Question

如何在没有％20％27的情况下传递变量并准确检索传递的内容我尝试了urlencode和解码但是没有工作需要一些帮助

传入网址http://localhost/civilhelp/enquiry.php?cat=%27PLUMBING%20FITTINGS%20&%20SANITARY%20WARE%27 我需要删除％27％20使用过的urlencode但没有使用

只追回PLUMBING FITTINGS并留下任何人帮助解决方案

我将其作为

传递

 $cate = 'PLUMBING FITTINGS & SANITARY WARE'; it is from db
    <a href="enquiry.php?cat='<?php echo urlencode($cate)  ;?>'">

尝试检索为

$category = urldecode($_GET['cat']); 
$category = str_replace("'"," ",$category);
print_r($category); 
$category = trim($category);

print_r($category);它只打印PLUMBING FITTINGS

我没有得到确切的通过注意：我使用过urlencode和urldecode也修剪了

但我仍然不能希望我的问题是什么!!

Answer 1

这是不对的我尝试了你的代码并显示正确的数据试试这个

$url = 'http://localhost/civilhelp/enquiry.php?cat=%27PLUMBING%20FITTINGS%20&%20SANITARY%20WARE%27';
$category = urldecode($url);
$category = str_replace("'","",$category);
echo $category; // result: http://localhost/civilhelp/enquiry.php?cat=PLUMBING FITTINGS & SANITARY WARE

编辑：
我只是在cat =＆＃39;＆＃39;变化与至％26

Answer 2

<?php  if(!empty($r['category'])){ $cat = $r['category']; $cate=substr($cat,0,25) . '';
 //$cat = str_replace(' ', '', $cate);
  $cat = str_replace('&', '%26', $cat);  after changing it works fine // print_r($cat);?>
  <a href="enquiry.php?cat='<?php echo $cat  ;?>'">

enter code here

如何在没有特殊字符的get中传递变量并返回

2 个答案: