Question

我需要一个匹配所有以*开头的字符串的正则表达式，直到遇到＆lt;

所以在这个文本块中：bob went to the *store and he bought a toy and <then he went outside *and went to his car for <a ride.

它会匹配bob went to the *store and he bought a toy and和and went to his car for

如果没有“＆lt;”它将匹配所有直到行的结尾

Answer 1

试试这个：

<pre>
<?
$s = 'bob went to the *store and he bought a toy and <then he went outside *and went to his car for <a ride.';

preg_match_all("/\*([^<]+)/", $s, $matched); 
print_r($matched);
?>

输出：

Array
(
    [0] => Array
        (
            [0] => *store and he bought a toy and 
            [1] => *and went to his car for 
        )

    [1] => Array
        (
            [0] => store and he bought a toy and 
            [1] => and went to his car for 
        )

)

Answer 2

应该是这样的：

使用PHP：

preg_match_all("#\*(.+?)<#", $stringWithText, $matches, PREG_SET_ORDER);
$mCount = count($matches);

foreach ($matches as $match)
    echo "Matched: " . $match[1] . "<br/>";

如果你想跳过结尾“＆lt;”将表达式更改为#\*(.+?)<?#，如果要允许换行，请使用以下标志：

preg_match_all("#\*(.+?)<#si", $stringWithText, $matches, PREG_SET_ORDER);

注意si标志尾随表达式

希望有所帮助

Answer 3

尝试\*(.+?)<

您可以使用此工具试用正则表达式：http://gskinner.com/RegExr/

修改要保持与字符串末尾的< OR匹配，请使用：

\*(.+?)[<|$.*]

Answer 4

我会用这样的东西

(?<=\*).*?(?=<|$)

在Regexr

上查看

(?<=\*)是一个背后的外观，它不匹配任何字符，但它确保前面的字符是*

.*?匹配非贪婪的一切

(?=<|$)是展望未来，它不匹配任何字符，但它确保以下字符为<或$（行结束）==＆gt;使用m（多行）修饰符，否则$将只匹配字符串的结尾而不是行的结尾。

正则表达式匹配字符串中的所有* texttexttext

4 个答案: