PHP sprintf转义%

我想要以下输出:

将从您的充值帐户中扣除27.59欧元的50%。

当我做这样的事情时:

$variablesArray[0] = '€';

$variablesArray[1] = 27.59;

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

echo vsprintf($stringWithVariables, $variablesArray);

但这给了我这个错误,vsprintf() [function.vsprintf]: Too few arguments in

...因为它也考虑使用%in 50%进行替换。我该如何逃脱?

回答:

与另一个转义%

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

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