Flask/Python：保存前修改上传的文件数据

2024-06-17 14:57:25 发布

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我想能够上传一个csv文件，然后用python脚本对文件做一些修改，最后在更改后将文件保存到特定文件夹。我有这样的东西，但我不知道为什么它不起作用：

import os
from flask import Flask, render_template, request, redirect, url_for, send_from_directory
from werkzeug.utils import secure_filename

app = Flask(__name__)
UPLOAD_FOLDER = 'C:/Users/tkp/Desktop/uploads_files'
app.config['UPLOAD_EXTENSIONS'] = ['.csv']
app.config['UPLOAD_PATH'] = UPLOAD_FOLDER

@app.route('/')
def index():
    files = os.listdir(app.config['UPLOAD_PATH'])
    return render_template('index.html', files=files)

@app.route('/', methods=['POST'])
def upload_files():
    uploaded_file = request.files['file']
    
    filename = secure_filename(uploaded_file.filename)
    
    if filename != '':
        
        uploaded_file.stream.seek(0)
        f = uploaded_file.read()
        
            #some change in the file
        
        f.save(os.path.join(app.config['UPLOAD_PATH'], filename))
        
    return redirect(url_for('index'))

@app.route('/Users/tkp/Desktop/uploads_files/<filename>')
def upload(filename):
    return send_from_directory(app.config['UPLOAD_PATH'], filename)

和HTML文件：

<!doctype html>
<html>
  <head>
    <title>File Upload</title>
  </head>
  <body>
    <h1>File Upload</h1>
    <form method="POST" action="" enctype="multipart/form-data">
      <p><input type="file" name="file"></p>
      <p><input type="submit" value="Convert"></p>
    </form>
    <hr>
  </body>
</html>

是否可以动态执行此类操作，或者您必须先保存上载的文件


Tags： 文件pathfromimportconfigapposdef
2条回答
网友
1楼 · 编辑于 2024-06-17 14:57:25

除了f.save(...)定义之外，您所做的一切都是正确的

执行f = uploaded_file.read()时，f是.read()操作的结果，它是字节，而不是文件。 您必须打开另一个文件并将内容保存到其中。 不要忘记.decode()字节，使其成为字符串

下面是一个工作片段：


@app.route('/', methods=['POST'])
def upload_files():

    uploaded_file = request.files['file']

    filename = secure_filename(uploaded_file.filename)

    if filename != '':

        uploaded_file.stream.seek(0)
        f = uploaded_file.read().decode()

        # WE don't want any failures
        f = f.replace("FAIL", "SUCCESS")

        filename_to_save = os.path.join(app.config['UPLOAD_PATH'], filename)
        with open(filename_to_save, "w") as file_to_save:
            file_to_save.write(f)

    return {"status": "OK"}
网友
2楼 · 编辑于 2024-06-17 14:57:25

这可能是动作属性中的输入错误。更改此行：

<form method="POST" action="" enctype="multipart/form-data">

为此：

<form method="POST" action="/" enctype="multipart/form-data">

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