awk数值比较

Question

I am confused with the awk return code...

Command :

cat /etc/redhat-release |awk '{print $3 >= 6.5;}'

result : 1 (should be 0 or I am looking for 0)

Here is my file /etc/redhat-release

cat /etc/redhat-release ->
CentOS release 6.5 (Final)

I am bit confused with this...

Help ...!

---

cat /etc/redhat-release |awk '{print $3 >= 6.5}'

1

cat /etc/redhat-release |awk '{print $3}'

6.5 

Answer 1

Your confusion may stem from the result you are expecting from a true or false comparison.

From the gawk manual;

In awk, any nonzero numeric value or any nonempty string value is true. Any other value (zero or the null string, "") is false.

In other words, if the field is not "0" or "", it is true. So if you are looking for (expecting) a 0, you are expecting the comparison to be false.

The following may help you control your comparisons, but still won't make "0" the result of a true comparison.

To force a field value into a numeric context, you can add zero or multiply by one. ($3 + 0) or ($3 * 1) before comparing to a numeric value. To force a field value into a string context, you can concatenate with an empty string. Remember awk uses white space to concatenate. ($3 "")

Answer 2

Looks like your intention is to print the whole line when the version (field 3) is >= 6.5. In an awk expression, the condition should precede action, like this (no need for cat ... | ):

awk '$3 >= 6.5 {print}' /etc/redhat-release

since print is the default action, just awk '$3 >= 6.5' /etc/redhat-release would suffice (as Inian pointed out).

The command

awk '{print $3 >= 6.5}' /etc/redhat-release

gives the output of 1 which is the value of expression, as $3 >= 6.5 evaluates to true.