如何在PHP中找到一个结果后停止继续执行strpos() - How to stop continuing a strpos() after it has found a result in PHP

How to stop continuing a strpos() after it has found a result in PHP

嘿，伙计们，我正在使用以下内容:

$pos1 = strpos($currentStatus, '#');
$pos2 = strpos($currentStatus, '#', $pos1 + strlen('#'));

如果它发现了第一个标签，则获取第二个标签，然后查找第二个......我把它存储到变量中，然后打印出来.....这个问题呢?当我打印出来的时候，我得到了字符串的其余部分例如:

$code = "Hi lets have #funfgs and than more #funny yup yup";
$pos1 = strpos($code , '#');
$pos2 = strpos($code , '#', $pos1 + strlen('#'));
echo substr($code , $pos2);

结果:#funny yup yup

所以我想要带连接词的标签，其余的都扔掉…我该怎么做呢?

大卫

编辑:

我想要的:#funny

您担心的答案是使用preg_match函数

所以，对于你的使用

$code = "Hi lets have #funfgs and than more #funny yup yup";
$pos1 = preg_match( "/.*#('S+)/", $code , $match );
print_r( $match[1] );

您也可以为您的匹配包含#。你可以这样做:

$pos1 = preg_match( "/.*(#'S+)/", $code , $match );
echo $match[1];

试试这个:

$code = "Hi lets have #funfgs and than more #funny yup yup";
preg_match_all('/#(?P<hash>'w+)/',$code,$match);
echo "<pre>";
print_r($match['hash']);

这里你会得到#之后的所有单词，你可以从数组$match['hash']中选择任何单词

对于您的问题中提到的情况，使用echo $match['hash'][1];

你可以使用爆炸函数:

$code = "Hi lets have #funfgs and than more #funny yup yup";
$pos1 = strpos($code , '#');
$pos2 = strpos($code , '#', $pos1 + strlen('#'));
$hashtag = explode(' ', substr($code , $pos2));
echo $hashtag[0];