Question

我有一个关于url_for Unable to open files in python flask- 404 not found的问题。但被标记为重复。

我的要求非常简单。在主页面中创建指向输出文件夹中的文件的href链接。我在SO中尝试了几乎所有的线程，答案似乎对我不起作用。我是python的新手。请帮忙。以下是我试过的示例代码

    from flask import Flask, redirect, url_for,send_from_directory
    app = Flask(__name__)    

    @app.route('/main')
    def index():
        print 'test'
        return '''<a href="{{ url_for('uploaded_file', filename='a.txt') }}">Open file</a>'''

    @app.route('/out/<filename>')
    def uploaded_file(filename):
        print filename
        return send_from_directory('./out/',
                                   filename)
   @app.route('/out/<filename>')
   def uploaded_file2(filename):
      print filename
      return './out/'+filename

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

Answer 1

你的app目录结构应该是这样的，所以这段代码可行。如果在out文件夹中找不到文件名，那么您将看到“url not found”：

app/
  app.py
  static/
        out/
           a.txt
  templates/
    index.html

from flask import Flask, render_template, redirect, url_for,send_from_directory

app = Flask(__name__)

app.config.update(
        UPLOAD_FOLDER = "static/out/"
)

@app.route('/main')
def index():
    print ('test')
    return render_template('index.html')

@app.route('/out/<filename>')
def uploaded_file(filename):
    print (filename)
    return send_from_directory(app.config["UPLOAD_FOLDER"], filename)

@app.route('/showfilepath/<filename>')
def uploaded_file2(filename):
    print (filename)
    return app.config["UPLOAD_FOLDER"] + filename

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

# index.html
<a href="{{url_for('uploaded_file', filename='a.txt')}}">open file</a>

# a.txt
hey there ...

使用url_for创建输出文件夹中excel文件的链接

1 个答案: