如何在 php 中查找、增加和替换? [英] How can I find, increment and replace in php?

问题描述

我有 \d+_\d+ 形式的字符串，我想在第二个数字上加 1.由于我的解释非常清楚，让我举几个例子:

I have strings in the form \d+_\d+ and I want to add 1 to the second number. Since my explanation is so very clear, let me give you a few examples:

• 1234567_2 应该变成 1234567_3
• 1234_10 应该变成 1234_11

这是我的第一次尝试:

$new = preg_replace("/(\d+)_(\d+)/", "$1_".((int)$2)+1, $old);

这会导致语法错误:

解析错误:语法错误，意外T_LNUMBER，期待 T_VARIABLE 或 '$'在 [...] 第 201 行

Parse error: syntax error, unexpected T_LNUMBER, expecting T_VARIABLE or '$' in [...] on line 201

这是我的第二次尝试

$new = preg_replace("/(\d+)_(\d+)/", "$1_".("$2"+1), $old);

这将 $old = 1234567_2 转换为 $new = 1234567_1，这不是预期的效果

This transforms $old = 1234567_2 into $new = 1234567_1, which is not the desired effect

我的第三次尝试

$new = preg_replace("/(\d+)_(\d+)/", "$1_".((int)"$2"+1), $old);

这产生了相同的结果.

通过这些尝试，我意识到我不明白新的 $1, $2, $3, .. 变量是如何真正起作用的，所以我真的不知道还能尝试什么，因为这些变量似乎不再退出 preg_replace 函数时存在...

By making these attempts, I realized I didn't understand how the new $1, $2, $3, .. variables really worked, and so I don't really know what else to try because it seems that these variables no longer exist upon exiting the preg_replace function...

有什么想法吗?

推荐答案

$new = preg_replace("/(\d+)_(\d+)/e", '"$1_" . ("$2" + 1)', $old);

$1 等术语实际上不是变量，它们是 preg_replace 将在替换文本中解释的字符串.所以没有办法使用直接基于文本的 preg_replace 来做到这一点.

The $1 etc terms are not actually variables, they are strings that preg_replace will interpret in the replacement text. So there is no way to do this using straight text-based preg_replace.

然而，正则表达式上的 /e 修饰符要求 preg_replace 将替换解释为代码，其中标记 $1 等实际上将被视为变量.您将代码作为字符串提供，preg_replace 将在适当的上下文中eval()，使用其结果作为替换.

However, the /e modifier on the regular expression asks preg_replace to interpret the substitution as code, where the tokens $1 etc will actually be treated as variables. You supply the code as a string, and preg_replace will eval() it in the proper context, using its result as the replacement.

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