PHP sprintf escaping %

I want the following output:- About to deduct 50% of € 27.59 from your Top-Up account. when I do something like this:- $variablesArray[0] = ‘€’; $variablesArray[1] = 27.59; $stringWithVariables=”About to deduct 50% of %s %s from your Top-Up account.”; echo vsprintf($stringWithVariables, $variablesArray); But it gives me this error vsprintf() [function.vsprintf]: Too few arguments in … … Read more