Passing Variables in a Variable That Is Inside another Function

[GargantulaKon]

OK, this is confusing so here is what I am trying to do:

 

I am trying to pass a variable that I want to resolve inside another function. How do I do so? Doing the code below actually outputs the PHP code. I tried using the code without escaping, but then the variables are not resoloved. Am I doing it the wrong way?

function Select_Star($Star_Type)
{
switch ($Star_Type)
{
	case "E":
		$HTML = '<img id="Rating_\' . $i . \'" \' . $OnClick . \' onmouseover="Rating(\' . $i . \');" onmouseout="Rating();" src="/Images/Rating_Empty.png" alt="\' . $i . \' Star\' . Add_S($i) . \'" title="\' . $i . \' Star\' . $s . \'" />';
	break;
}
return $HTML;
}

function Rating_HTML($Rating_Number, $User_Mood_ID, $Activated)
{
	for ($i = 1; $i <= 5; $i++)
	{
		$Return .= Select_Star("E");
	}
}

[abazoskib]

what exactly is your goal

[GargantulaKon]

I want $i to be resolved when it is inside the loop. I am trying to prevent having to retype the same HTML code over and over and instead pick from a switch depending on the case.

[trq]

You would need to make your Select_Star() function except another param. eg;

 

function Select_Star($Star_Type, $i)

 

Then call it.....

 

for ($i = 1; $i <= 5; $i++)
{
    $Return .= Select_Star("E", $i);
}

[GargantulaKon]

Thanks! Wow, great idea! I just had a duh  moment! It must be because I am been programming all day today...

[GargantulaKon]

I got it working now. Thanks thorpe for helping with my code. I will be sure to keep this in mind for future issues.