PHP sprintf转义％ [英] PHP sprintf escaping %

问题描述

我想要以下输出： -

I want the following output:-

关于从您的充值帐户扣除€27.59的50％。 >

About to deduct 50% of € 27.59 from your Top-Up account.

当我做这样的事情： -

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);

但是它给了我这个错误 vsprintf（）[function.vsprintf]因为它在 50％中也考虑了％ 中的参数太少替代。如何逃避它？

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?

推荐答案

使用另一个％

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

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