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;
}
}
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>示例强>
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>示例强>
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。