0
所以我有一個小的鏈接器,我試圖讓工作,當用戶輸入他們的用戶名和密碼它將連接到這個網站,而沒有實際離開頁面,但有一些困難這樣做。我正在使用jQuery中的.post()發送到aspx.page。問題與.post()不正確閱讀
當我點擊提交我會發送到該網站,我不希望出現這種情況,再加上它不會登錄
碼 -
<div class="linkdimesAccount">
<form name="aspnetForm" method="post" action="http://1betvegas.com/default.aspx" onkeypress="javascript:return WebForm_FireDefaultButton(event, 'ctl00_MainContent_ctlLogin_BtnSubmit')" id="aspnetForm" style="margin: 0 0 0 0;">
<script type="text/javascript">
//<![CDATA[
var theForm = document.forms['aspnetForm'];
if (!theForm) {
theForm = document.aspnetForm;
}
function __doPostBack(eventTarget, eventArgument) {
if (!theForm.onsubmit || (theForm.onsubmit() != false)) {
theForm.__EVENTTARGET.value = eventTarget;
theForm.__EVENTARGUMENT.value = eventArgument;
theForm.submit();
}
}
//]]>
</script>
<div>
<input type="hidden" name="__EVENTTARGET" id="__EVENTTARGET" value="">
<input type="hidden" name="__EVENTARGUMENT" id="__EVENTARGUMENT" value="">
<input type="hidden" name="__VIEWSTATE" id="__VIEWSTATE" value="eYjk1H7gGdxNJmuevBY9L0vo1lS2NWgmApCJFEQznQr+AwZaPf/hYV4iGSnPCSknyHjESN/BgHueeChVuGgmN4wtWFCdUaYPJRItirNF0nIBHe9Q">
</div>
<script src="/WebResource.axd?d=FrT3YL7-WvrI_DKD4vsDo2d0Al_8j_u_HUym76C9Z5ggdJlIe1yu5cWI_jZDYcizjTU0SkefrocS8ATa0&t=634604245351482412" type="text/javascript"></script>
<div>
<input type="hidden" name="__EVENTVALIDATION" id="__EVENTVALIDATION" value="yepc3zJ4Kjr5ru/aXm+s9SqYXyFoM+cwIlWPI69lDMnO4eybZJ1cM1pfzowM47Ggezne5505JNUx/VGe3XO8OCmvZnghqv8ZCuJm+yffjs4inz2n6ctjK/0F/qER0ARSznB8iJsMIZ7HxPXA/Stv+0ubH0U=">
</div>
<h2>Link your 5dimes account</h2>
<table>
<tr>
<?php
if ($dimesaccount == 1){
echo "<h2 style='color:red;'> Please note your account is already linked. If password or other information has changed please re-enter your username and password";
}
?>
<td>
Username:
</td>
<td>
<input name="ctl00$MainContent$ctlLogin$_UserName" type="text" size="15" id="ctl00_MainContent_ctlLogin__UserName" accesskey="u" tabindex="60" class="login_input">
</td>
</tr>
<tr>
<td>
Password:
</td>
<td>
<input name="ctl00$MainContent$ctlLogin$_IdBook" type="hidden" id="ctl00_MainContent_ctlLogin__IdBook">
<input name="ctl00$MainContent$ctlLogin$Redir" type="hidden" id="ctl00_MainContent_ctlLogin_Redir" value="wager/welcome.aspx">
<input name="ctl00$MainContent$ctlLogin$_Password" type="password" size="15" id="ctl00_MainContent_ctlLogin__Password" accesskey="p" tabindex="61" class="login_input">
</td>
</tr>
<tr>
<td>
<input type="submit" name="ctl00$MainContent$ctlLogin$BtnSubmit" value="Link" id="ctl00_MainContent_ctlLogin_BtnSubmit" class="login_input" style="text-transform: uppercase;">
</td>
</tr>
</table>
</div><!-------.linkdimesAccount ------>
</form>
<div id="result"></div>
<script>
/* attach a submit handler to the form */
$("#aspnetForm").submit(function(event) {
/* stop form from submitting normally */
event.preventDefault();
/* get some values from elements on the page: */
var $form = $(this),
term = $form.find('input[name="__EVENTTARGET"]').val(),
term1 = $form.find('input[name="__EVENTARGUMENT"]').val(),
term2 = $form.find('input[name="__VIEWSTATE"]').val(),
term3 = $form.find('input[name="ctl00$MainContent$ctlLogin$_UserName"]').val(),
term4 = $form.find('input[name="ctl00$MainContent$ctlLogin$_IdBook"]').val(),
term5 = $form.find('input[name="ctl00$MainContent$ctlLogin$Redir"]').val(),
term6 = $form.find('input[name="ctl00$MainContent$ctlLogin$_Password"]').val(),
term7 = $form.find('input[name="ctl00$MainContent$ctlLogin$BtnSubmit"]').val(),
url = $form.attr('action');
/* Send the data using post and put the results in a div */
$.post(url, { s: term, s: term1, s: term2, s: term3, s: term4, s: term5, s: term6, s: term7 },
function(data) {
var content = $(data).find('#content');
$("#result").empty().append(content);
}
);
});
</script>
感謝您的幫助!
除了僅提供jQuery和提及ASP(至少更合適的標籤)之外,您需要提供有關上下文的更多信息。還有其他影響JS的東西。唯一直接的建議是嘗試'event.stopPropagation();' –
我被CORS拒絕 –
是的,你會被SOP拒絕...... CORS可以提供幫助,或者你必須做完整的請求。或者你可以做一些事情,比如使用在頂部窗口中觸發事件的iframe。 –