Question

我有一个小脚本来检查用户提供的链接的有效性，使其可以安全存储在数据库中（至少更安全）并确认它是指向facebook的链接。

现在我想将此代码转换为另一个链接，在需要时更改参数，以便链接到这些网站上的用户个人资料，我不想复制并粘贴代码5次，然后尝试和如果有更好的方法来解决这个问题，请调整Ajax以使用它。

这是我的代码，可以在www.vwrx_project.co.uk/test.php上查看。它希望只接受facebook.com/(something here）。

link_checker.php

<?php

function check_url($dirty_url) {
//remove anything before facebook.com using strstr() 
//clean url leaving alphanumerics : / . only -  required to remove facebook link format with /#!/
$clean_url = strstr(preg_replace('#[^a-z0-9:/.?=]#i', '', $dirty_url), 'facebook.com'); 

$parsed_url = parse_url("http://www.".$clean_url); //parse url to get brakedown of components

$safe_host = $parsed_url['host']; // safe host direct from parse_url

// str_replace to switch any // to a / inside the returned path - required due to preg_replace process above
$safe_path = str_replace("//", "/", ($parsed_url['path']));

if ($parsed_url['host'] == 'www.facebook.com' && $parsed_url['path'] != '' && $parsed_url['path'] != '/') {
    echo "<a href=\"http://$safe_host$safe_path\" alt=\"facebook\" target=\"_new\">Facebook</a>";
} else if ($parsed_url['host'] == 'www.facebook.com' && $parsed_url['path'] == '') {
    echo "missing_profile1";
} else if ($parsed_url['host'] == 'www.facebook.com' && $parsed_url['path'] == '/') {
    echo "missing_profile2";
} else {            
    echo "invalid_url";
}
}
?>

test.php的

<?php
include_once ("includes/check_login_status.php");
include_once ("includes/link_checker.php");

// AJAX CALLS THIS LOGIN CODE TO EXECUTE
if(isset($_POST["L"])){
$dirty_url = $_POST["L"]; //user supplied link
//$dirty_url = "http://www.facebook.com/profile.php?id=4"; 
    // if $dirty_url is blank
if($dirty_url == ""){
    echo "no link supplied";
    exit();     
} else {
    check_url($dirty_url);
}
exit();
}
?>

<html>
<head>
<title>testing</title>
<script type="text/javascript" src="js/main.js"></script>
<script src="js/main.js"></script>
<script src="js/ajax.js"></script>
<script>

function emptyElement(x){
_(x).innerHTML = "";
}

function cleanURL(){
var user_url = _("user_link").value;
var func = _("hidden").value;
if(user_url == ""){
        _("status").innerHTML = "Please provide a link before clicking submit";
    } else {
        _("submitbtn").style.display = "none";
        _("status").innerHTML = 'please wait ...';
        var ajax = ajaxObj("POST", "test.php");
        ajax.onreadystatechange = function() {
            if(ajaxReturn(ajax) == true) {
                if(ajax.responseText == "no link supplied"){
                    _("status").innerHTML = "Submitted blank form data.";
                    _("submitbtn").style.display = "block";
                } else if(ajax.responseText == "invalid_url"){
                    _("status").innerHTML = "The url supplied is invalid";
                    _("submitbtn").style.display = "block";
                } else if(ajax.responseText == "missing_profile1"){
                    _("status").innerHTML = "Please supply a link to your profile";
                    _("submitbtn").style.display = "block";
                } else if(ajax.responseText == "missing_profile2"){
                _("status").innerHTML = "Please supply a link to your profile";
                _("submitbtn").style.display = "block";
            } else{
                _("status").innerHTML = ajax.responseText;
            }
        }
    }
    ajax.send("L="+user_url);
}
}
</script>
</head>
<body>
    <p id="status"></p>

 <form id="linkform" onSubmit="return false;">
    <input type="text" id="user_link">
    <input type="hidden" id="hidden" value="Facebook">
    <button id="submitbtn" onClick="cleanURL()">Submit</button>
</form>

Answer 1

为什么不添加一个您想要允许的网站的附加参数？

function check_url($dirty_url, $websiteURL)

然后更新您的功能以使用$websiteURL变量而不是硬编码的＆＃39; facebook.com＆＃39;

然后当你想要几个不同的网址时，你可以这样做

check_url($dirty_url, 'facebook.com');

或

check_url($dirty_url, 'twitter.com');

或者您是否希望能够在单个功能中检查多个站点？例如facebook.com和twitter.com

我如何以智能方式Ajax / PHP编写代码

1 个答案: