正則表達式（的preg_match）

Question

這是不工作代碼：

<?php 
$matchWith = " http://videosite.com/ID123 "; 
preg_match_all('/\S\/videosite\.com\/(\w+)\S/i', $matchWith, $matches); 
foreach($matches[1] as $value) 
{ 
    print '<a href="http://videosite.com/'.$value.'">Hyperlink</a>';   
} 
?> 

我要的是，它不應該顯示的鏈接，如果它之前或之後有一個空格。 所以現在它應該什麼都不顯示。但它仍然顯示鏈接。

Answer 1

這也可以匹配ID12，因爲3不是空格，而http：/不是空格。您可以嘗試：

preg_match_all('/^\S*\/videosite\.com\/(\w+)\S*$/i', $matchWith, $matches); 

Answer 2

所以，如果有空白，你不希望它顯示。像這樣的東西應該工作，沒有測試。

preg_match_all('/^\S+?videosite\.com\/(\w+)\S+?$/i', $matchWith, $matches); 

Answer 3

您可以試試這個。它的工作原理：

if (preg_match('%^\S*?/videosite\.com/(\w+)(?!\S+)$%i', $subject, $regs)) { 
    #$result = $regs[0]; 
} 

但我肯定我張貼此之後，您將更新你的問題:)

說明：

" 
^   # Assert position at the beginning of the string 
\S   # Match a single character that is a 「non-whitespace character」 
    *?   # Between zero and unlimited times, as few times as possible, expanding as needed (lazy) 
\/   # Match the character 「/」 literally 
videosite # Match the characters 「videosite」 literally 
\.   # Match the character 「.」 literally 
com   # Match the characters 「com」 literally 
\/   # Match the character 「/」 literally 
(   # Match the regular expression below and capture its match into backreference number 1 
    \w   # Match a single character that is a 「word character」 (letters, digits, etc.) 
     +   # Between one and unlimited times, as many times as possible, giving back as needed (greedy) 
) 
(?!   # Assert that it is impossible to match the regex below starting at this position (negative lookahead) 
    \S   # Match a single character that is a 「non-whitespace character」 
     +   # Between one and unlimited times, as many times as possible, giving back as needed (greedy) 
) 
\$   # Assert position at the end of the string (or before the line break at the end of the string, if any) 
" 

Answer 4

它很可能是簡單的使用這個正則表達式：

'/^http:\/\/videosite\.com\/(\w+)$/i' 

我相信你指的是th在http之前的一個空格，以及目錄之後的空格。因此，您應該使用^字符來指示該字符串必須以http開頭，並在末尾使用$字符來指示該字符串必須以單詞字符結尾。