printf - Frustrated with simple awk command -

printf - Frustrated with simple awk command -

i trying list out contents of field 1 using function:

help(){ if [[ $# -eq 0 ]] ; echo '######################################' echo '' echo 'argument run run name must given: ./report.sh name' echo 'report names are:' allnames=$(cut -d '|' -f 1 $configfile | awk '{printf $0"\n"}') echo $allnames echo '######################################' exit 0 fi }

the output :

$ bin/report.sh ###################################### argument run run name must given: ./report.sh name study names are: itema itemb ######################################

whereas want:

$ bin/report.sh ###################################### argument run run name must given: ./report.sh name study names are: itema itemb ######################################

if run cutting command get:

[david@kallibu]$ cutting -d '|' -f 1 conf/report.conf itema itemb

whatdo need alter newline ?

your code be,

help(){ if [[ $# -eq 0 ]] ; echo '######################################' echo '' echo 'argument run run name must given: ./report.sh name' echo 'report names are:' allnames=$(awk -f'|' '{print $1}' $configfile) echo "$allnames" echo '######################################' exit 0 fi }

you seek awk -f'|' '{print $1}' $configfile command value of first column | delimiter.

you need set allnames within double quotes. only, allnames variable got expanded.

awk printf newline