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Top Forums Shell Programming and Scripting Retain quotes from bash script arguments Post 302216141 by zaxxon on Friday 18th of July 2008 05:06:31 AM
Old 07-18-2008
Try escaping them:
Code:
myscript.sh grep \"foo bar\"

 

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PRINT(3)								 1								  PRINT(3)

print - Output a string

SYNOPSIS
int print (string $arg) DESCRIPTION
Outputs $arg. print is not actually a real function (it is a language construct) so you are not required to use parentheses with its argument list. PARAMETERS
o $arg - The input data. RETURN VALUES
Returns 1, always. EXAMPLES
Example #1 print examples <?php print("Hello World"); print "print() also works without parentheses."; print "This spans multiple lines. The newlines will be output as well"; print "This spans multiple lines. The newlines will be output as well."; print "escaping characters is done "Like this"."; // You can use variables inside a print statement $foo = "foobar"; $bar = "barbaz"; print "foo is $foo"; // foo is foobar // You can also use arrays $bar = array("value" => "foo"); print "this is {$bar['value']} !"; // this is foo ! // Using single quotes will print the variable name, not the value print 'foo is $foo'; // foo is $foo // If you are not using any other characters, you can just print variables print $foo; // foobar print <<<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; ?> NOTES
Note Because this is a language construct and not a function, it cannot be called using variable functions. SEE ALSO
echo(3), printf(3), flush(3), Heredoc syntax. PHP Documentation Group PRINT(3)
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