使用字符串作为模式，但不将其删除

发布于 2024-08-30 19:12:56 字数 834 浏览 8 评论 0原文

我对正则表达式还很陌生，我顺便学到了一些东西，但仍然是知识！

所以我想请您澄清它是如何工作的！

假设我有以下字符串，正如您所看到的，它们的格式可能略有不同，但它们非常相似！

DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE

现在我想替换第一个AZ块和冒号之间的所有内容，例如我会保留我

DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART

已经掌握的菜鸟知识这个该死的正则表达式！ :-(

preg_replace( '/^[A-Z](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );

但为什么我确定这个正则表达式不起作用！？:-)

请帮助我！

PS：问题的标题已经解释得很清楚了，我还想知道如何使用众所周知的字符串块来匹配另一个...

preg_replace( '/^[DTSTART](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );

..无需删除 DTSTART

谢谢您的时间！

问候卢卡·菲洛索菲

原文

i'm pretty new on regex, i have learned something by the way, but is still pour knowledge!

so i want ask you for clarification on how it work!

assuming i have the following strings, as you can see they can be formatted little different way one from another but they are very similar!

DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE

now i want replace everything between the first A-Z block and the colon so for example i would keep

DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART

so on my very noobs knowledge i have worked out this shitty regex! :-(

preg_replace( '/^[A-Z](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );

but why i'm sure this regex will not work!? :-)

Pls help me!

PS: the title of question is pretty explaned, i want also know how for example use a well know string block for match another...

preg_replace( '/^[DTSTART](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );

..without delete DTSTART

Thanks for the time!

Regards
Luca Filosofi

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枉心 2024-09-06 19:12:56

您可以使用相对简单的正则表达式，如下所示。

$subject = 'DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE';

echo preg_replace('/^[A-Z]+\K[^:\n]*/m', '', $subject) . PHP_EOL;

它在行的开头查找一系列大写字母，将匹配起点（这就是 \K 所做的）重置为这些字母的末尾，并匹配除冒号或换行符之外的任何内容（即您想要删除的部分）。然后，这些匹配的部分将被替换为空字符串。

上面的输出将是

DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART

如果您感兴趣的行仅以 DTSTART 或 DTEND 开头，那么我们可以更准确地了解要匹配的内容（例如 < code>^DT(?:START|END)) 但 [AZ] 显然涵盖了这两者。

You could use a relatively simple regex like the following.

$subject = 'DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE';

echo preg_replace('/^[A-Z]+\K[^:\n]*/m', '', $subject) . PHP_EOL;

It looks for a series of capital letters at the start of a line, resets the match starting point (that's what \K does) to the end of those and matches anything not a colon or newline (i.e. the parts you want to remove). Those matched parts are then replaced with an empty string.

The output from the above would be

DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART

If the lines that you are interested in will only ever start with DTSTART or DTEND then we could be more precise about what to match (e.g. ^DT(?:START|END)) but [A-Z] obviously covers both of those.

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没有你我更好 2024-09-06 19:12:56

如果您想在替换中保留部分匹配模式，请在其两侧加上括号，然后通过 $1 （或任何分组）引用它。

例如：

s/^(this is a sentence) to edit/$1/

给出“这是一个句子”

If you want to retain part of the matched pattern in a substitution, you put parentheses around it and then refer to it by $1 (or whichever grouping it is).

For example:

s/^(this is a sentence) to edit/$1/

gives "this is a sentence"

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找回味觉 2024-09-06 19:12:56

您可以查看此示例，其工作方式与您的问题类似，

\w+): (?P\d+)/', $str, $matches);

/* This also works in PHP 5.2.2 (PCRE 7.0) and later, however
* the above form is recommended for backwards compatibility */
// preg_match('/(?\w+): (?\d+)/', $str, $matches);

print_r($matches);

The above example will output:

Array
(
    [0] => foobar: 2008
    [name] => foobar
    [1] => foobar
    [digit] => 2008
    [2] => 2008
)

因此如果您只需要数字，则需要打印 $matches[digit]

You can check out this example work similarly as your problem

\w+): (?P\d+)/', $str, $matches);

/* This also works in PHP 5.2.2 (PCRE 7.0) and later, however
* the above form is recommended for backwards compatibility */
// preg_match('/(?\w+): (?\d+)/', $str, $matches);

print_r($matches);

The above example will output:

Array
(
    [0] => foobar: 2008
    [name] => foobar
    [1] => foobar
    [digit] => 2008
    [2] => 2008
)

so if u need only digit u need to print $matches[digit]

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夏夜暖风 2024-09-06 19:12:56

您想要删除分号和冒号或行尾之间的所有内容，对吧？所以用它作为你的表达方式。你把事情搞得太复杂了。

preg_replace('/(?:;.+?:)|(?:;.+?$)/m','',$data);

这是一个非常简单的表达。匹配 (?:;.+?:) 或 (?:;.+?$)，它们仅在终止符上有所不同（第一个匹配最多冒号，第二个匹配到行尾）。

每个都是一个非捕获组，以分号开头，勉强读入所有字符，然后在终止符处停止。根据您的描述，与此匹配的所有内容都可以移除。

You want to remove everything between a semicolon and either a colon or the end of the line, right? So use that as your expression. You're overcomplicating things.

preg_replace('/(?:;.+?:)|(?:;.+?$)/m','',$data);

It's a pretty simple expression. Either match (?:;.+?:) or (?:;.+?$), which differ only by their terminator (the first one matches up to a colon, the second one matches up to the end of the line).

Each is a non-capturing group that starts with a semicolon, reluctantly reads in all characters, then stops at the terminator. Everything matched by this is removable according to your description.

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