How to take value from file (bash script)

teguh.purnama · 09-08-2008, 07:11 AM

The content of the file is (file name is varlist):

Quote:

list1: 235.214 -6532.211 0.00214
list2(addition): -6524.214 23.125 0.00234

how to take those value using bash script. I tried to use

Quote:

$var1=`cat varlist | grep "list1" | expr substr \([0-9]......)\`

it result only error.

matthewg42 · 09-08-2008, 07:22 AM

Do not prefix a variable name with $ when assigning to it, only when taking the value.
Unnecessary use of cat, you can pass the input file name to grep directly as an argument.
You can use cut with -d option to split lines by some character, which might be cleaner than using expr.

david1941 · 09-08-2008, 07:23 AM

var1 = `awk '/list1/{print $2}' varlist`

Dave

More clearly var1 = $(awk '/list1/{print $2}' varlist)

teguh.purnama · 09-08-2008, 07:32 AM

awk script seems easier for me.
thanks for the answer.