我正在嘗試在 VSCode 中撰寫一個燒瓶代碼，該代碼讀取一個 CSV 檔案并將其作為表格顯示在瀏覽器上。這是.py代碼：
from flask import Flask,render_template,request
import os
import pandas as pd
app=Flask(__name__)
app.secret_key="123"
app.config["UPLOAD_FOLDER1"]="static/csv"
@app.route("/upload",methods=['GET','POST'])
def upload():
return render_template("UploadCsv.html")
@app.route("/display",methods=["GET","POST"])
def display():
upload_file = request.files['upload_csv']
if upload_file.filename != '':
file_path = os.path.join(app.config["UPLOAD_FOLDER1"], upload_file.filename)
upload_file.save(file_path)
data=pd.read_csv(upload_file,sep=",")
return render_template("ExcelContent.html",data=data.to_html(index=False))
if __name__=='__main__':
app.run(debug=True)
我在代碼中使用了兩個 HTML 檔案，這些是： UploadCsv.html
<html>
<head>
<title>Upload CSV File</title>
</head>
<body>
<div class="col-md-offset-3 col-md-5" style="margin-top:70px">
<form method="POST" action="http://127.0.0.1:5000/display" enctype="multipart/form-data">
<h3 class="page-header text-primary">Upload CSV File</h3>
<div class="form-group">
<label>Browse CSV File</label>
<input type="file" class="form-control" name="upload_csv">
</div>
<div class="form-group">
<button type="submit" class="btn btn-success btn-block">Upload CSV</button>
</div>
</form>
</div>
</body>
</html>
CSVContent.html
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>CSV File</title>
</head>
<body>
<h2>Here's your uploaded CSV file :</h2>
{{data|safe}}
</body>
</html>
現在，當我運行 .py 并打開本地主機并上傳 csv 時，會彈出上述錯誤。當我單擊“提交”時，我的 csv 檔案必須存盤到的檔案夾作業正常。顯示部分是彈出此錯誤的地方。我不知道如何糾正。我的檔案是一個簡單的 CSV 檔案（以逗號分隔）。這是錯誤訊息的圖片：
幫我解決錯誤，在此先感謝！！
uj5u.com熱心網友回復：
您可能應該傳遞filepathtopd.read_csv方法，如下所示：
def display():
upload_file = request.files['upload_csv']
if upload_file.filename != '':
file_path = os.path.join(app.config["UPLOAD_FOLDER1"], upload_file.filename)
upload_file.save(file_path)
data=pd.read_csv(file_path,sep=",")
return render_template("ExcelContent.html",data=data.to_html(index=False))
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