如何在 php 中查找、增加和替换？

发布于 2024-10-02 15:59:13 字数 777 浏览 2 评论 0原文

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：

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，需要 T_VARIABLE 或 '$' 在 [...] 第 201 行

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，这不是所需的效果

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、.. 变量是如何真正工作的，所以我真的不知道还能尝试什么，因为这些变量在退出时似乎不再存在preg_replace 函数...

有什么想法吗？

原文

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 should become 1234567_3
1234_10 should become 1234_11

Here is my first attempt:

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

This results in a syntax error:

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

Here is my second attempt

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

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

My third attempt

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

This yeilds the same result.

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...

Any ideas?

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鹤舞 2024-10-09 15:59:13

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

$1 等术语实际上并不是变量，它们是 preg_replace 将在替换文本中解释的字符串。因此，无法使用直接基于文本的 preg_replace 来做到这一点。

但是，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。您以字符串形式提供代码，preg_replace 将在适当的上下文中 eval() 它，并使用其结果作为替换。

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

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.

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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扛刀软妹 2024-10-09 15:59:13

这是 PHP 5.3 的解决方案（现在 PHP 支持 lambda）

$new = preg_replace_callback("/(\d+_)(\d+)", function($matches)
{
    return $matches[1] . (1 + $matches[2]);
}
, $new);

Here's the solution for the PHP 5.3 (now when PHP supports lambdas)

$new = preg_replace_callback("/(\d+_)(\d+)", function($matches)
{
    return $matches[1] . (1 + $matches[2]);
}
, $new);

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柠檬色的秋千 2024-10-09 15:59:13

使用 explode （逐步）：

$string = "123456_2";

echo $string;

$parts = explode("_", $string);

$lastpart = (int)$parts[1];

$lastpart++;

$newstring = $parts[0] . "_" . (string)$lastpart;

echo $newstring;

这会分隔“_”字符上的字符串，并将第二部分转换为整数。递增整数后，将重新创建字符串。

Use explode (step-by-step):

$string = "123456_2";

echo $string;

$parts = explode("_", $string);

$lastpart = (int)$parts[1];

$lastpart++;

$newstring = $parts[0] . "_" . (string)$lastpart;

echo $newstring;

This separates the string on the "_" character and converts the second part to an integer. After incrementing the integer, the string is recreated.

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