我給自己買了一個小腳本,檢查由用戶提供的鏈接的有效性,使得它的安全數據庫來存儲(至少更安全)和確認這是一個鏈接到Facebook。我怎麼這樣聰明的辦法代碼的Ajax/PHP
現在我想將此代碼轉出另一個鏈接,在需要時更改參數,以便鏈接到這些網站上的用戶個人資料的工作,我不想複製和粘貼代碼5次,然後嘗試並使Ajax適應它,如果這是更好的方法來解決這個問題。
這是我的代碼,它可以被看作在www.vwrx_project.co.uk/test.php工作。它希望只接受facebook.com/(這裏有些東西)。
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>
我也不太清楚,這就是爲什麼我想諮詢一下。我希望在的地方,我可以檢查Facebook,微博,谷歌+,Instagram的,YouTube和他們的個人網站用戶提供的URL,如果他們有一個,確保了主要的匹配環節的正確的格式來獲取系統。 我的想法是提供一個形式6場和提交表單時,檢查,看看是否有被填補的字段,然後運行該腳本 –
你可以創建你可以列出所有允許的URL函數內部的數組。然後,您的代碼可以檢查粘貼的URL是否與陣列中保存的任何URL匹配。你可以使用'''in_array()'''函數來做到這一點。 –