在Django中传递数据view-template-view
[英]Passing data view-template-view in django
问题描述
I'm handling input file in my project and passing processed text to another html for user approvement 
我正在处理项目中的输入文件，并将处理后的文本传递给另一个HTML以供用户批准
def vm_to_csv_upload(request):
if request.method == "POST":
some code here
br_work_string = re.sub('\\n', '<br>', work_string)
page_text = 'Here\'s what gonna be uploaded to crowdin:<br><br><br>' + br_work_string + '<br><br><br>Is that okay?'
return render(request, 'upload_crowdin_check.html', {'page_text': page_text, 'csv_text': work_string})
On this page i show passed data: 在此页面上，我显示传递的数据：
<body>
<div id="text_check">
{% autoescape off %}
{{ page_text }}
{% endautoescape %}
</div>
<form action="/crowdin_approve/" method="POST" value={{ csv_text }} id="csv">
{% csrf_token %}
<input type="submit" class="btn btn-xs btn-danger" value="Okay">
</form>
</body>
And trying to pass it to another view to process it again. 并尝试将其传递给另一个视图以再次对其进行处理。
def crowdin_approve(request):
if request.method == "POST":
return HttpResponse(request.POST)
else:
return HttpResponse('nope')
But only thing i get is csrfmiddlewaretoken. 但是我唯一得到的是csrfmiddlewaretoken。 What am i doing wrong? 我究竟做错了什么？ How should i pass {{ csv_text }} value to view? 如何传递{{csv_text}}值进行查看？ Thx. 谢谢。
1 个解决方案
解决方案1
1 已采纳 2015-07-29 14:10:38
You can pass the content of csv_text using an hidden input field: 您可以使用隐藏的输入字段传递csv_text的内容：
<form action="/crowdin_approve/" method="POST" id="csv">
{% csrf_token %}
<input type="hidden" name="csv_text" value="{{ csv_text }}">
<input type="submit" class="btn btn-xs btn-danger" value="Okay">
</form>
Then you can access it in the view this way: 然后，您可以通过以下方式在视图中访问它：
def crowdin_approve(request):
if request.method == "POST":
csv_text = request.POST.get('csv_text', None)
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