PHP sprintf escaping %

php escaping printf
Sandeepan Nath picture Sandeepan Nath · Sep 8, 2010 · Viewed 81.6k times · Source

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 ... because it considers the % in 50% also for replacement. How do I escape it?

Answer

BoltClock picture BoltClock · Sep 8, 2010

Escape it with another %:

$stringWithVariables = 'About to deduct 50%% of %s %s from your Top-Up account.';

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