PHP過濾字符串之前和之後的特定詞用不同長度

Question

我有一個類似的字符串....

$string = "order apples oranges pears and bananas from username"; 

我怎樣才能得到它弄成這個樣子？這裏

$products = "apples oranges pears and bananas"; 
$username = "username"; 

Answer 1

$string = "order apples oranges pears and bananas from username"; 
list($products,$username) = explode(" from ", $string); 
$products = str_replace('order ', '', $products); 


//print results for verification 
var_dump(array($products,$username)); 

輸出：

array(2) { [0]=> string(32) "apples oranges pears and bananas" [1]=> string(8) "username" } 

但這不會處理不同的 「命令」 比 「命令」 等，也假設所有空白之間的空白ds是單個空格字符。您需要更復雜的代碼來處理多個案例，但這需要您提供更多信息。

Answer 2

list($products,$username) = explode("from", $string); 

Answer 3

<?php 

$string = "order apples oranges pears and bananas from username"; 

$part = explode("order",$string); 

$order = explode("from",$part[1]); 

echo $order[0]; 

echo $order[1]; 

?> 

檢查現場演示http://codepad.org/5iD02e6b

Answer 4

答案已被接受，但仍然..我的答案的優點是，如果在原始字符串中存在錯誤，則結果變量將爲空。這是很好的，因爲那樣很容易測試一切是否按照計劃進行:)

<?php 
    $string = "order apples oranges pears and bananas from username"; 
    $products_regex = "/^order\\s(.*)\\sfrom/i"; 
    $username_regex = "/from\\s(.*)$/i"; 
    $products_matches = array(); 
    $username_matches = array(); 
    preg_match($products_regex, $string, $products_matches); 
    preg_match($username_regex, $string, $username_matches); 
    $products = ""; 
    $username = ""; 

    if(count($products_matches) === 2 && 
      count($username_matches) === 2) 
    { 
     $products = $products_matches[1]; 
     $username = $username_matches[1]; 
    } 

    echo "products: '$products'<br />\nuser name: '$username'";