小编Lou*_*ero的帖子

TemplateSyntaxError:期望令牌':',得到'}'

这是我的源代码:

<ul class="filebase">
        {% for file in finance %}
        <li class="filelist">
            <div class="file-author"><a href="http://127.0.0.1:5000/uploads/files/{{file. filename}}">{{ file.filename }}</a></div>
            <div class="file-body">Description: {{ file.description }}</div>
            <div class="file-date">Date posted: {{ moment(file.date).fromNow() }}</div>
            <div class="delete-file"><a href="{{ url_for('.delete_file', file_name={{ file.filename }} ) }}">Delete File</a></div><br>
        </li>
        {% endfor %}
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起初,我的代码工作正常,我突然得到这样的错误:

TemplateSyntaxError:期望令牌':',得到'}'

这是我的追溯(如果你需要):

Traceback (most recent call last):
  File "C:\Users\LouieCubero\Documents\GitHub\flasky\venv\lib\site-packages\flask\app.py", line 1836, in __call__
    return self.wsgi_app(environ, start_response)
  File "C:\Users\LouieCubero\Documents\GitHub\flasky\venv\lib\site-packages\flask\app.py", line 1820, in wsgi_app
    response = self.make_response(self.handle_exception(e))
  File "C:\Users\LouieCubero\Documents\GitHub\flasky\venv\lib\site-packages\flask\app.py", line 1403, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "C:\Users\LouieCubero\Documents\GitHub\flasky\venv\lib\site-packages\flask\app.py", line …
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python jinja2 flask

8
推荐指数
2
解决办法
2万
查看次数

如何在url中传递带有*斜杠*的参数?

我想要做的是在函数的参数中传递日期,然后处理输入。这是它的功能

@HR.route('/confirm_sickLeave/<date>/<user>', methods=['GET', 'POST'])
def confirm_sickLeave(user,date):
    u = User.query.filter_by(username=user.username).first()
    print u
    us = UserStatistics.filter_by(userId=u.id).first()
    temp = us.slDates
    dates = temp.keys()
    for a in dates:
        if a == date:
            temp['date'] = True
            flash('Date Confirmed.')
            return redirect(url_for('.approval_of_leaves'))


    return redirect(url_for('.approval_of_leaves'))
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现在,我的问题是我无法在我的函数中传递值。原因是我的输入中dates有斜杠 ( /)。让我演示给你看:

HTML:

{% for u in all_users %}
## Returns all the dates applied by the user (it's in dict format)
{% set user_info = u.return_sickLeaves(u.username) %}  
{% for us in user_info %}
<tr>

        <td>     {{ u.username }} …
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python flask

3
推荐指数
1
解决办法
2475
查看次数

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