本文介绍了PHP sprintf转义%的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想要以下输出: -
I want the following output:-
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
中的参数太少替代。如何逃避它? 50%
中也考虑了%
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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