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因此,多虧了response to my previous question,我試着讓代碼通過電子郵件發送給我,如果代碼位於另一個站點上。Ajax調用PHP函數
這裏是我的javascript其目的是爲潛在的小偷代碼採取與他們自己的網站:
<script type="text/javascript">
var mypostrequest=new ajaxRequest()
mypostrequest.onreadystatechange=function(){
if (mypostrequest.readyState==4){
if (mypostrequest.status==200 || window.location.href.indexOf("http")==-1){
document.getElementById("result").innerHTML=mypostrequest.responseText
}
else{
alert("An error has occured making the request")
}
}
}
var url = document.domain;
var joel="www.joelhoskin.net76.net";
if (url!=joel)
{
mypostrequest.open("POST", "http://www.joelhoskin.net76.net/email.php", true)
mypostrequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded")
mypostrequest.send(url)
}
</script>
,這裏是在joelhoskin.net76.net/email.php的PHP
`<?php
$url=$_POST['url'];
if(isset($url))
{
$to = '[email protected]';
$from = '[email protected]';
$subject = 'Stolen Page';
$content = $url."Site Stolen";
$result = mail($to,$subject,$content,'From: '.$from."\r\n");
die($result);
}
?>`
像它不是我發送電子郵件它應該
謝謝我回頭看看我使用的參考,事實證明,我確實錯過了那一點 但是,它仍然沒有給我發電子郵件 – Joel