如何创建一个页面来显示我从 Google 表格创建的网站的 html 源代码（仅）？

Question

我有一个使用 API 从 Google Sheets 创建的日历。 我想创建另一个页面（路由）来显示日历的 html 代码（不是呈现的模板）。 这是我的代码：

from googleapiclient.discovery import build
from google.oauth2.credentials import Credentials
from google.oauth2 import service_account

from flask import Flask
from flask import Flask, flash, jsonify, redirect, render_template, request, url_for, 
session

import requests

app = Flask(__name__)

SERVICE_ACCOUNT_FILE = 'keys.json'
SCOPES = ['https://www.googleapis.com/auth/spreadsheets']

creds = None
creds = service_account.Credentials.from_service_account_file(
     SERVICE_ACCOUNT_FILE, scopes=SCOPES)

# The ID and range of a sample spreadsheet.
SAMPLE_SPREADSHEET_ID = 'removed for security'

service = build('sheets', 'v4', credentials=creds)

# Call the Sheets API
sheet = service.spreadsheets()
result = sheet.values().get(spreadsheetId=SAMPLE_SPREADSHEET_ID,
                            range="new!A2:D9").execute()

header = sheet.values().get(spreadsheetId=SAMPLE_SPREADSHEET_ID,
                            range="new!A1:D1").execute()
                                                                    

request = sheet.values().update(spreadsheetId=SAMPLE_SPREADSHEET_ID, 
         range="new!A10", valueInputOption="USER_ENTERED", body= 
         {"values":newvalues}).execute()


@app.route('/', methods=["GET", "POST"])
def index():
        values = result.get('values', [])
        headers = header.get('values', [])

          return render_template("index.html", values=values, headers=headers)


 @app.route('/text', methods=["GET", "POST"])
 def text():
           t = requests.get('http://127.0.0.1:5000')
           return (t.content)

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

'/text' 路由显示的页面与 '/' 页面完全一样。 我希望它只返回 html 代码而不是呈现的模板。

Answer 1

尝试使用make_response将响应的 mime 类型设置为text/plain ：

from flask import app, make_response, render_template

app = Flask(__name__)

@app.route('/')
def index():
    resp = make_response(render_template('template.html'))
    resp.mimetype = 'text/plain'
    return resp