这是我第一次涉足Google脚本,我有一个表单可以调用两个不同的Google应用脚本(两者都在.gs文件中)。 One上传文件,而另一个文件将表单数据保存到Google电子表格。出于某种原因,我在调用文件上传脚本时遇到错误
(未捕获的TypeError:无法读取属性' closure_lm_407634'为null)
虽然上传数据的脚本运行正常。 将表单数据保存到电子表格(有效):
google.script.run.withUserObject(data).saveToSheet(data);
- 调用:
function saveToSheet(data) {
var date = new Date();
var sheet = SpreadsheetApp.openById(submissionSSKey);
sheet
.appendRow([date, data.name, data.email, data.headline,
data.location, data.newsContent, data.websiteContent, data.expiry, data.fileUrl]);
}
上传文件(不工作):
google.script.run
.withUserObject(theForm)
.withSuccessHandler(processForm)
.uploadFile(theForm);
- 调用:
function uploadFile(form) {
var folder = DriveApp.getFolderById(folderId), doc = '', file = form.uploadedFile;
if (file.getName()) { doc = folder.createFile(file); }
return doc;
}
我不能为我的生活弄清楚为什么一个电话有效而另一个电话无效。我已经尝试过各种方式来调用上传脚本,但没有任何作用。我已经尝试删除用户对象和成功处理程序。
HTML:
<?!= include('styles'); ?>
<div id="container" class="col-lg-12 col-md-12 col-sm-12">
<header class="col-lg-offset-3 col-md-offset-3"></header>
<div class="col-lg-offset-3 col-lg-6 col-md-6 col-sm-12" id="form">
<h1 class="text-center">
SAS News Submission
</h1>
<span id="required-content">
<sup>*</sup>
Required
</span>
<br>
<br>
<form name="sas-form">
<label for="name" class="required">Contact Person/ Source of News</label>
<input type="text" name="name" value="test" class="form-control" id="name" required="required">
<br>
<label for="email" class="required">Contact Person's email (in case we have questions regarding your News content)</label>
<input type="email" name="email" value="me@me.com" id="email" class="form-control" required="required">
<br>
<label for="headline" class="required">Headline (try to keep below 10 words if possible) </label>
<input type="text" name="headline" value="headline" id="headline" class="form-control" required="required">
<br>
<label for="newsContent" class="required">News Content *Note all content must be fully edited to ensure posting</label>
<textarea rows="5" cols="0" name="newsContent" class="form-control" id="newsContent" required="required">
Content
</textarea>
<br>
<label class="required">Where would you like the News to be shared? (You may choose more than one)</label>
<ul id="social-list">
<li>
<input type="checkbox" name="newsletter" id="newsletter" value="newsletter">
<label for="newsletter"> Newsletter</label>
</li>
<li>
<input type="checkbox" name="social" id="social" value="social">
<label for="social"> Social Media (Facebook, LinkedIn, Twitter)</label>
</li>
<li>
<input type="checkbox" name="website" id="website" value="website" checked>
<label for="website"> Website </label>
</li>
</ul>
<br>
<label for="websiteContent">If you chose the website, please provide specific instructions on where you would like the content to be posted.</label>
<br>
<small>News and Events Page, Volunteer Page, Student Page, etc. Ex: Please post in the News and Events Page and send the link and headline out on social media.</small>
<textarea rows="5" cols="0" name="websiteContent" id="websiteContent" class="form-control">website content</textarea>
<br>
<label for="expiry">If your content has an expiration date, please share that date below.</label>
<input type="date" name="expiry" id="expiry" class="form-control">
<br>
<label>If you have files that need to be attached, pick them below.</label>
<input type="file" name="uploadedFile" id="file">
<br>
<div id="not-valid"><span></span></div>
<div id="error"><span>
An error occurred, please try submitting again.
</span></div>
<div id="success"><span>
Form submission was successful! Thank you!
</span></div>
<input type="button" value="Submit" class="btn btn-primary" id="submit"
onclick="validateForm(this.parentNode)">
</form>
</div>
</div>
<footer>
<?!= include('javascript'); ?>
</footer>
<script>
var validateForm = function(theForm)
{
var valid = true;
$('#not-valid span').empty();
$('input').removeClass('warning');
if($('#name').val() == '')
{
$('#name').addClass('warning');
$('#not-valid span').append('Please enter a name <br>');
valid = false;
}
if($('#email').val() == '')
{
$('#email').addClass('warning');
$('#not-valid span').append('Please enter an email <br>');
valid = false;
}
if($('#headline').val() == '')
{
$('#headline').addClass('warning');
$('#not-valid span').append('Please enter a headline <br>');
valid = false;
}
if($('#newsContent').val() == '')
{
$('#newsContent').addClass('warning');
$('#not-valid span').append('Please enter news content <br>');
valid = false;
}
if(!$('#social').is(':checked') && !$('#website').is(':checked') && !$('#newsletter').is(':checked'))
{
$('#not-valid span').append('Please choose where you would like the news to be shared. <br>');
$('#social-list').addClass('warning');
valid = false;
}
if(valid)
{
google.script.run.withSuccessHandler(processForm).uploadFile(theForm)
}
};
function processForm(file)
{
var fileUrl = file ? file.getUrl() : 'No file uploaded',
location = [];
if($('#social').is(':checked'))
{
location.push('social');
}
if($('#newsletter').is(':checked'))
{
location.push('newletter');
}
if($('#website').is(':checked'))
{
location.push('website');
}
var data = {
name: $('#name').val(),
email: $('#email').val(),
headline: $('#headline').val(),
location: location.toString(),
newsContent: $('#newsContent').val(),
websiteContent: $('#websiteContent').val(),
expiry: $('#expiry').val() ? $('#expiry').val() : 'No expiration date selected',
fileUrl: fileUrl
};
google.script.run.saveToSheet(data);
clearForm();
success();
};
var clearForm = function()
{
$("input[type=text], input[type=date], textarea, input[type=email], input[type=file]").val("");
$("input[type=checkbox]").attr('checked', false);
enableSubmit();
};
var success = function()
{
$('#success').show()
};
var enableSubmit = function()
{
$("#submit").prop('disabled', false);
};
</script>
答案 0 :(得分:1)
我能够重现你的错误。我不知道为什么会出现这种错误,但我找到了一种方法让它发挥作用。
以下是您需要做的事情:
将id属性放入上部表单标记:
<form id="myForm">
在表单中添加<{1}}标记 。必须在表格之外。并使用<button>
document.getElementById('myForm')
我已经测试了这个。它获取了文件,并将其发送到表单元素内的服务器。
您可以在服务器代码中使用<form id="myForm">
<input type="file" name="uploadedFile">
</form>
<button onclick="validateForm(document.getElementById('myForm'))">submit</button>
而无需使用调试器。
Logger.log()