Re: Variable variables

If I'm reading your snippet correctly, then : $foo = 'bar'; $bar = array('apple','banana'); print ${$foo}[0]; // apple Note the use of {braces}. The last paragraph in the manual describes this a bit : http://www.php.net/manual/en/language.variables.variable.php Although I don't see the point of $newvar as you're keeping track of $a in the array itself. Looks like the below code will create a ton of variables (arrays) each with one $a as the single key. Just using $finalresult may be more appropriate, hard to say. And some tips : $arr[foo] will create an error/warning here, $arr['foo'] will not. "$foo" isn't as pretty as $foo regards, Philip Olson On Wed, 26 Sep 2001, Richard Baskett wrote: > I can not figure out why this is not working! > > for ($j=0; $j<$resultNum; $j++) { > $newvar = "finalresult".$a; > $$newvar[$a][name] = $resultRow[name]; > $$newvar[$a][title] = $resultRow[title]; > $$newvar[$a][descript] = $resultRow[descript]; > $$newvar[$a][countkey] = substr_count("$resultRow[keywords]", > "$keyword"); > a++; > } > > This is how you use variable variables, is it not? Is it because it's a > multidimensional array? or am I missing something else? > > Rick > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net > For additional commands, e-mail: php-general-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net >