Question

创建了一个Web应用程序，并成功构建了一个功能来清除Google我的商家导出的CSV文件。但是，当我使用编写的代码运行该函数时，会收到以下错误消息：

未找到

在服务器上找不到请求的URL。如果输入URL   手动，请检查您的拼写，然后重试。

不确定我要去哪里哪里

mport os
import pandas as pd
from flask import Flask, request, redirect, url_for
from flask import Flask, make_response
from werkzeug.utils import secure_filename

UPLOAD_FOLDER = './Downloads/gmbreports'
if not os.path.exists(UPLOAD_FOLDER):
    os.makedirs(UPLOAD_FOLDER)

ALLOWED_EXTENSIONS = 'csv'

app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
def allowed_file(filename):
    return '.' in filename and \
           filename.rsplit('.', 1)[1].lower() in ALLOWED_EXTENSIONS

def transform(text_file_contents):
    disc = open('clean.csv')
    disc2 = open('clean_two.csv','w')
    #cleaning up csv
    for row in disc:
        row = row.strip()
        row = row[1:-1]
        row = row.replace('""','"')
        disc2.write(row+'\n')
    disc2.close()
    disc.close()
    discovery = pd.read_csv('clean_two.csv')
    discovery_clean = discovery.iloc[1:]
    cols = list(discovery_clean.columns[4:])
    discovery_clean[cols] = discovery_clean[cols].apply(pd.to_numeric,errors='coerce')
    return discovery_clean

@app.route('/', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        if 'file' not in request.files:
            flash('No file part')
            return redirect(request.url)
        file = request.files['file']
        if file.filename == '':
            flash('You need to upload a csv file')
            return redirect(request.url)
        if file and allowed_file(file.filename):
            filename = secure_filename(file.filename)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
            return redirect(url_for('uploaded_file',
                                    filename=filename))
    return '''
    <!doctype html>
    <title>Google My Business Discovery Report Builder</title>
    <h1>Upload GMB Discovery csv</h1>
    <form action="\transform" method="post" enctype="multipart/form-data">
      <p><input type="file" name="file">
         <input type="submit" value=Upload>
    </form>
    '''
@app.route('/transform',methods=["POST"])
def transform_view():
    request_file=request.files['file']
    request_file.save('clean.csv')
    if not request_file:
        return "No file"
    result = transform()
    print(result)

    response = make_response(result)
    response.headers["Content-Disposition"] ="attachment; filename=result.csv"
    return response


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

注意：运行脚本并上传csv后，我得到此错误消息。理想的结果是上传一个csv，并将其作为清理后的数据表显示在我的屏幕上

Answer 1

这里有多个问题。首先，Wondercricket关于表单操作是正确的。需要将其更改为“ / transform”。

完成后，您仍然会收到内部服务器错误。在这种情况下，您的transform函数被定义为采用单个参数，但没有被一个参数调用。只需更改

def transform(text_file_contents):

到

def transform():

这应该允许进行上传，并且转换现在将运行。但是，还有最后一个问题。 transform返回一个熊猫DataFrame，Flask不能将其用作响应。更改

response = make_response(result)

到

response = make_response(result.to_csv())

你应该很好。

运行Flask代码时出现未找到请求的URL错误

1 个答案: