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);
但它给了我这个错误 vsprintf() [function.vsprintf]: Too Fewarguments in ... 因为它考虑了 % 50% 也可用于更换。我该如何逃脱呢？
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?
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（1）
用另一个
%转义它：Escape it with another
%: