如何在 preg_replace() 的替换参数中引用完整的字符串匹配？
好的，我有这个 PHP 脚本：
<?php
$stringz = "Dan likes to eat pears and his favorite color is green!";
$patterns = array("/pears/","/green/");
$string = preg_replace($patterns, '<b>\\1</b>', $stringz);
echo "<textarea rows='30' cols='100'>$string</textarea>";
?>
当我运行它时，我得到这个： Dan 喜欢吃 他最喜欢的颜色是 !
应该有一个单词......但它没有......
OK so I have this PHP script:
<?php
$stringz = "Dan likes to eat pears and his favorite color is green!";
$patterns = array("/pears/","/green/");
$string = preg_replace($patterns, '<b>\\1</b>', $stringz);
echo "<textarea rows='30' cols='100'>$string</textarea>";
?>
and when I run it I get this:Dan likes to eat <b></b> and his favorite color is <b></b>!
The is suppose to have a word in it... but it doesn't...
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那是因为您没有明确捕获任何内容。当然，
\\0捕获整个匹配，但是为了捕获特定部分，如果要使用\\1，则需要使用捕获组，\\2、\\3等。将$patterns更改为：()表示捕获组，它们内部捕获的任何值都存储在引用\\n中，其中\\n指的是第 1 个索引的n捕获组。That's because you aren't explicitly capturing anything.
\\0captures the entire match, of course, but in order to capture specific parts, you need to use capturing groups if you want to use\\1,\\2,\\3, etc. Change$patternsto this:()'s denote capturing groups, and whatever value is captured inside of them is stored in the reference\\n, where\\nrefers the the 1-indexednth capturing group.将
\\1更改为\\0。Change
\\1for\\0.怎么样
？
\\1适用于主题，即括号中的任何内容。How about
?
\\1applies to a subject, which is whatever you have in parentheses.