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Top Forums Shell Programming and Scripting How to pass nawk variable to shell within the same script? Post 302766101 by radoulov on Monday 4th of February 2013 07:54:38 AM
Old 02-04-2013
Try this:

Code:
nawk 'BEGIN {
        print "|========================================|"
        print "|   SI No     CHECKS           STATUS    |"
        print "|========================================|"
 } /^[0-9]/ {
        c++;
        print  "|------------------------------------------|"
        printf "| %s ",$0; getline; s=$0; sub(/:.*/,"",s);
        if (s == "Passed") { ++p; _s = "\033[1;32m" s "\033[0m" }
        if (s == "Failed") { ++f; _s = "\033[1;31m" s "\033[0m" }
        if((s!="Passed")&& (s!="Failed")||(s==" ") ) ++n;
        printf " : %s   |\n", _s;
        print "|------------------------------------------|"
 } END {
           print "|========================================|"
           print "| Total Executed: " c "                      |"
           print "|========================================|"
           printf "| Total \033[1;35;40m Executed: %s\033[0m                 |\n", c
}' "$1"

This User Gave Thanks to radoulov For This Post:
 

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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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