我可以链接到不是烧瓶中模板文件夹的html文件吗

Question

*我正在尝试制作烧瓶，动态生成指向每个“1 .index.html”( *存在于 中的“项目文件夹”的链接，它本身就在“项目”文件夹 )

*并且链接必须是项目文件夹的名称，并指向其中的“1 .index.html”

*我已经做了一切，一切正常，唯一的问题是烧瓶不允许链接到模板文件夹中不存在的 html 文件，所以当我单击链接时出现错误

jinja2.exceptions.TemplateNotFound: projects\0 。市场\1 . index.html

这是文件夹和文件的结构：

/flask-app
   app.py
   /templates
      0 . base.html
      1 . index.html
      2 . home.html
      3 . about.html
      4 . contact.html
      6 . projects.html
      /Projects
         /0 . market
            1 . index.html
            ..other files...
         /1 . chat
            1 . index.html
            ..other files...
         /2 . stocks
            1 . index.html
            ..other files...
 ..other files...

这是 app.py 文件

    from flask import Flask, request, render_template
import os


class fodler_prop:
    def __init__(self, file_name):
        self.real_name = file_name          #exemple ("0 . Flask")
        self.folder_number = file_name[0]          #exemple ("0")
        self.url = file_name[4 :]      #exemple ("Flask")


class file_prop:
    def __init__(self, file_name):
        self.real_name = file_name          #exemple ("0 . index.html")
        self.folder_number = file_name[0]          #exemple ("0")
        self.url = file_name[4 : -5]      #exemple ("index")


class project_prop:
    def __init__(self, real_name, url, folder_number, x, v):
        self.real_name = real_name
        self.url = url
        self.folder_number = folder_number 
        self.position = x       # position in table
        self.validation = v     # True if there is an 'index.html" file for that project


app = Flask(__name__)


@app.route("/")
def index():
    return render_template("1 . index.html")


@app.route("/home")
def home():
    return render_template("2 . home.html")


@app.route("/about")
def about():
    return render_template("3 . about.html")


@app.route("/contact")
def contact():
    return render_template("4 . contact.html")


@app.route("/projects")
def projects():
    projects_folder_path = "E:\\Mbarki\\My Projects\\Web development\\0 . Flask-Projects-Manager\\templates\\projects"
    projects_names = os.listdir(projects_folder_path)
    projects_names_list = []        # all projects folders names are here
    for project_name in projects_names:
        project_name = fodler_prop(project_name)
        projects_names_list.append(project_name)

    valid_projects_list = []        # all valid projects names are here (projects that contains and index.html file)
    all_projects_final_form = []     # all projects in their final form
    c = -1                          # this will count the number of projects folders
    x = -1                          # this save the position of the valid projects in the table "valid_projects_list"
    for project_name in projects_names_list:
        c += 1
        x += 1
        project_path = f"E:\\Mbarki\\My Projects\\Web development\\0 . Flask-Projects-Manager\\templates\\projects\\{project_name.real_name}"
        project_files_names = os.listdir(project_path)
        for project_file_name in project_files_names:
            project_file_name = file_prop(project_file_name)
            if f"{project_file_name.url}.html" == "index.html":
                project_file_name = project_prop(project_name.real_name, project_name.url, project_name.folder_number, x, True)
                valid_projects_list.append(project_file_name)
                project_file_name = project_prop(project_name.real_name, project_name.url, project_name.folder_number, c, True)
                all_projects_final_form.append(project_file_name)
                break
            else:
                project_file_name = project_prop(project_name.real_name, project_name.url, project_name.folder_number, c, False)
                all_projects_final_form.append(project_file_name)
                break
    
    return render_template("5 . projects.html", all_projects_final_form = all_projects_final_form)


@app.route("/projects/<link>")                                     # THE PROBLEM IS HERE
def routes(link):                                                  # THE PROBLEM IS HERE
    return render_template(f'projects\\{link}\\1 . index.html ')   # THE PROBLEM IS HERE



if __name__ == "__main__":
    app.run(debug=True)

这是“6 .projects.html”文件：

[{% extends '0 . base.html' %}


{% block title %}
Projects
{% endblock %}


{% block content %}
<div class="content">
    <h1 class="tha">Projects Table</h1>
    <table class="table1">
        <thead>
            <tr>
                <!-- Your Columns HERE -->
                <th scope="col">Projects Names</th>
                <th scope="col">Discription</th>
            </tr>
        </thead>
        <tbody>
            <!-- Your rows inside the table HERE: -->
            {% for project in all_projects_final_form %}
            {% if project.validation == True %}
            <tr>
                <td><a href="/projects/{{project.real_name}}">{{ project.real_name }}</a></td>
                <td>None</td>
            </tr>
            {% else %}
            <tr></tr>
            <td>
                <p class="not_active">{{ project.real_name }}</p>
            </td>
            <td>
                <p class="not_active">(- we couldn't find the "index.html" file for this project -)</p>
            </td>
            </tr>
            {% endif %}
            {% endfor %}


        </tbody>
    </table>
</div>
{% endblock %}

这是“6 .projects.html”文件的图片：

https://drive.google.com/file/d/1oXQtvXPJTRt8ebqmP6Ktvq4_tiK4Y_Uh/view?usp=sharing

当我点击其中一个链接时，我得到的错误图片是：

jinja2.exceptions.TemplateNotFound: projects\0 。市场\1 . index.html

https://drive.google.com/file/d/18NcEJPF-Dy9i4TFHs44OQVne3JyUucdO/view?usp=sharing

提醒：这段代码一切正常，运行良好，唯一的问题是我无法链接不在模板目录中的 html 文件 p>

Answer 1

我在自己的项目中使用子目录做了类似的事情，对我有用的内容类似于以下内容：

return render_template('projects/' + link + '/1 . index.html')

如果有帮助请告诉我！