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Full Discussion: calling a aliased variable
Top Forums Shell Programming and Scripting calling a aliased variable Post 24612 by RTM on Monday 15th of July 2002 05:35:13 PM
Old 07-15-2002
Quote:
now if by hand i say
doh=$testme
echo $doh
(it will then print the correct output) but useing the awk statement to fill the contents of $doh does not seem to work.
Realize you are not setting DOH to a value of another variable - you are setting it to "$" $1$2. This is two different things. Just as I stated earlier, if B is nothing setting A=$B give you nothing.

You are stating DOH to be equal to a string "$TESTIT" not a value of the variable $TESTIT. I'm not sure the work around (I've been looking...). Will get back to it tomorrow if no one else has (must go home as the EOD is here).
 

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PARSE_STR(3)								 1							      PARSE_STR(3)

parse_str - Parses the string into variables

SYNOPSIS
void parse_str (string $str, [array &$arr]) DESCRIPTION
Parses $str as if it were the query string passed via a URL and sets variables in the current scope. Note To get the current QUERY_STRING, you may use the variable $_SERVER['QUERY_STRING']. Also, you may want to read the section on vari- ables from external sources. Note The magic_quotes_gpc setting affects the output of this function, as parse_str(3) uses the same mechanism that PHP uses to populate the $_GET, $_POST, etc. variables. PARAMETERS
o $str - The input string. o $arr - If the second parameter $arr is present, variables are stored in this variable as array elements instead. RETURN VALUES
No value is returned. EXAMPLES
Example #1 Using parse_str(3) <?php $str = "first=value&arr[]=foo+bar&arr[]=baz"; parse_str($str); echo $first; // value echo $arr[0]; // foo bar echo $arr[1]; // baz parse_str($str, $output); echo $output['first']; // value echo $output['arr'][0]; // foo bar echo $output['arr'][1]; // baz ?> SEE ALSO
parse_url(3), pathinfo(3), http_build_query(3), get_magic_quotes_gpc(3), urldecode(3). PHP Documentation Group PARSE_STR(3)
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