正則表達式匹配字符串中的所有* texttexttext

Question

我需要一個正則表達式，它將匹配所有以*開頭的字符串，直到遇到<

所以在這段文字中： bob went to the *store and he bought a toy and <then he went outside *and went to his car for <a ride.

它會匹配bob went to the *store and he bought a toy and ， and went to his car for

如果沒有“ <”，它將匹配所有行，直到行尾

Answer 1

嘗試這個：

<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);
?>

輸出：

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

應該是這樣的：

使用PHP：

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

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

如果要跳過結尾“ <”，請將表達式更改為#\\*(.+?)<?#並且如果要允許更改行，請使用以下標志：

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

注意表達式后面的si標志

希望能幫助到你

Answer 3

嘗試\\*(.+?)<

您可以使用此工具來測試正則表達式： http : //gskinner.com/RegExr/

編輯要匹配到<或匹配到字符串的末尾，請使用：

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

Answer 4

我會用這樣的東西

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

在Regexr上查看

(?<=\\*)是后面的字符，它不匹配任何字符，但可以確保之前的字符是*

.*? 匹配所有非貪婪的事物

(?=<|$)是向前看的字符，它不匹配任何字符，但可以確保后面的字符是<或$ （行尾）==>使用m（多行）修飾符，否則$將僅匹配字符串的末尾，而不匹配行的末尾。