Question

我想编写一个简单的Greasemonkey脚本，每次加载页面时都要填写并提交此表单。

在脚本中输入预定义的电子邮件地址（email@gmail.com），然后每次都提交

<div class="col-md-4">
    <label for="Content_C001_LI_02_txtEmail">Send proof of delivery email to:</label>
</div>
<div class="col-md-5">
    <input name="ctl00$Content$C001$LI_02_txtEmail" type="text" value="email@gmail.com" id="Content_C001_LI_02_txtEmail" class="form-control" onkeypress="return clickButton(event,'Content_C001_LI_02_btnSend')"><span id="Content_C001_ctl23" style="display:none;"></span>
</div>
<div class="col-md-3">
    <button onclick="if (typeof(Page_ClientValidate) == 'function') Page_ClientValidate('LI_02'); __doPostBack('ctl00$Content$C001$LI_02_btnSend','')" id="Content_C001_LI_02_btnSend" type="button" class="btn btn-primary btn-inline-label" validationgroup="LI_02">Send</button>
</div>

表单提交后，这段代码会发生变化。

所以我需要弄清楚如果以前还没有提交过该脚本的运行方法。

<div id="Content_C001_ctl00" class="alert alert-success" style="display:none;">

到

<div id="Content_C001_ctl00" class="alert alert-success">
Proof of delivery has been successfully emailed to email@gmail.com <br>

非常感谢您的帮助！：）

Answer 1

这样的事可能吗？

document.addEventListener ("DOMContentLoaded", checkSubmitEmail);

function checkSubmitEmail(){
    var myEmail = 'vns@gmail.com';

    // 1- check if email has been submitted
        var submitted = document.getElementById('Content_C001_ctl00').innerHTML.startsWith('\nProof of delivery has been successfully emailed to ' + myEmail);

    // 2- if not continue
        if (!submitted){
            // 3- fill in email
                document.getElementById('Content_C001_LI_02_txtEmail').value='' + myEmail;

            // 4- click button
                document.getElementById('Content_C001_LI_02_btnSend').click();
        }

}

Answer 2

很难准确地说出来，但饼干可能就是你在这里寻找的。

MDN's documentation on the JS Cookie interface

如何填写表单并在页面加载时按提交

2 个答案: