perl using xml passed from stdin using XML::Simple

Question

i want to read an xml from stdin.

For parsing the xml I would use $data = $xml->XMLin(<STDIN>), anyway which is the most efficient way for evaluating all the data passed from stdin without rewriting it into a file?

If I use $data = $xml->XMLin(<STDIN>) it does not work.

Any suggestion?

Main goal is to use the data passed from the stdin in the most efficient way.

gangabass · Accepted Answer · 2011-11-24 09:05:59Z

5

 $data = $xml->XMLin('-');

Note, the filename '-' can be used to parse from STDIN.

answered Nov 24, 2011 at 9:05

gangabass

10.7k2 gold badges26 silver badges36 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Kerby82 Over a year ago

what's the meaning of "-" is?

gangabass Over a year ago

It's just a special symbol which XML::Simple understand as STDIN. This is XML::Simple feature described in docs.

Grant McLean Over a year ago

It's quite common for Unix utilities and Perl functions that expect a filename to recognise '-' as meaning read from STDIN.

stevenl · Accepted Answer · 2011-11-24 09:23:38Z

4

Passing a reference to the builtin filehandle also works using typeglob notation.

my $data = XMLin(\*STDIN);

answered Nov 24, 2011 at 9:23

stevenl

6,79828 silver badges33 bronze badges

Collectives™ on Stack Overflow

perl using xml passed from stdin using XML::Simple

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related