PHP preg_replace以$开头的任何数字xxx by xxx

-1

我想用xxx替换文本中每个出现的$（任何数字）。例如，如果在文本中出现$ 923，那么它应该是xxx。

模式我都试过

    $string = "Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. $350 remaining essentially unchanged. It was popularised in the 1960s with the release of $90 required Letraset sheets containing Lorem Ipsum passages, and more recently with desktop publishing software like Aldus PageMaker including versions of Lorem Ipsum."; 

        $pattern = '/^\$[0-9]+\.?[0-9]?[0-9]/i'; 

        $pattern = '/($\[0-9]+)/'; 

        $replacement = "xxx"; 
        echo preg_replace($pattern, $replacement, $string);

来源

2016-02-05 dr10

欢迎SO。请阅读[我可以问哪些主题]（http://stackoverflow.com/help/on-topic）和[如何提出一个好问题]（http://stackoverflow.com/help/how-to - 问）和[完美的问题]（http://codeblog.jonskeet.uk/2010/08/29/writing-the-perfect-question/） SO是**不是免费的编码或教程服务* *你必须表明你已经努力解决你自己的问题。 – RiggsFolly

试试这个 -

$s = "$434 dsvsv $567 fsdgjfb"; 

echo preg_replace('/\$\d+/', 'xxx', $s);

输出

xxx dsvsv xxx fsdgjfb

来源

2016-02-05 12:14:14

谢谢！ :) – dr10

PHP preg_replace以$开头的任何数字xxx by xxx

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