Outputting a variable name with its value (in a function) - php

Question

I would like to be able to output the name of a variable, along with that variable's value. My use case is something close to a debug situation, but I'm actually building a proof of concept for other developers and managers so we can talk about things like input filtering. Anyway...
The output HTML is currently a table, but that shouldn't matter. I can of course, just print out the name and then the contents in the HTML, but that gets tedious and is prone to typing errors. I'd like a function that I could call with the variable name, or a string with the variable name as the argument, and have that function generate the appropriate HTML for display. This doesn't appear to work.
This works:
<tr><td>$variable</td><td><?php print $variable?></td></tr>
This doesn't:
function rowFromVar($varname) {
$result = "<tr>";
$result .= "<td>\$$varname</td>";
$result .= "<td>" . $$varname . "</td>";
$result .= "</tr>";
return $result;
}
// now in the HTML...
<table><?php print rowFromVar("variable");?></table>
The variable variable $$varname is always empty. Notice that I'm passing in the name of the variable as a string, rather than trying to pass in the variable itself and work with it. I know that there be dragons that way. Instead, I'm trying to pass in the name of the variable as a string, and then use a variable variable (yeah, I know) to return the value of the original variable. That appears not to work either, and I'm betting it's a scope issue.
Is there any good way to accomplish what I'm looking for?

Answer 1

The variable variable $$varname is always empty. Notice that I'm
passing in the name of the variable as a string, rather than trying to
pass in the variable itself and work with it
It should be empty
Since you are passing variable as value, so:
$$varname
becomes:
$variable
and there is no value set for it.
In other words, you are just creating a variable with the name you pass as argument to function but you are not creating a value for it.
Solution:
This would work fine though if that's what you are looking to do:
function rowFromVar($varname) {
$result = "<tr>";
$result .= "<td>\$$varname</td>";
$result .= "<td>" . $varname . "</td>";
$result .= "</tr>";
return $result;
}
// now in the HTML...
<table><?php print rowFromVar("variable");?></table>
Simply remove $ from the variable on this line as done above:
$result .= "<td>" . $$varname . "</td>";
--------------------^

Answer 2

Did you try using $GLOBALS[$varname] instead of $$varname? Your variable in this example is at global scope, and you cannot directly access globals within a function.

Answer 3

You need declare your variable global within the function so that it references the global version. Adding this line at the start of your function should fix the problem:
function rowFromVar($varname) {
global $$varname;
$result = "<tr>";
$result .= "<td>\$$varname</td>";
$result .= "<td>" . $$varname . "</td>";
$result .= "</tr>";
return $result;
}

Answer 4

The following code should work, as you suspected scope is the main problem here.
function rowFromVar($varname, $contents) {
$result = "<tr>";
$result .= "<td>\$$varname</td>";
$result .= "<td>" . $contents . "</td>";
$result .= "</tr>";
return $result;
}
// now in the HTML...
<table><?php print rowFromVar("variable", $variable);?></table>

Answer 5

Please try this code:
function rowFromVar($varname) {
$result = "<tr>";
$result .= "<td>{$varname}</td>";
$result .= "<td>".$varname."</td>";
$result .= "</tr>";
return $result;
}
<table><?php echo rowFromVar("variable");?></table>

Answer 6

I had a similar problem, which I solved as follows.
function contents_of($string)
{
eval("global $string; \$value = $string;");
return "Variable $string has value $value";
}
$testing = 123;
echo contents_of('$testing');
// produces...
// Variable $testing has value 123