How do I insert $ in front of variable?

[dalton]

Hi, I have a SQL column called cost of field type INT with two decimals. Before I output the numbers in my script via printf, I want to insert a $ in front of each number above 0.00 and replace all 0.00 numbers with N/A. Something like the following but I don't know how to insert a $ in front of numbers over 0.00

if ($myrow["cost"] != 0.00) {
$myrow["cost"] = insert $ in front of variable
}
else {
$myrow["cost"] = "N/A";
}
}

[spider_man]

You can escape it with a leading backslash; "\$$foo"

Or you can cocatenate it;
'$' . $foo

Is that what you meant, isn't it?

[dalton]

Hi Pablo, sorry but I am a bit new to this. In my code below, what code exactly would I replace the word HERE with to add the $ to the beginning of the string? For example, if the variable $cost is anything other than 0.00, say 1.25, then I want to add the $ to it, like, $1.25 - otherwise change it to N/A.

if ($myrow["cost"] != 0.00) {
$myrow["cost"] = HERE
}
else {
$myrow["cost"] = "N/A";
}
}

[dalton]

Ok, I get it now! Thanks!

if ($myrow["cost"] != 0.00) {
$myrow["cost"] = "$" . $myrow["cost"];
}
else {
$myrow["cost"] = "N/A";
}

[Anon]

Try This

if ($myrow["cost"] != 0.00) {
$myrow["cost"] = '$'.$myrow["cost"];
}
else {
$myrow["cost"] = "N/A";
}

[Anon]

You can also get this effect by using
${$somevar}

[dalton]

Cool, thanks for the tip!