在 preg_replace 函数期间删除字符

发布于 2024-11-01 18:54:27 字数 687 浏览 2 评论 0原文

我需要从字符串中提取一些文本，然后用在一个实例中而不是在另一个实例中删除的字符替换该文本。希望这个例子能告诉你我的意思（这是我到目前为止所得到的）：

$commentEntry = "@Bob1990 I think you are wrong...";
$commentText = preg_replace("/(@[^\s]+)/", "<a target=\"_blank\" href=\"http://www.youtube.com/comment_search?username=${1}$1\">$1</a>", $commentEntry);

我希望结果是：

<a href="http://www.youtube.com/comment_search?username=Bob1990">@Bob1990</a> I think you are wrong...

但是我得到：

 <a href="http://www.youtube.com/comment_search?username=@Bob1990">@Bob1990</a> I think you are wrong...

我已经在这个问题上工作了至少一个小时，几乎放弃了希望，所以非常感谢任何帮助！

原文

I need to extract some text from a string, then replace that text with a character removed in one instance and not in another. Hopefully this example will show you what I mean (this is what I have so far):

$commentEntry = "@Bob1990 I think you are wrong...";
$commentText = preg_replace("/(@[^\s]+)/", "<a target=\"_blank\" href=\"http://www.youtube.com/comment_search?username=${1}$1\">$1</a>", $commentEntry);

I want the result to be:

<a href="http://www.youtube.com/comment_search?username=Bob1990">@Bob1990</a> I think you are wrong...

But am getting:

 <a href="http://www.youtube.com/comment_search?username=@Bob1990">@Bob1990</a> I think you are wrong...

I have been working on this one problem for at least an hour and nearly given up hope, so any help is greatly appreciated!

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凉风有信 2024-11-08 18:54:27

可以尝试这样的事情

$commentText = preg_replace("/(@)([^\s]+)/", "<a target=\"_blank\" href=\"http://www.youtube.com/comment_search?username=$2\">$1$2</a>", $commentEntry);

could try something like this

$commentText = preg_replace("/(@)([^\s]+)/", "<a target=\"_blank\" href=\"http://www.youtube.com/comment_search?username=$2\">$1$2</a>", $commentEntry);

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妥活 2024-11-08 18:54:27

您可以做的是调整捕获。将 @ 从大括号中移出：

preg_replace("/@([^\s]+)/",

然后您可以编写替换字符串，例如

'<a href="...$1">@$1</a>'

注意第一个 $1 如何重新插入文本，第二个 $1 > 以逐字 @ 为前缀以将其重新输入。

What you can do is adapt the capturing. Move the @ out of the braces:

preg_replace("/@([^\s]+)/",

Then you can write your replacement string like

'<a href="...$1">@$1</a>'

Note how the first $1 just reinserts the text, and the second $1 is prefixed by a verbatim @ to get it back in.

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北方的韩爷 2024-11-08 18:54:27

您正在捕获模式中的 @，因此当您使用 $1 时它将始终输出。试试这个：

$commentText = 
  preg_replace(
    "/@([^\s]+)/", 
    "<a target=\"_blank\" href=\"http://www.youtube.com/comment_search?username=$1\">@$1</a>", 
    $commentEntry
  );

这里的区别是 @ 不再被捕获为 $1 的一部分（即它只会捕获 Bob1990。由于它是一个文字值，所以它不会不需要成为任何模式的一部分。相反，我只是将其更改为在元素文本中直接在捕获的名称之前输出（即它现在是 @$1而不仅仅是 $1）。

You're capturing the @ in your pattern, so it'll always be output when you use $1. Try this instead:

$commentText = 
  preg_replace(
    "/@([^\s]+)/", 
    "<a target=\"_blank\" href=\"http://www.youtube.com/comment_search?username=$1\">@$1</a>", 
    $commentEntry
  );

The difference here is that the @ is no longer captured as part of $1 (i.e. it'll capture only Bob1990. Since it's a literal value, it doesn't need to be part of any pattern. Instead, I just changed it to output as a literal value in the element text, directly before the captured name. (i.e. it now does <a>@$1</a> rather than just <a>$1</a>).

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