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Top Forums Shell Programming and Scripting scripting newbie... some help please? Post 302293676 by jmd9qs on Tuesday 3rd of March 2009 03:34:59 PM
Old 03-03-2009
Quote:
Originally Posted by cfajohnson

Why do you need an "overwrite" option? What do you want to overwrite?

Why would you need find?

To copy $FILENAME into $SAVEPATH, you use:

Code:
cp "$FILENAME" "$SAVEPATH"

i guess i just don't know what i'm doing. I thought i'd need an 'overwrite' option because of output i have recieved from programs before... thinking about it now, i guess it was a silly question, seeing as the filename has a timestamp on it Smilie...

now i have a different issue... i have put all of my changes that you guys have helped me figure out into the script. i also added a little bit of code to create a directory if the $SAVEPATH wasn't one already. here is that section:

Code:
press_enter
        echo ""
        echo ""
        echo "Where would you like the new .tar.gz archive for $FILENAME stored?"
        echo -n "(just specify a directory with / at the end, filename is automatic) >  "
        echo ""
        read SAVEPATH
        if [ -z $SAVEPATH ]; then
                echo "No directory specified. Exiting..."; exit 1;
        fi
        if [ ! -d $SAVEPATH ]; then
                echo -n "$SAVEPATH does not exist. Would you like to create it? (y/n) >  ";
                read $answer
                        if [ "$answer" != "y" ]; then
                                echo "O.K. Exiting..."; exit 1;
                        fi
        fi
        echo "Making directory $SAVEPATH..."
        mkdir $SAVEPATH
        echo ""
        echo "$SAVEPATH created..."

i know i have to be overlooking a syntax error (or i just don't have it set up correctly) because, when i test the program with a directory that doesn't exist, it asks if I want to create it, i signal "y", and then it immediately exits... what am i doing wrong?

edit -

also, when i give it a directory that does exist, it does the mkdir $SAVEPATH command... i must have them switched up somehow. what should this look like?
 

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ECHO(3) 								 1								   ECHO(3)

echo - Output one or more strings

SYNOPSIS
void echo (string $arg1, [string $...]) DESCRIPTION
Outputs all parameters. echo is not actually a function (it is a language construct), so you are not required to use parentheses with it. echo (unlike some other language constructs) does not behave like a function, so it cannot always be used in the context of a function. Additionally, if you want to pass more than one parameter to echo, the parameters must not be enclosed within parentheses. echo also has a shortcut syntax, where you can immediately follow the opening tag with an equals sign. Prior to PHP 5.4.0, this short syn- tax only works with the short_open_tag configuration setting enabled. I have <?=$foo?> foo. PARAMETERS
o $arg1 - The parameter to output. o $... - RETURN VALUES
No value is returned. EXAMPLES
Example #1 echo examples <?php echo "Hello World"; echo "This spans multiple lines. The newlines will be output as well"; echo "This spans multiple lines. The newlines will be output as well."; echo "Escaping characters is done "Like this"."; // You can use variables inside of an echo statement $foo = "foobar"; $bar = "barbaz"; echo "foo is $foo"; // foo is foobar // You can also use arrays $baz = array("value" => "foo"); echo "this is {$baz['value']} !"; // this is foo ! // Using single quotes will print the variable name, not the value echo 'foo is $foo'; // foo is $foo // If you are not using any other characters, you can just echo variables echo $foo; // foobar echo $foo,$bar; // foobarbarbaz // Some people prefer passing multiple parameters to echo over concatenation. echo 'This ', 'string ', 'was ', 'made ', 'with multiple parameters.', chr(10); echo 'This ' . 'string ' . 'was ' . 'made ' . 'with concatenation.' . " "; echo <<<END This uses the "here document" syntax to output multiple lines with $variable interpolation. Note that the here document terminator must appear on a line with just a semicolon. no extra whitespace! END; // Because echo does not behave like a function, the following code is invalid. ($some_var) ? echo 'true' : echo 'false'; // However, the following examples will work: ($some_var) ? print 'true' : print 'false'; // print is also a construct, but // it behaves like a function, so // it may be used in this context. echo $some_var ? 'true': 'false'; // changing the statement around ?> NOTES
Note Because this is a language construct and not a function, it cannot be called using variable functions. SEE ALSO
print(3), printf(3), flush(3), Heredoc syntax. PHP Documentation Group ECHO(3)
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