Shell Programming and Scripting

Hi All,

Can some one help me in identifying the significance of character "$" ,Which is playing critical role in matching decimal point numbers as below.

Code:

[sys@localhost ~]$ echo "01#.01"|awk '{if ($0 ~ /^[0-9]+(\.[0-9]*)?$/) print}'
[sys@localhost ~]$ echo "01#.01"|awk '{if ($0 ~ /^[0-9]+(\.[0-9]*)?/) print}'
01#.01
[sys@localhost ~]$

Regards,
Rmkganesh.

end of record/line

Hi,

Yes it is ,but wanna know how its eliminating special chars and alphabets.

Can u let me know.

Regards,
Ganesh G.

Rmkganesh,
The awk utility uses extended regular expressions (EREs). The ERE ^[0-9]+(\.[0-9]*)? will match any string in which the 1 or more (+) decimal digits ([0-9]) followed by zero or one occurrences (\(expression\)?) of a period (\.) followed by zero or more (*) decimal digits ([0-9]) appearing at the start of the string (^). In this case, it matches the 01 at the start of the given string (01#.01). (I.e., two decimal digits at the start of the string followed by zero occurrences of a decimal point followed by any number of decimal digits.)

When you add a dollar sign to the end of the ERE (i.e., ^[0-9]+(\.[0-9]*)?$), the match only succeeds if the match specified above appears at the end of the string. In other words, an ERE with a circumflex (^) at the start and a dollar sign ($) at the end only matches if the entire string is matched by the the ERE between the circumflex and the dollar sign. Since nothing in the ERE [0-9]+(\.[0-9]*)? matches the octothorpe (#) in the middle of the string 01#.01, the ERE cannot match this string.