App-Scripts如何根据条件删除和保留一些行

Question

我将使用谷歌表而不是Excel。所以我面临一些编码问题。我之前曾经问过这个问题。

假设我有两天的数据（早，晚）有25列和超过50K行我想保留一些符合标准的数据并删除其他我不需要的数据。

例。我想从早期删除'AAA'，'BBB'，'DDD'，'FFF'并保留其他'CCC'，'EEE'，'GGG'。我希望保留'AAA'，'BBB'，'DDD'，'FFF'并删除其他人

enter image description here

结果就像这个enter image description here

这是我尝试过的代码，但它没有运行以获得结果。

//text data in D, dates to evaluate earlyDay/lateDay in C

var Offset = 1;

function deleteRows() {
 var app = SpreadsheetApp.getActiveSpreadsheet();
 var targetSheet = app.getSheetByName('Sheet1');
 var r = targetSheet.getRange('C:C');
 var v = r.getValues();
 var r1 = targetSheet.getRange('D:D');
 var f = r1.getValues(); 
 var lateDay = new Date(getLateDay()).getTime();
 var earlyDay = new Date(getEarlyDay()).getTime();
  
 for(var i = f.length-1; i>=Offset; i--){
  var tmp = new Date(v[0,i]).getTime();
   if(tmp==earlyDay && (f[0,i]=="AAA" || f[0, i]=="BBB")){
    targetSheet.deleteRow(i+1);
   }
 }
  for(var i = f.length-1; i>=Offset; i--){
  var tmp = new Date(v[0,i]).getTime();
   if(tmp==lateDay && (f[0,i]!="AAA" && f[0, i]!="BBB")){
    targetSheet.deleteRow(i+1);
   }
 }
}

function getLateDay() {
 var app = SpreadsheetApp.getActiveSpreadsheet();
 var targetSheet = app.getSheetByName('Sheet1');
 var r = targetSheet.getRange('C:C');
 var v = r.getValues();
  for(var i = v.length-1; i>=Offset; i--){
  var tmp = new Date(v[0,i]).getTime();
  var tmp1 = new Date(v[0,i-1]).getTime();
   if(tmp>tmp1){
    return tmp;
   }
  }
  return null;
}

function getEarlyDay() {
 var app = SpreadsheetApp.getActiveSpreadsheet();
 var targetSheet = app.getSheetByName('Sheet1');
 var r = targetSheet.getRange('C:C');
 var v = r.getValues();
  for(var i = v.length-1; i>=Offset; i--){
  var tmp = new Date(v[0,i]).getTime();
  var tmp1 = new Date(v[0,i-1]).getTime();
   if(tmp<tmp1){
    return tmp;
   }
  }
  return null;
}

Answer 1

OP代码有一些基本的语法和逻辑缺陷。使用Logger命令进行故障排除可能有助于OP识别问题。

1）使用错误的语法来访问数组的值。例如，getLateDay使用v[0,i];这里有两个问题。

• 数组的两个元素必须在方括号中（[0][i]），
• 它正在查询错误的元素。而不是[0] [i]，它应该是[i] [0]。

2）getLateDay和getEarlyDay正在查询C列中的值。相反，它们应该查询D列（日期列）。

3）getLateDay和getEarlyDay都应该评估if(tmp>tmp1)。但是getLateDay应该返回“tmp”而getEarlyDay应该返回“tmp1”。

4）getLateDay和getEarlyDay重复了大部分主要代码，并且大致相同。为了性能，它们可以很容易地合并到主代码中。

5）仅需要单个循环来评估EarlyDay / LateDay与“tmp”，但是当然需要分别评估适用于EarlyDay和LateDay的alpha代码条件。我还发现，Lateday保留的值最好在一系列IF语句中进行评估。可以删除整个第二个循环。

6）如果不是使用C列和D列的单独范围，而是为组合列声明单个范围和值，则可能不那么容易混淆，并且节省了一些处理时间。随着OP的任务增长并且它们使用25列，这将更加重要。

我修改了OP的代码来解决这些问题。我在代码中留下了几个Logger语句，以便OP可以在代码的不同状态下评估变量。

//text data in D, dates to evaluate earlyDay/lateDay in C
var Offset = 1;

function deleteRows() {

  // set up the spreadsheet
  var app = SpreadsheetApp.getActiveSpreadsheet();
  var targetSheet = app.getSheetByName('Sheet1');

