PHP过滤字符串之前和之后的特定词用不同长度

Question

我有一个类似的字符串....

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

我怎样才能得到它弄成这个样子？这里

$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)); 

输出：

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

但这不会处理不同的 “命令” 比 “命令” 等，也假设所有空白之间的空白ds是单个空格字符。您需要更复杂的代码来处理多个案例，但这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]; 

?> 

检查现场演示http://codepad.org/5iD02e6b

Answer 4

答案已被接受，但仍然..我的答案的优点是，如果在原始字符串中存在错误，则结果变量将为空。这是很好的，因为那样很容易测试一切是否按照计划进行:)

<?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'";