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Full Discussion: why am I losing line-end chars
Top Forums Shell Programming and Scripting why am I losing line-end chars Post 302408546 by bakunin on Monday 29th of March 2010 10:13:05 PM
Old 03-29-2010
The reason is that several characters - newline among them - have special meaning to the shell and are therefore interpreted instead of used like "normal" characters.

The general way to protect these special characters - that is, make the shell treat them like they were normal ones - is to enclose a string into double quotes:

Code:
myvar="foo bar"      # works
myvar=foo bar        # will not work because the space after foo is interpreted
                     # as end-of-statement and that leaves "bar", which the shell
                     # cannot understand.

In your case you have to protect your variable two times: first when you fill it with content, second when you print it out. The reason is that the shell interprets the commands in a certain step-by-step process: in one step all the variables are replaced by their content and in the next step the command is interpreted. To understand what i mean make the following test:

Code:
myvar="-lisa a*"
ls ${myvar}
ls "${myvar}"

With the first command you will see the same directory listing format as if you had entered "ls -lisa a*" directly. The reason is that first the shell replaces "${myvar}" with its contents and only then executes the resulting line as the string "-lisa a*" was not protected it splits into two distinct strings, the first one being "-lisa" and the second one being "a*". This is exactly what the ls command expects.

The second command will give a strange error message. Because we protected the string it is still one string, the space character included. The ls command doesn't have any option "a*" (this can only be a parameter) and it doesn't have an option " " (space) either! This is where the error message will come from.

You see this mechanism of protecting (or not protecting) variables can be used both ways. Here is your solution:

Code:
OUTFILE="$(egrep -e 'matchstring' infile.txt)"    # notice the double quotes

echo "$OUTFILE"            # with newlines
echo $OUTFILE              # newlines replaced by spaces

I hope this helps.

bakunin
 

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LEARN ABOUT DEBIAN

template::plugin::yaml

Template::Plugin::YAML(3pm)				User Contributed Perl Documentation			       Template::Plugin::YAML(3pm)

NAME

       Template::Plugin::YAML - Plugin interface to YAML

SYNOPSIS

	   [% USE YAML %]

	   [% YAML.dump(variable) %]
	   [% YAML.dump_html(variable) %]
	   [% value = YAML.undump(yaml_string) %]
	   [% YAML.dumpfile(filename, variable) %]
	   [% value = YAML.undumpfile(filename) %]

DESCRIPTION

       This is a simple Template Toolkit Plugin Interface to the YAML module.  A YAML object will be instantiated via the following directive:

	   [% USE YAML %]

       As a standard plugin, you can also specify its name in lower case:

	   [% USE yaml %]

METHODS

       These are the methods supported by the YAML object.

   dump( @variables )
       Generates a raw text dump of the data structure(s) passed

	   [% USE Dumper %]
	   [% yaml.dump(myvar) %]
	   [% yaml.dump(myvar, yourvar) %]

   dump_html( @variables )
       Generates a dump of the data structures, as per "dump", but with the characters <, > and & converted to their equivalent HTML entities,
       spaces converted to &nbsp; and newlines converted to <br>.

	   [% USE yaml %]
	   [% yaml.dump_html(myvar) %]

   undump( $string )
       Converts a YAML-encoded string into the equivalent data structure.  Here's a way to deep-copy a complex structure by completely serializing
       the data.

	 [% USE yaml;
	    yaml_string = yaml.dump(complex_data_structure);
	    complex_copy = yaml.undump(yaml_string);
	 %]

   dumpfile( $file, @variables )
       Converts the data to YAML encoding, and dumps it to the specified filepath.

	 [% USE yaml; yaml.dumpfile(".storage", my_data) %]

   undumpfile( $file )
       Loads the YAML-encoded data from the specified filepath

	 [% USE yaml; my_data = yaml.undumpfile(".storage") %]

AUTHORS

       Richard Clamp <richardc@unixbeard.net>, with undump, undumpfile, and dumpfile implementation by Randal L. Schwartz <merlyn@stonehenge.com>

       based on Simon Matthews' Template::Plugin::Dumper

COPYRIGHT

       Copyright 2003, 2008 Richard Clamp All Rights Reserved.

       This module is free software; you can redistribute it and/or modify it under the same terms as Perl itself.

SEE ALSO

       Template::Plugin::Dumper

perl v5.10.0							    2008-12-15					       Template::Plugin::YAML(3pm)
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