[SOLVED] awk print dos not print text value

jozelo · 10-23-2013, 04:22 AM

I Have a simple shell-awk program called "nu" to print a list of numbers as follows:

#Genero una lista de números
#recibo 1, 2, 3 o 4 parametros Si me dan 1 la genero desde 0 hasta ese numero de 1 en 1
#Si me dan 2 la genero desde $1 hasta $2 de uno en uno
# si me dan tres desde $1 hasta $2 de $3 en $3
# si me dan cuatro desde $1 hasta $2 de $3 en $3 uso como prefijo el parametro $4

echo $# $1 $2 $3 $4
case $#
in
"1") awk "BEGIN {i=0;j=$1;k=1 ; for (x=i;x<=j;x=x+k) {print x}}";;
"2") awk "BEGIN {i=$1;j=$2;k=1 ; for (x=i;x<=j;x=x+k) {print x}}";;
"3") awk "BEGIN {i=$1;j=$2;k=$3 ; for (x=i;x<=j;x=x+k) {print x}}";;
"4") awk "BEGIN {i=$1;j=$2;k=$3 ; for (x=i;x<=j;x=x+k) {print $4 x}}";;
esac

when i run it as:
/jml58z/ShLinux# nu 1320 1339 1 CMUR0
4 1320 1339 1 CMUR0
1320
1321
1322
1323
1324
1325
1326
1327
1328
1329
1330
1331
1332
1333
1334
1335
1336
1337
1338
1339

It dos not print the variable $4 "CMUR0"
I have tried with printf("%s %d \n", $4 x) but not work either

TenTenths · 10-23-2013, 04:33 AM

Your use of $4 in this line is incorrect:

Code:

awk "BEGIN {i=$1;j=$2;k=$3 ; for (x=i;x<=j;x=x+k) {print $4 x}}";;

In that context it tells awk to print the 4th field, which doesn't exist.

You could pass your value to awk as a parameter:

Code:

awk "BEGIN {i=$1;j=$2;k=$3 ; for (x=i;x<=j;x=x+k) {print ARGV[1] x}}" $4;;

Which will give you:

#./nu 1320 1339 1 CMUR0
4 1320 1339 1 CMUR0
CMUR01320
CMUR01321
CMUR01322
CMUR01323
CMUR01324
CMUR01325
CMUR01326
CMUR01327
CMUR01328
CMUR01329
CMUR01330
CMUR01331
CMUR01332
CMUR01333
CMUR01334
CMUR01335
CMUR01336
CMUR01337
CMUR01338
CMUR01339

jozelo · 10-23-2013, 04:37 AM

It worked perfectly thanks a lot