我正在尝试从谷歌表格中提取单元格图像,以便稍后粘贴到谷歌文档中。使用此方法时我收到错误,并且找不到任何有关此的文档。
function onClick(){
var app = SpreadsheetApp;
var sheet = app.getActive().getSheetByName('FILE');
var image = sheet.getRange('D4').getValue();
var url = image.getContentUrl()
Logger.log(url)
}
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异常:获取对象 SpreadsheetApp.CellImage 上的方法或属性 getContentUrl 时出现意外错误
我也尝试过使用DocsServiceApp ->使用此方法
DocsServiceApp.openBySpreadsheetId(spreadsheetId)
.getSheetByName("Sheet1")
.getImages();
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但这导致了一个不同的错误:
Exception: Request failed for https://www.googleapis.com returned code 403. Truncated server response: {
"error": {
"errors": [
{
"domain": "global",
"reason": "exportSizeLimitExceeded",
"message": "This file is too large to be expo... (use muteHttpExceptions option to examine full response)
at gToM(SpreadsheetAppp:171:25)
at getImages(SpreadsheetAppp:34:14)
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有谁知道如何修复这些错误之一,或者从谷歌表格中提取图像的不同方法?
javascript image google-sheets http-status-code-403 google-apps-script
我正在向某个 URL 发送请求。我将curl命令复制到python中。因此,所有标头都包含在内,但我的请求不起作用,并且我在 HTML 输出中收到状态代码 403 和错误代码 1020。
代码是
import requests
headers = {
'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:106.0) Gecko/20100101 Firefox/106.0',
'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,*/*;q=0.8',
'Accept-Language': 'en-US,en;q=0.5',
# 'Accept-Encoding': 'gzip, deflate, br',
'DNT': '1',
'Connection': 'keep-alive',
'Upgrade-Insecure-Requests': '1',
'Sec-Fetch-Dest': 'document',
'Sec-Fetch-Mode': 'navigate',
'Sec-Fetch-Site': 'none',
'Sec-Fetch-User': '?1',
}
response = requests.get('https://v2.gcchmc.org/book-appointment/', headers=headers)
print(response.status_code)
print(response.cookies.get_dict())
with open("test.html",'w') as f:
f.write(response.text)
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我也收到了 cookie,但没有得到所需的响应。我知道我可以用硒做到这一点,但我想知道这背后的原因。
注意:
我已经安装了所有库并检查了版本,但它仍然无法工作并抛出 403 错误。
我正在使用wikipedia-api插件从维基百科获取内容的页面.从某些天开始,当我尝试获取内容时,我收到此错误:
page = Wikipedia.find_by_titles('Foo')
OpenURI::HTTPError: 403 Forbidden
from /usr/lib/ruby/1.8/open-uri.rb:277:in `open_http'
from /usr/lib/ruby/1.8/open-uri.rb:616:in `buffer_open'
from /usr/lib/ruby/1.8/open-uri.rb:164:in `open_loop'
from /usr/lib/ruby/1.8/open-uri.rb:162:in `catch'
from /usr/lib/ruby/1.8/open-uri.rb:162:in `open_loop'
from /usr/lib/ruby/1.8/open-uri.rb:132:in `open_uri'
from /usr/lib/ruby/1.8/open-uri.rb:518:in `open'
from /usr/lib/ruby/1.8/open-uri.rb:30:in `open'
from /home/luca/Programming/nerto_5/vendor/plugins/wikipedia-api/lib/mediawiki.rb:139:in `get_xml'
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那是什么?
我能怎么做?
谢谢
奇怪的是,我正在尝试阅读许多不同网站的<Head>部分,而一种特定类型的服务器Apache有时会禁止代码403.并非所有apache服务器都这样做,因此它可能是配置设置或服务器的特定版本.
