preg_replace:添加回引用后的数字

Situation

情况

I want to use preg_replace() to add a digit '8' after each of [aeiou].

Example

我想使用preg_replace()在每个[aeiou]之后添加一个数字“8”。例子

from

从

abcdefghij

来

a8bcde8fghi8j

Question

问题

How should I write the replacement string?

我应该如何写替换字符串?

// input string
$in = 'abcdefghij';

// this obviously won't work ----------↓
$out = preg_replace( '/([aeiou])/', '\18',  $in);

This is just an example, so suggesting str_replace() is not a valid answer.
I want to know how to have number after backreference in the replacement string.

这只是一个示例，因此建议str_replace()不是有效答案。我想知道如何在替换字符串中有回引用后的数字。

2 个解决方案

#1

The solution is to wrap the backreference in ${}.

解决方案是用${}包装回引用。

$out = preg_replace( '/([aeiou])/', '${1}8',  $in);

which will output a8bcde8fghi8j

将输出a8bcde8fghi8j

See the manual on this special case with backreferences.

请参阅关于此特殊情况的手册，其中包含反向引用。

#2

You can do this:

你可以这样做:

$out = preg_replace('/([aeiou])/', '${1}' . '8', $in);

Here is a relevant quote from the docs regarding backreference:

以下是文件的相关引用:

When working with a replacement pattern where a backreference is immediately followed by another number (i.e.: placing a literal number immediately after a matched pattern), you cannot use the familiar \1 notation for your backreference. \11, for example, would confuse preg_replace() since it does not know whether you want the \1 backreference followed by a literal 1, or the \11 backreference followed by nothing. In this case the solution is to use \${1}1. This creates an isolated $1 backreference, leaving the 1 as a literal.

当使用一个替换模式时，一个反向引用紧接着是另一个数字(即::在匹配的模式之后立即放置一个文字数字)，您不能使用熟悉的\1表示法进行反向引用。例如，\11会混淆preg_replace()，因为它不知道您是想要用后跟文字1的\1的backreference，还是用不带文字1的\11的backreference。在这种情况下，解决方案是使用\${1}1。这将创建一个单独的$1回引用，将1保留为文字。

#1