Questions relating replacing a variable and file parse.

thanawala27

Hi,

I'm trying to replace a character in a string. was facing problem in it.

$string = "D:\Folder\TEST\abc.txt"

i wanted to transform this string to "D:/Folder/TEST/abc.txt"

this is wot i tried, but doesn't work

$string = ~ s/\\/\//;

Another Q is related to fileparse.

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$value = D:/Folder/TEST/abc.txt

my ( $name, $path, $extension ) = fileparse ( $value, '\..*' );

    $filename = $name . $extension;

I am getting the values as follows:

$name = abc
$path = /
$extension =

Where am I going wrong. Do i need to install anything thing for this. I have included File::Basename in the file.

Any help is appreciated.

Ravi

KevinADC

Look at this:

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 $string = "D:\Folder\TEST\abc.txt";

print $string;

the output is:

D:FolderTEST.bc.txt

notice there are no backslashes to substitute with forward slashes. The backslash in a double-quoted string is an escape character, not a literal backslash. You ned to use single-quotes:

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$string = 'D:\Folder\TEST\abc.txt';

Are you trying to parse Windows filepaths on a Linux/Unix box?

From File::Basname:

<start quote>
fileparse_set_fstype

my $type = fileparse_set_fstype();
my $previous_type = fileparse_set_fstype($type);

Normally File::Basename will assume a file path type native to your current operating system (ie. /foo/bar style on Unix, \foo\bar on Windows, etc...). With this function you can override that assumption.

Valid $types are "MacOS", "VMS", "AmigaOS", "OS2", "RISCOS", "MSWin32", "DOS" (also "MSDOS" for backwards bug compatibility), "Epoc" and "Unix" (all case-insensitive). If an unrecognized $type is given "Unix" will be assumed.
<end quote>

numberwhun

Ravi,

The slash in the closing code tag goes the other way. It is a forward slash and not a back slash.

Please follow the example in the Reply Guidelines box as it shows the correct format.

Regards,

Jeff

Icecrack

just another note

when using Single quotes ' ' variables, arrays, etc. do not get printed/parsed eg.

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 $echo = "hello";
 
print ' $echo world';

this will print $echo world

KevinADC

Quote:

Originally Posted by Icecrack

just another note

when using Single quotes ' ' variables, arrays, etc. do not get printed/parsed eg.

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 $echo = "hello";
 
print ' $echo world';

this will print $echo world

True but thats not the case here. This is a lack of knowing the basics of perl beforing trying to write programs.

eWish

Quote:

Originally Posted by KevinADC

This is a lack of knowing the basics of perl beforing trying to write programs.

Are you talking about me again?

--Kevin

KevinADC

Quote:

Originally Posted by eWish

Are you talking about me again?

--Kevin

hehehe... were your ears buring?

Questions relating replacing a variable and file parse.

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