

I've seen people talk about buffering speed, and I've tried flushing; I've fed awk the data file with cat; I've tried printing all variables from awk after declaring them with further -v flags; nothing seems to work. Eventually they all do the same thing, print the first three columns in $outfile and not the fourth.

Thanks in advance for any help, this is driving me nuts! The script's been practically ready to run for over a day now...

Upvotes: 0

Views: 1119

Answers (2)

tripleee
tripleee

Reputation: 189638

The set command doesn't do what you are hoping.

The Bourne shell syntax for an assignment is simply

variable=value

with no spaces (use quotes if the value needs to contain whitespace).

So,

volume=$(awk -v side="$side" -v struct="$struct" '$5 ~ (side "-" struct) {print $4}' filename.stats)

Notice also how the variables side and struct are now actually used by the script.

However, I'm wondering if you should not be doing all your processing in Awk instead.

Upvotes: 2

Izaaz Yunus
Izaaz Yunus

Reputation: 2828

Try this

volume=`awk -v side="$side" -v struct="$struct" '$5 ~ /$side-$struct/ {print $4}' filename.stats
echo "${subj}    ${struct}    ${side}    ${volume}" >> $outfile

I've removed the spaces on either side of the '=' sign. I've seen that having a space around the equal sign throws errors when assigning value to a variable

Upvotes: 0

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