Get list of files.

[Binny V A]

Hello Everyone,

I want to get a list of all HTML files in a folder. How do I do this?

I tried
$list = <*.htm>;
but it gave only one html file. The same result with glob("*.htm")

I am currently doing this with
$list = `dir *.htm`;
and then spliting it. But that is not portable.

Other questions - How do I get the names of folders?
How do I relusivly list all files in a folder and within folders in that folder?

Thanking You,
Binny V A
http://www.geocities.com/binnyva
Perl Special http://www.geocities.com/binnyva/code/perl

 

[A. Sinan Unur]

(e-mail address removed) (Binny V A) wrote in

Hello Everyone,

I want to get a list of all HTML files in a folder. How do I do this?

I tried
$list = <*.htm>;
but it gave only one html file. The same result with glob("*.htm")

I am currently doing this with
$list = `dir *.htm`;
and then spliting it. But that is not portable.

Other questions - How do I get the names of folders?
How do I relusivly list all files in a folder and within folders in
that folder?

perldoc -f opendir
perldoc -f readdir
perldoc -f closedir

or

use File::Find::Rule;

Sinan

 

[Paul Lalli]

I want to get a list of all HTML files in a folder. How do I do this?

I tried
$list = <*.htm>;
but it gave only one html file. The same result with glob("*.htm")

You should always read the documentation for the function you're using:
perldoc -f glob
glob EXPR
glob In list context, returns a (possibly empty) list of
filename expansions on the value of EXPR such as the
standard Unix shell /bin/csh would do. In scalar
context, glob iterates through such filename
expansions, returning undef when the list is
exhausted.

You're using a scalar context, and not attempting to iterate. You need
to do one or the other to get the full list.

Other questions - How do I get the names of folders?

Open a directory, read the directory one entry at a time, and test each
entry to determine whether or not it is a directory or a file:
perldoc -f opendir
perldoc -f readdir
perldoc -f -X

How do I relusivly list all files in a folder and within folders in

that folder?

Assuming you mean "recursively", use the standard File::Find module, or
the available-on-CPAN File::Find::Rule module.

Paul Lalli

 

[gnari]

$list = <*.htm>;
but it gave only one html file. The same result with glob("*.htm")

actually, <*.htm> does return a list of files, but you are assigning it to
a scalar, as if you had done : $x=('a','b','c');
try @list instead

gnari

 

[Jürgen Exner]

I want to get a list of all HTML files in a folder. How do I do this?

I tried
$list = <*.htm>;
but it gave only one html file. The same result with glob("*.htm")

1: You are asking for only one file. If you want a list of all files, then
ask for a list:
@list = <*.htm>;

2: Your verbal specification conflicts with your code. You said you want
HTML files, but in your code you are globbing htm files.

Other questions - How do I get the names of folders?

See "perldoc -f -X" and check in particular the -d operator.

How do I relusivly list all files in a folder and within folders in
that folder?

I guess you meant 'recursively'? See "perldoc File::Find".

jue

 

[Leon]

Hello Everyone,

I want to get a list of all HTML files in a folder. How do I do this?

I tried
$list = <*.htm>;
but it gave only one html file. The same result with glob("*.htm")

Try @list = glob("*.htm");

# Just to note that this is for *.htm files, it does not include *.html.

I am currently doing this with
$list = `dir *.htm`;
and then spliting it. But that is not portable.

Other questions - How do I get the names of folders?

Check opendir, readdir, closedir command.

How do I relusivly list all files in a folder and within folders in that

folder?
Not sure on this one, sure someone will point you in the right direction

 

[Paul Lalli]

actually, <*.htm> does return a list of files, but you are assigning it to
a scalar, as if you had done : $x=('a','b','c');

Well, not quite the same since the OP's code will assign $list to the
*first* element otherwise returned by the glob, whereas this code will
assing $x to the last element of the 'list'....

Paul Lalli

 

[gnari]

Well, not quite the same since the OP's code will assign $list to the
*first* element otherwise returned by the glob, whereas this code will
assing $x to the last element of the 'list'....

true.

$x=('a','b','c');
is more an example of the comma operator in action than an list
assignment.

gnari