  // define the data ranges and get values
  var r = targetSheet.getRange('C:C');
  var v = r.getValues();
  var r1 = targetSheet.getRange('D:D');
  var f = r1.getValues();
  //Logger.log("DEBUG: code: "+v[0][0]+", date: "+f[0][0].getTime());//DEBUG
  //Logger.log("DEBUG: length of f: "+f.length);//DEBUG

  // get the respective values for lateDay and earlyDay
  var lateDay = new Date(getLateDay()).getTime();
  var earlyDay = new Date(getEarlyDay()).getTime();
  //Logger.log("DEBUG: LateDay: "+lateDay+", EarlyDay: "+earlyDay); //DEBUG

  // Loop through the data, starting at te bottom.
  for (var i = f.length - 1; i > -1; i--) {

    // get the date for this row
    var tmp = new Date(f[i][0]).getTime();

    // evaluate if this row is early and the values of possible codes
    if (tmp == earlyDay && (v[i][0] == "AAA" || v[i][0] == "BBB" || v[i][0] == "DDD" || v[i][0] == "FFF")) {
      Logger.log("DEBUG: Option#1 i: " + i + ", row: " + (i + 1) + "- delete row , earlyDay: " + earlyDay + ", tmp: " + tmp + ", code: " + v[i][0]); //DEBUG

      // Note the row deleted is "i+1" because the deleteRow command matchs the actual row number, whereas the loop works on a zero-basis.
      targetSheet.deleteRow(i + 1);
    } else {
      //Logger.log("DEBUG: Option#1 - do nothing");//DEBUG
    }

    // evaluate if this row is late and the values of possible codes
    // Note this is an either or; a day cannot be both early and late
    if (tmp == lateDay) {
      Logger.log("today is Late Day");
      if (v[i][0] != "FFF") {
        if (v[i][0] != "DDD") {
          if (v[i][0] != "BBB") {
            if (v[i][0] != "AAA") {
              Logger.log("DEBUG: code <> AAA or BBB or DDD or FFF"); //DEBUG
              Logger.log("DEBUG: Option#2 i: " + i + ", row: " + (i + 1) + "- delete row , lateDay: " + lateDay + ", tmp: " + tmp + ", code: " + v[i][0]); //DEBUG
              targetSheet.deleteRow(i + 1);
            } else {
              //Logger.log("DEBUG: Option#2 - do nothing");//DEBUG
            }
          }
        }
      }
    }
  }
}

function getLateDay() {

  var app = SpreadsheetApp.getActiveSpreadsheet();
  var targetSheet = app.getSheetByName('Sheet1');
  var r1 = targetSheet.getRange('D:D');
  var f = r1.getValues();
  for (var i = f.length - 1; i > -1; i--) {
    var tmp = new Date(f[i][0]).getTime();
    var tmp1 = new Date(f[i - 1][0]).getTime();
    //Logger.log("DEBUG: LateDay: i: "+i+", this day: "+tmp+", yesterday: "+tmp1);//DEBUG
    if (tmp > tmp1) {
      //Logger.log("DEBUG: return this day: "+tmp);//DEBUG
      return tmp;
    }
  }
  //Logger.log("DEBUG: return null");
  return null;
}

function getEarlyDay() {

  var app = SpreadsheetApp.getActiveSpreadsheet();
  var targetSheet = app.getSheetByName('Sheet1');
  var r1 = targetSheet.getRange('D:D');
  var f = r1.getValues();
  for (var i = f.length - 1; i > -1; i--) {
    var tmp = new Date(f[i][0]).getTime();
    var tmp1 = new Date(f[i - 1][0]).getTime();
    //Logger.log("DEBUG: EarlyDay: i: "+i+", tmp: "+tmp+", tmp1: "+tmp1);//DEBUG
    if (tmp > tmp1) {
      //Logger.log("DEBUG: return this day: "+tmp1);//DEBUG
      return tmp1;
    }
  }
  //Logger.log("DEBUG: return null");//DEBUG
  return null;
}

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屏幕截图之前和之后