Bash AWK 和正则表达式应用于特定列

Question

I have the following dataset

Name,quantity,unit
car,6,6
plane,7,5
ship,2,3.44
bike,8,7.66

I want to print only the names which has unit in whole numbers.

I have done the following which does not give out the result

#!/bin/bash
awk 'BEGIN {
 FS=","
}
/^[0-9]*$/ {
 print "Has Whole numbers: " $1
} 
' file.csv

The result should be

Has Whole numbers: car
Has Whole numbers: plane

Answer 1

Added a couple of lines to your test data:

Name,quantity,unit
car,6,6
plane,7,5
ship,2,3.44
bike,8,7.66
Starship,1,1.0
Super Heavy,2,0
null,0,

And awk:

$ awk -F, 'int($3)==$3 ""' file

Output:

car,6,6
plane,7,5
Super Heavy,2,0

int($3) makes an integer of $3 and $3 "" turns $3 to a string.

Answer 2

Try changing /^[0-9]*$/ to $3 ~ /^[0-9]*$/ && $3 != 0 once in your tried attempt it should work then.

In case you DO NOT want to hard code field number and want to find out unit field number automatically then try following.

awk -F="," -v field_val="unit"  '
FNR==1{
  for(j=1;j<=NF;j++){
    if($j==field_val){
       field_number=j
       next
    }
  }
}
$field_number ~ /[0-9]*$/ && $field_number!=0{
    print "Has whole numbers: " $1
}'  Input_file

Answer 3

If you are sure 3rd column is a number:

awk -F, '(NR != 1 && $3 !~ /\./){print "Has Whole numbers:", $1}' file.csv

or well actually its better the way you did it:

awk -F, '$3 ~ /^[0-9]$/{print "Has Whole numbers:", $1}' file