title: PHP sprintf escaping %
body: <p>I want the following output:-</p>
  
  <blockquote>
    <p>About to deduct 50% of € 27.59 from your Top-Up account.</p>
  </blockquote>
  
  <p>when I do something like this:-</p>
  
  <pre><code>$variablesArray[0] = '€';
  $variablesArray[1] = 27.59;
  $stringWithVariables = 'About to deduct 50% of %s %s from your Top-Up account.';
  echo vsprintf($stringWithVariables, $variablesArray);
  </code></pre>
  
  <p>But it gives me this error <code>vsprintf() [function.vsprintf]: Too few arguments in ...</code> because it considers the <code>%</code> in <code>50%</code> also for replacement. How do I escape it?</p>