当我使用网络浏览器(例如Firefox)检查网址时,页面加载正常.代码sorta看起来像这样:
var client = new WebClient();
var stream = client.OpenRead(new Uri("http://en.wikipedia.org/wiki/Barack_Obama"));
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通常,403是一种访问权限失败的东西,但这些通常是不安全的页面.我认为Apache正在过滤请求标头中的某些东西,因为我不打算创建任何东西.
也许对Apache有更多了解的人可以给我一些关于标题中缺少什么的想法.我想保持标头尽可能小,以尽量减少带宽.
谢谢
我有一个脚本,支付处理器附带付款确认.为了使页面安全,因为它可以访问订单信息和其他用户相关的东西,我不得不通过ip(/ 24)来限制访问,如下所示:
$ipAllowed = array(
'192.192.192',
'172.172.172'
);
$ipAllowed = str_replace(".", "\.", implode("|", $ipAllowed));
if(!preg_match("/^($ipAllowed)\.[0-9]{1,3}$/", $_SERVER['REMOTE_ADDR'])){
header('HTTP/1.0 403 Forbidden');
die('You are not allowed to access this file.');
}
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*ip就是一个例子
在我用之前:
if(!in_array(@$_SERVER['REMOTE_ADDR'], array('ips here'))); //only works with full ip
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!in_array比我现在使用的那个更整洁,但我需要一些适用于/ 24 ips的东西,甚至两者都适用!
你知道更好/更快,可靠和更整洁的东西吗?
@ rap-2-h如你所说,这是适用于全ip,/ 24甚至/ 16的整洁版本
$ipAllowed = array( '192.168.1.153' '172.172.172');
$allowed = false;
foreach($ipAllowed as $ip):
if(strpos($_SERVER['REMOTE_ADDR'], $ip) === 0) $allowed = true;
endforeach;
if (!$allowed) {
header('HTTP/1.0 403 Forbidden');
die('You are not allowed to access this file.');
}
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$ curl -A "Mozilla/5.0 (X11; Linux x86_64; rv:18.0) Gecko/20100101 Firefox/18.0 (compatible;)" http://fifa-infinity.com/robots.txt
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并打印robots.txt.省略user-agent选项会导致服务器出现403错误.检查robots.txt文件表明允许抓取http://www.fifa-infinity.com/board下的内容.但是,以下失败(python代码):
import logging
import mechanize
from mechanize import Browser
ua = 'Mozilla/5.0 (X11; Linux x86_64; rv:18.0) Gecko/20100101 Firefox/18.0 (compatible;)'
br = Browser()
br.addheaders = [('User-Agent', ua)]
br.set_debug_http(True)
br.set_debug_responses(True)
logging.getLogger('mechanize').setLevel(logging.DEBUG)
br.open('http://www.fifa-infinity.com/robots.txt')
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我的控制台上的输出是:
No handlers could be found for logger "mechanize.cookies"
send: 'GET /robots.txt HTTP/1.1\r\nAccept-Encoding: identity\r\nHost: www.fifa-infinity.com\r\nConnection: close\r\nUser-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:18.0) Gecko/20100101 Firefox/18.0 (compatible;)\r\n\r\n'
reply: 'HTTP/1.1 403 Bad Behavior\r\n'
header: Date: Wed, 13 Feb 2013 …Run Code Online (Sandbox Code Playgroud) 我是数据库工作的新手,我想知道是否有人可以帮助我理解为什么当我只是在本地开发时我一直得到403 Forbidden错误.我正在使用codeigniter尝试创建一个简单的登录程序.前几天我让程序在不同的计算机上运行,但是当我尝试在浏览器中打开"视图"或"控制器"文件时,现在我在这台计算机上的所有内容都是403错误.它必须在某处设置,但我根本不知道在哪里看.有任何想法吗?
