awk script for finding smallest value from column

Question

I am beginner in AWK, so please help me to learn it. I have a text file with name snd and it values are

1 0 141
1 2 223
1 3 250
1 4 280

I want to print the entire row when the third column value is minimu

Answer 1

This should do it:

awk 'NR == 1 {line = $0; min = $3}
     NR > 1 && $3 < min {line = $0; min = $3}
     END{print line}' file.txt

EDIT:

What this does is:

• Remember the 1st line and its 3rd field.
• For the other lines, if the 3rd field is smaller than the min found so far, remember the line and its 3rd field.
• At the end of the script, print the line.

Note that the test NR > 1 can be skipped, as for the 1st line, $3 < min will be false. If you know that the 3rd column is always positive (not negative), you can also skip the NR == 1 ... test as min's value at the beginning of the script is zero.

EDIT2:

This is shorter:

awk 'NR == 1 || $3 < min {line = $0; min = $3}END{print line}' file.txt