在PHP中獲取兩個字符之間的字符串

Question

我在php中有一個字符串說

$string = 'All I want to say is that <#they dont really care#> about us.
    I am wasted away <#I made a million mistake#>, am I too late. 
    Theres a storm in my head and a race on my bed, <#when you are not near#>' ;


$expected_output = array(
    'they dont really care',
    'I made a million mistake',
    'when you are not near'
);

我如何使用PHP regex實現這一點？ 謝謝閱讀 ：）

Answer 1

該代碼將完成您想要的

<?php

$string = 'All I want to say is that <#they dont really care#> about us.
    I am wasted away <#I made a million mistake#>, am I too late. 
    Theres a storm in my head and a race on my bed, <#when you are not near#>' ;


preg_match_all('/<#(.*)#>/isU', $string, $matches);

var_dump($matches[1]);

Answer 2

您可以使用此正則表達式：

'/<#(.*?)#>/s'

在preg_match_all函數調用中。

我不想給您完整的代碼，但這應該足以讓您繼續前進。

Answer 3

通過向前和向后看，

(?<=<#).*?(?=#>)

最后，調用preg_match_all函數以打印匹配的字符串。

您的PHP代碼是

<?php
$data = 'All I want to say is that <#they dont really care#> about us.
    I am wasted away <#I made a million mistake#>, am I too late. 
    Theres a storm in my head and a race on my bed, <#when you are not near#>' ;
$regex =  '~(?<=<#).*?(?=#>)~';
preg_match_all($regex, $data, $matches);
var_dump($matches);
?>

輸出：

array(1) {
  [0]=>
  array(3) {
    [0]=>
    string(21) "they dont really care"
    [1]=>
    string(24) "I made a million mistake"
    [2]=>
    string(21) "when you are not near"
  }
}

Answer 4

更緊湊的版本：

$regex = '~<#\K.*?(?=#>)~';
preg_match_all($regex, $string, $matches);
print_r($matches[0]);

在正則表達式演示中查看匹配項。

火柴

they dont really care
I made a million mistake
when you are not near

說明

• ^錨斷言我們在字符串的開頭
• <#匹配左定界符
• \\K告訴引擎放棄與最終匹配相距甚遠的匹配，它返回
• .*? 懶惰地匹配字符，直到...
• 先行(?=#>)可以斷言以下是#>
• $錨斷言我們在字符串的末尾

參考

• 零長度斷言的前瞻性和后瞻性
• 掌握前瞻性和后向性