Google脚本删除重复的行，保留最新的 [英] Google script delete duplicate row, leave most recent

问题描述

我正在使用脚本中的GmailApp从gmail抓取票证数据.但是，当票证状态发生变化时，我会收到一封新电子邮件，并会在新行中附加最新的时间戳.

I am scraping ticket data from gmail using the GmailApp in scripts. However, when a ticket status changes, I get a new email and we get a new row appended with a more recent time stamp.

我要搜索重复的票证编号(A)，并删除较早的时间戳(J).

I want to search for the duplicate ticket number (A) and delete the older timestamp (J).

我遇到的问题是oldTime实际上是最新的条目，因此什么也没发生.

The issue I am having is that oldTime is actually the latest entry so nothing happens.

13007 | B | C | D | E | F | G | H | I | 2/25/2019  
13007 | B | C | D | E | F | G | H | I | 2/26/2019  

A是票号.B-I根据票证信息而变化.J从电子邮件日期中拉出.

A is the ticket number. B-I change depending on the ticket information. J is pulled from the email date.

 function removeDuplicates() {
var sheet = SpreadsheetApp.getActiveSheet();
var data = sheet.getDataRange().getValues();
var newData = [];
for(i in data){
var row = data[i];
var duplicate = false;
for(j in newData){
  //If Column A in the old entry matches Column A in the new entry
  if(row[0] == newData[j][0]){
    //Pull New Timestamp and Old Timestamp
    var newTime = Date.parse(newData[j][9]);
    var oldTime = Date.parse(row[9]);
    if (newTime>oldTime) duplicate=true; // number is milliseconds in 24 hours      
  }
}
if(!duplicate){
  newData.push(row);
}
}
sheet.clearContents();
sheet.getRange(1, 1, newData.length, newData[0].length).setValues(newData);
}

推荐答案

用较新的行替换较旧的行

好的，这正在处理我的数据，让我们看看它是否对您的数据有效.

Replacing Older Rows with newer Rows

Okay this is working on my data let's see if it works on yours.

function outWithTheOldInWithTheNew() {
  var sh=SpreadsheetApp.getActiveSheet();
  var eA=sh.getRange(2,1,sh.getLastRow()-1,sh.getLastColumn()).getValues();
  var tA=sh.getRange(2,1,sh.getLastRow()-1,1).getValues();
  var dA=sh.getRange(2,10,sh.getLastRow()-1,10).getValues();
  var uA=[];//ticket numbers
  var uB=[];//dates
  var uC=[];//entire rows
  for(var i=0;i<tA.length;i++) {
    var idx=uA.indexOf(tA[i][0]);
    if(idx==-1) {
      uA.push(tA[i][0]);
      uB.push(dA[i][0]);
      uC.push(eA[i]);
    }else if(new Date(dA[i][0]).valueOf() > new Date(uB[idx]).valueOf()) {
      uB.splice(idx,1,dA[i][0]);//replace newer dates
      uC.splice(idx,1,eA[i]);//replace newer rows
    }
  }
  sh.getRange(2,1,sh.getLastRow()-1,sh.getLastColumn()).clearContent();
  sh.getRange(2,1,uC.length,uC[0].length).setValues(uC);//newest rows
}

之前的电子表格:

以下电子表格:

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