Can I declare a variable name like this $(variable name) in PHP?

Question

I want to get the resulting string(field name) of the mysql function, mysql_field_name() and declare that as a variable,so that I can use it later in my program.
For example, like this:

$(mysql_field_name($result,2)) = 2;

Here, can I use this statement like this : $(mysql_field_name($result,2));
Is it a correct variable declaration?
Will it work and will it declare a variable with field name as the variable name?

docs: [Variable variables](http://us.php.net/manual/en/language.variables.variable.php). — Wiseguy, Jun 18 '12 at 15:00
http://stackoverflow.com/questions/9257505/dynamic-variable-names-php — prodigitalson, Jun 18 '12 at 15:02

score 4 · Answer 1 · answered Jun 18 '12 at 14:59

4

${mysql_field_name($result,2)} = 2

http://codepad.org/NRok2Cha

answered Jun 18 '12 at 14:59

Esailija

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this one seems more elegant to me :) – Rifat Jun 18 '12 at 15:01

J. Bruni · Answer 2 · 2012-06-18T15:02:23.680

3

This is how you can do what you want in PHP:

$varname = mysql_field_name($result, 2);
$$varname = 2;

More info in the PHP manual: Variable variables

edited Jun 18 '12 at 15:02

answered Jun 18 '12 at 14:58

J. Bruni

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Can I declare a variable name like this $(variable name) in PHP?

2 Answers2