PHP獲取整個字符串，而不是URL中的單個參數

Question

如果我有一個域名，例如www.example.com/w/

我希望能夠將整個文本字符串附加在URL后面，我知道如何獲取？key = value格式的參數，這不是我想要的。

但我想在/ w /前綴之后獲取所有內容，所以整個字符串都應該完整，所以如果有人在上述內容之后附加

www.example.com/w/ https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html

我將能夠獲得https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html

我正在考慮在服務器上安裝codeigniter，如果有幫助，但此刻我只是在使用核心php

Answer 1

您只需要使用str_replace()

$str = "www.example.com/w/https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html";
$str2 =  str_replace('www.example.com/w/', '', $str);
echo $str2;

產量

https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html 

閱讀有關str_replace（）的更多信息

Answer 2

嘗試使用strpos和substr

$str = "www.example.com/w/https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html";
echo $str.'<pre>';
$start =  strpos($str, '/w/');
echo substr($str, $start + 3);die;

輸出：

www.example.com/w/https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html

https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html

strpos（）將使您第一次出現/w/然后可以用+3進行substr刪除/w/

或者嘗試使用strstr和str_replace

$str = "www.example.com/w/https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html";
echo $str.'<pre>';
$str1 =  strstr($str, '/w/');
echo $str1.'<pre>';
$str2 = str_replace('/w/', '', $str1);
echo $str2.'<pre>';die;

輸出：

www.example.com/w/https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html

/w/https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html

https://www.nytimes.com/2019/04/17/us/politics/trump-mueller-report.html

strstr（）將為您提供給定/w/子字符串，並使用str_replace（）從新字符串中刪除/w/