使用url_for創建鏈接到輸出文件夾中的excel文件

Question

我有一個關於url_for Unable to open files in python flask- 404 not found的問題。但被標記爲重複。

我的要求很簡單。在主頁面中創建指向輸出文件夾中文件的href鏈接。我嘗試了幾乎所有的線程，因爲答案似乎不適合我。我非常新的python。請幫忙。下面是一個示例代碼，我已經試過

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

您的應用程序的目錄結構應該是這樣的所以這段代碼會工作。如果在out文件夾中找不到文件名，那麼你會看到「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 ...