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Top Forums Shell Programming and Scripting bash hell , removing " and adding from a strings Post 302329565 by bwhitehd on Sunday 28th of June 2009 11:12:25 PM
Old 06-29-2009
I'm not sure that I understand correctly what you're asking, but I think you are assigning a value that start with $ to a variable and want that value output. You will need to escape the character as \$value.

Here is an example:
Code:
foo=a
bar=b
dog=c

orange=\$foo
chicken=\$bar
cat=\$dog

echo "--------------"
enable="$orange $chicken $cat"
echo $enable

echo "--------------"
deref="$orange $chicken $cat"
eval echo $deref

echo "--------------"
alist="foo bar dog"
for var in $alist
do
    eval echo \$$var
    eval echo \$var
done

echo "--------------"
blist="orange chicken cat"
for var in $blist
do
    eval echo \$$var
    eval echo \$var
done

One other thing I am showing here is how to dereference a variable reference with eval.

Code:
$ ./mytest.sh 
--------------
$foo $bar $dog
--------------
a b c
--------------
a
foo
b
bar
c
dog
--------------
$foo
orange
$bar
chicken
$dog
cat

I hope this gives you an idea to solve your question.

-B
 

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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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