PHP函數刪除

Question

我創造了這個函數從字符串中刪除特定變量的URL指定的變量

if(!function_exists("remove_variable")) { 
    function remove_variable($remove = array(), $url) { 
     if($url == '') { 
      $url = $_SERVER["REQUEST_URI"]; 
     } 

     foreach($remove as $r) { 
      echo $r.'<br>'; 
      $url = preg_replace('/([?&])'.$r.'=[^&]+(&|$)/','$1', $url); 
      echo $url.'<br><br>'; 
     } 

     return $url; 
    } 
} 

我與測試：

<?php echo remove_variable(array("productsearch_name", "productsearch_type", "productsearch_supplier"), $_SERVER["REQUEST_URI"].uri_glue()); ?> 

這是返回：

productsearch_name 
/companies/customers/pricelist?productsearch_name=&productsearch_type=Broadband%20&productsearch_supplier=& 

productsearch_type 
/companies/customers/pricelist?productsearch_name=&productsearch_supplier=& 

productsearch_supplier 
/companies/customers/pricelist?productsearch_name=&productsearch_supplier=& 

/companies/customers/pricelist?productsearch_name=&productsearch_supplier=& 

因此，它不會按預期去除變量

Answer 1

可以實現這種方式，

$str = "http://google.com/companies/customers/pricelist?productsearch_name=asd&productsearch_type=Broadband%20&productsearch_supplier=test"; 
parse_str(parse_url($str)['query'],$output); 
print_r($output); 

parse_str - 解析字符串到變量。
parse_url - 解析一個URL並返回其組件

我希望這能解決您的問題。