dongxinyue2817 2011-11-09 13:17
浏览 11
已采纳

PHP http链接中的变量

I'm trying to get the cureent url and send that through in a link and I'm stuck. The link displays but it's missing the url that I want to include (content in the code sample here)

'href' => ( "http://chart.apis.google.com/chart?cht=qr&chs=300x300&choe=UTF-8&chld=H&chl='.$content .'")
  • 写回答

3条回答 默认 最新

  • douqian9729 2011-11-09 13:22
    关注

    Your quotes aren't correct:

    'href' => ( "http://chart.apis.google.com/chart?cht=qr&chs=300x300&choe=UTF-8&chld=H&chl=".$content)
    

    You should also use urlencode() and urldecode() to ensure the $content variable is properly formatted

    Edit Try this then:

    $currentUrl = rawurlencode($_SERVER['PATH_INFO']);
    $newUrl = "http://chart.apis.google.com/chart?cht=qr&chs=300x300&choe=UTF-8&chld=H&chl=";
    
    header("Location: $newUrl.$currentUrl");
    

    //Not sure if path_info will always contain the full url but there are lots of functions on the web to grab the current url that you can google for.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(2条)

报告相同问题?