正則表達式字符

Question

我試圖提取字在最後/和倒數第二個/之間的最後和倒數第二次數之間提取字符。

1. $string = https://ss1.xxx/img/categories_v2/FOOD/fastfood（想更換$string到food）
2. $string = https://ss1.xxx/img/categories_v2/SHOPS/barbershop（想更換$string到shops）

我是新來的正則表達式，並試圖/[^/]*$ - 當然，這是後返回萬萬最後/ ..任何幫助將不勝感激..謝謝！

我正在使用PHP。

Answer 1

您可以使用此：

$result = preg_replace_callback('~(?<=/)[^/]+(?=/[^/]*$)~', function ($m) { 
    return strtolower($m[0]); }, $string); 

模式的細節：

~   # pattern delimiter 
(?<=/)  # zero width assertion (lookbehind): preceded by/
[^/]+  # all characters except/one or more times 
(?=/[^/]*$) # zero width assertion (lookahead): followed by /, 
      # all that is not a/zero or more times, and the end of the string 
~   # pattern delimiter 

Answer 2

用途：

preg_match('#/([^/]*)/[^/]*$#', $string, $match); 
echo $match[1]; 

您還可以使用：

$words = explode('/', $string); 
echo $words[count($words)-2]; 

Answer 3

正則表達式：

(\w+)(/[^/]+)$ 

PHP代碼：

<?php 
    $string = "https://ss1.xxx/img/categories_v2/FOOD/fastfood"; 
    echo preg_replace("@(\w+)(/[^/]+)$@", "food$2", $string); 
    $string = "https://ss1.xxx/img/categories_v2/SHOPS/barbershop"; 
    echo preg_replace("@(\w+)(/[^/]+)$@", "shops$2", $string); 
?>