如何使用 Strpos() 查找针后的下一个字符串

发布于 2024-10-15 19:29:35 字数 589 浏览 2 评论 0原文

我正在使用 PHP strpos() 在文本段落中查找针。我正在努力解决如何在找到针后找到下一个单词。

例如，请考虑以下段落。

$description = "Hello, this is a test paragraph.  The SCREENSHOT mysite.com/screenshot.jpg and the LINK mysite.com/link.html is what I want to return.";

我可以使用 strpos($description, "SCREENSHOT") 来检测 SCREENSHOT 是否存在，但我想获取 SCREENSHOT 之后的链接，即 mysite.com/screenshot.jpg 。以类似的方式，我想检测描述是否包含 LINK，然后返回 mysite.com/link.html。

如何使用 strpos() 然后返回以下单词？我假设这可以通过正则表达式完成，但我不确定。下一个单词将是“针后一个空格，后跟任何内容，后跟一个空格”。

谢谢！

原文

I'm using PHP strpos() to find a needle in a paragraph of text. I'm struggling with how to find the next word after the needle is found.

For example, consider the following paragraph.

$description = "Hello, this is a test paragraph.  The SCREENSHOT mysite.com/screenshot.jpg and the LINK mysite.com/link.html is what I want to return.";

I can use strpos($description, "SCREENSHOT") to detect if SCREENSHOT exists, but I want to get the link after SCREENSHOT, namely mysite.com/screenshot.jpg. In a similar fashion, I want to detect if the description contains LINK and then return mysite.com/link.html.

How can I use strpos() and then return the following word? I'm assuming this might be done with a RegEx, but I'm not sure. The next word would be "a space after the needle, followed by anything, followed by a space".

Thanks!

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话少情深 2024-10-22 19:29:35

或者“旧”方式......:-)

$word = "SCREENSHOT ";
$pos = strpos($description, $word);
if($pos!==false){
    $link = substr($description, $pos+strlen($word));
    $link = substr($link, strpos($link, " "));
}

Or the "old" way... :-)

$word = "SCREENSHOT ";
$pos = strpos($description, $word);
if($pos!==false){
    $link = substr($description, $pos+strlen($word));
    $link = substr($link, strpos($link, " "));
}

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別甾虛僞 2024-10-22 19:29:35

您可以使用单个正则表达式来完成此操作：

if (preg_match_all('/(SCREENSHOT|LINK) (\S+?)/', $description, $matches)) {
    $needles = $matches[1]; // The words SCREENSHOT and LINK, if you need them
    $links = $matches[2]; // Contains the screenshot and/or link URLs
}

You could do this with a single regular expression:

if (preg_match_all('/(SCREENSHOT|LINK) (\S+?)/', $description, $matches)) {
    $needles = $matches[1]; // The words SCREENSHOT and LINK, if you need them
    $links = $matches[2]; // Contains the screenshot and/or link URLs
}

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南街九尾狐 2024-10-22 19:29:35

我使用以下内容在我的网站上做了一些测试：

$description = "Hello, this is a test paragraph. The SCREENSHOT mysite.com/screenshot.jpg and the LINK mysite.com/link.html is what I want to return.";

$matches = array();
preg_match('/(?<=SCREENSHOT\s)[^\s]*/', $description, $matches);
var_dump($matches);
echo '<br />';
preg_match('/(?<=LINK\s)[^\s]*/', $description, $matches);
var_dump($matches);

我正在使用积极的后视来获得你想要的东西。

I did a little testing on my site using the following:

$description = "Hello, this is a test paragraph. The SCREENSHOT mysite.com/screenshot.jpg and the LINK mysite.com/link.html is what I want to return.";

$matches = array();
preg_match('/(?<=SCREENSHOT\s)[^\s]*/', $description, $matches);
var_dump($matches);
echo '<br />';
preg_match('/(?<=LINK\s)[^\s]*/', $description, $matches);
var_dump($matches);

I'm using positive lookbehind to get what you want.

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