这是我的database.php文件:
$active_group = 'default';
$active_record = TRUE;
$db['default']['hostname'] = 'localhost';
$db['default']['username'] = 'tester';
$db['default']['password'] = 'tester';
$db['default']['database'] = 'intranet';
$db['default']['dbdriver'] = 'mysql';
$db['default']['dbprefix'] = '';
$db['default']['pconnect'] = TRUE;
$db['default']['db_debug'] = TRUE;
$db['default']['cache_on'] = FALSE;
$db['default']['cachedir'] = '';
$db['default']['char_set'] = 'utf8';
$db['default']['dbcollat'] = 'utf8_general_ci';
$db['default']['swap_pre'] = '';
$db['default']['autoinit'] = TRUE;
$db['default']['stricton'] = FALSE;
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这是控制器文件welcome.php:
<?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');
class Welcome extends CI_Controller {
/**
* Index Page for this controller.
*
* Maps …Run Code Online (Sandbox Code Playgroud) 我想从网址打开一个链接:" http://www.kohls.com/search.jsp?search=jacket&submit-search=web-regular ",有时我得到:
java.io.IOException:服务器返回HTTP响应代码:403表示URL.但是使用浏览器打开网址时没关系.以下是我的代码的一部分:
URL url = new URL("http://www.kohls.com/search.jsp?search=jacket&submit-search=web-regular");
InputStream is = url.openConnection().getInputStream();
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错误细节
线程"main"中的异常java.io.IOException:服务器返回HTTP响应代码:403为URL:http://www.kohls.com/search.jsp? N = 0&search = jacket&WS = 96 at sun.net.www. Links.main上的protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1627)(Links.java:41)
好吧,我很难找到问题,因为它在本地工作,但在发布后,结果很简单:
错误代码:403禁止.服务器拒绝指定的统一资源定位符(URL).联系服务器管理员.(12202)
代码:
[RoutePrefix("api/v1/project")]
public class ProjectController : BaseApiController
{
[HttpGet]
public HttpResponseMessage GetProjects()
{
HttpResponseMessage resp = new HttpResponseMessage(HttpStatusCode.OK);
if(User.Identity.IsAuthenticated)
{
var model = new ModelFactory().CreateProjects();
resp = Request.CreateResponse(HttpStatusCode.OK, model);
}
return resp;
}
}
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public static class WebApiConfig
{
public static void Register(HttpConfiguration config)
{
// all actions under /project routes require authentication
config.Routes.MapHttpRoute(
name: "ProjectApi",
routeTemplate: "api/v1/{controller}/{action}/{apikey}",
defaults: new { apikey = RouteParameter.Optional },
constraints: new { controller = "project" },
handler: new BasicAuthHandler(config));
// all routes …Run Code Online (Sandbox Code Playgroud) 你可以帮我解决下面的问题.
我只是编写一个简单的代码来检索我的谷歌+域用户配置文件.1.我正在使用域名为spaceandhow.com的谷歌+域帐户
我已提供https://developers.google.com/identity/protocols/OAuth2ServiceAccount#delegatingauthority中列出的所有权限
该帐户已启用Google+ API和Google+域Api.
这是我写的功能.
func (p *GoogleAUTH) sendGoogleAuthReq(){
fmt.Println("GoogleAUTH package: Enter sendGoogleAuthReq")
data, err := ioutil.ReadFile("D:\\Cygwin\\home\\praprasa\\pragna2.json")
if err != nil {
fmt.Printf("ReadFile error: %s", err)
}
conf, err := google.JWTConfigFromJSON(data, "https://www.googleapis.com/auth/plus.me")
if err != nil {
fmt.Println("json error")
}
client := conf.Client(oauth2.NoContext)
resp, err := client.Get("https://www.googleapis.com/plusDomains/v1/people/me")
if err != nil {
fmt.Printf("GoogleAUTH package: request execution failed: %s", err)
return
}
defer resp.Body.Close()
fmt.Println("GoogleAUTH package: response Status:", resp.Status)
fmt.Println("GoogleAUTH package: response Headers:", resp.Header)
body, err := ioutil.ReadAll(resp.Body) …Run Code Online (Sandbox Code Playgroud) download go user-profile http-status-code-403 google-domain-api
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python ×2
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codeigniter ×1
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image ×1
java ×1
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