在 PHP 中的 preg_replace 中使用 $ 变量
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Using $ variables in preg_replace in PHP
提问by Steve
Ummm... how do I use variables in a call to preg_replace?
嗯...如何在调用 preg_replace 时使用变量？
This didn't work:
这不起作用：
foreach($numarray as $num => $text)
{
$patterns[] = '/<ces>(.*?)\+$num(.*?)<\/ces>/';
$replacements[] = '<ces><$text/></ces>';
}
Yes, the $numis preceeded by a plus sign. Yes, I want to "tag the $num as <$text/>".
是的，$num前面有一个加号。是的，我想要“ tag the $num as <$text/>”。
回答by Paul Dixon
Your replacement pattern looks ok, but as you've used single quotes in the matching pattern, your $num variable won't be inserted into it. Instead, try
您的替换模式看起来不错，但由于您在匹配模式中使用了单引号，因此不会将 $num 变量插入其中。相反，尝试
$patterns[] = '/<ces>(.*?)\+'.$num.'(.*?)<\/ces>/';
$replacements[] = '<ces><'.$text.'/></ces>';
Also note that when building up a pattern from "unknown" inputs like this, it's usually a good idea to use preg_quote. e.g.
另请注意，当从这样的“未知”输入构建模式时，使用preg_quote通常是个好主意。例如
$patterns[] = '/<ces>(.*?)\+'.preg_quote($num).'(.*?)<\/ces>/';
Though I guess given the variable name it's always numeric in your case.
虽然我猜给出了变量名，但在你的情况下它总是数字。
回答by Gumbo
Variables will only be expanded in strings declared with double quotes. So either use double quotes:
变量只会在用双引号声明的字符串中扩展。所以要么使用双引号：
$patterns[] = "/<ces>(.*?)\+$num(.*?)<\/ces>/";
$replacements[] = "<ces><$text/></ces>";
Or use string concatenation:
或者使用字符串连接：
$patterns[] = '/<ces>(.*?)\+'.$num.'(.*?)<\/ces>/';
$replacements[] = '<ces><'.$text.'/></ces>';
You should also take a look at preg_quoteif your variables may contain regular expression meta characters.
您还应该查看preg_quote您的变量是否可能包含正则表达式元字符。
