UrlFetchApp.fetch(url)间歇性“意外错误”

时间:2013-05-27 18:53:54

标签: google-apps-script

为了检查HTTP服务状态,我编写了通过电子表格编写的脚本,并且对于URL列表检查它们是Up还是Down,脚本是每五分钟驱动一次。

我有来自UrlFetchApp.fetch(url)的罕见间歇性“意外错误”错误 如果我在几秒钟后重复请求,会不时出现DNS和超时错误。

如果有人可以提供帮助,那么实际问题就不多了  我使用Utilities.sleep(5000)暂停5秒,  是的还是有更好的等待方式  尝试在几秒钟后再次获取?

为什么即使我在5秒后重复请求,当脚本是,我也会收到“意外错误”  五分钟后再次运行没有“意外错误”!?

我如何改进下面的代码?

实际脚本:

/* 
Periodically check status of web sites :-)
Google Apps for Busines UrlFetch daily limit is 100.000 requests,

Algorithm
  read site and old status from sheet
  check site and set new status  
  if status changed send email (+sms in future by using twilio)
  update status in spreadsheet

"Site, Status code, Time of last change, Last error description"
*/
function main() {
  var sheet = SpreadsheetApp.getActiveSheet() ;
  var currentRow, oldStatusCode, newStatusCode ;
  var url, response, err, subject, message ; 
  var today = new Date() ;

  currentRow = 2
  while ((url = sheet.getRange(currentRow, 1).getValue()) != "") {
    oldStatusCode = sheet.getRange(currentRow, 2).getValue() ;
    newStatusCode = "Ok"
    subject = "mCheck: " + url + " Up Status Change!" ;  
    message = url + " Up Status Change!" + "\n Time: " + today.toUTCString() ;

    var tries = 3 ; // Check at least three times that status changed and it is not a one time glitch
    do {
      try {
        response = UrlFetchApp.fetch(url) ;
      } catch (err) {
      newStatusCode = "Down"
      sheet.getRange(currentRow, 4).setValue(err.message) ;
      subject = "mCheck: " + url + " Down Status Change!" ;  
      message = url + " Down Status Change!" + "\n Error message: " + err.message + "\n Time: " + today.toUTCString() ;
      if (err.message.indexOf("Unexpected") > -1) { // If UrlFetch failed on Google side just ignore this iteration...
        newStatusCode = oldStatusCode ;
      }
    }
    if (oldStatusCode != newStatusCode) { // In case of status change wait 5 seconds before trying again
      Utilities.sleep(5000) ;
    }
    --tries ;
  } while ((oldStatusCode != newStatusCode) && tries >= 0)

  if (oldStatusCode != newStatusCode) {
    sheet.getRange(currentRow, 2).setValue(newStatusCode) ;
    sheet.getRange(currentRow, 3).setValue(today.toUTCString()) ;
    if (oldStatusCode != "") {
      MailApp.sendEmail(email_to, subject, message) ;
    }
  }
  ++currentRow;
 } 

}

1 个答案:

答案 0 :(得分:1)

您可以使用muteHttpExceptions parameter来捕获失败。如果响应代码指示失败,则fetch不会抛出异常,而是返回HTTPResponse。

<强>示例

var params = {muteHttpExceptions:true};
response = UrlFetchApp.fetch(url, params);

你可以use a cURL或者为一个社区进行分类。

<强>示例

卷曲。将此设置在您的操作系统的上层。

curl -L URL_OF_YOUR_SCRIPT

添加到脚本

function doGet(e) {
    var result = {'status': 'ok'};
    try{
        main();
    }catch(err){
        result.['status'] = err;
    }
  return ContentService.createTextOutput(JSON.stringify(result)).setMimeType(ContentService.MimeType.JSON);
 }

您应deploy your script as a Web App为此。

如果您要使用第二个样本,则不需要使用Utilities.sleep(5000)。

最佳。