Stupid regex problem

[Koos Pol]

Hi,

Can someone explain why

Pattern.matches(".*", "123\r\n");

yields false? (Removing \r\n makes it return true)

Thanks,
Koos

 

[Gordon Beaton]

Can someone explain why

Pattern.matches(".*", "123\r\n");

yields false? (Removing \r\n makes it return true)

Did you read what the Pattern class documentation says matching line
terminators?

/gordon

 

[Gordon Beaton]

Did you read what the Pattern class documentation says matching line
terminators?

....says *about* matching line terminators.

/gordon

 

[Roedy Green]

Can someone explain why

Pattern.matches(".*", "123\r\n");

see http://mindprod.com/jgloss/regex.html#QUOTING

 

[Koos Pol]

Did you read what the Pattern class documentation says matching line
terminators?

Aah, thanks Gordon. My Perl background stood in the way. In Perl the EOL
would be irrelevant.

Thanks again.
Koos - Using (?s:.) now

 

[Koos Pol]

see http://mindprod.com/jgloss/regex.html#QUOTING

If I'm not mistaken that's about the regex itself and not the search
string.

Koos

 

[Roedy Green]

If I'm not mistaken that's about the regex itself and not the search
string.

The regex IS the search string. The complexity comes from quoting for
two logical levels, the regex and Java string literals, both of which
use \ The entry covers both.

 

[Stefan Ram]

Aah, thanks Gordon. My Perl background stood in the way.
In Perl the EOL would be irrelevant.

It depends on the meaning of »the EOL would be irrelevant«.

For example, in Perl 5.8.3,

print "12345" =~ /^.*$/ ? 1 : 0;
print "123\r4\n5" =~ /^.*$/ ? 1 : 0;

prints

10

 

[Koos Pol]

It depends on the meaning of »the EOL would be irrelevant«.

No it does not. In the described problem the EOL is simply irrelevant.

print "12345" =~ /^.*$/ ? 1 : 0;
print "123\r\n" =~ /^.*$/ ? 1 : 0;
print "12345 " =~ /./ ? 1 : 0; # or even this
print "123\r\n" =~ /./ ? 1 : 0; #

prints 1111

Cheers,
Koos

 

[Stefan Ram]

No it does not. In the described problem the EOL is simply irrelevant.
print "12345" =~ /^.*$/ ? 1 : 0;
print "123\r\n" =~ /^.*$/ ? 1 : 0;
print "12345 " =~ /./ ? 1 : 0; # or even this
print "123\r\n" =~ /./ ? 1 : 0; #
prints 1111

Insofar you are right. But the following example for Perl
5.8.3 shows that "\r\n" is not always matched by the ".*":

print "12345" =~ /^.*$/ ? 1 : 0;
print "123\r\n5" =~ /^.*$/ ? 1 : 0;

It prints »10« here. It seems as if there was a special rule
for a trailing "\r\n", that does not apply anymore when "\r\n"
appears in the middle of a text.

What I sometimes use to match it, is:

print "12345" =~ /^[^\001]*$/ ? 1 : 0;
print "123\r\n5" =~ /^[^\001]*$/ ? 1 : 0;

This could be used in Java as well, I assume.