Question

第1页

if let featured = dict["featured"] as? String {
    print("Success")
}
else {
    print("Failure")
}

由于这是一张google工作表，我试图运行以下脚本来删除与Name和Date Columns匹配的行。例如。 Laura在01AUG的同一天有2个条目。因此，我希望删除这两行。我正在尝试以下脚本，但它不起作用。

Timestamp           |Name | Shift.....| Date....| Letter..|    
19/05/2016 15:44:46|Mark | Swap      |  02/AUG | S       |    
Timestamp2           |Ivon | Give away | 25/ AUG | G       |    
Timestamp3           |Laura| Take      | 01/AUG  | A       |     
Timestamp4           |Ann  | Take      | 02/AUG  | A       |    
Timestamp5           |Ann  | Give away | 03/AUG  | G       |
Timestamp6           |Laura| Delete    | 01/AUG  | D       |

Answer 1

V4。最后（我想）。

如果Cx是＆＃34;删除＆＃34;或＆＃34;删除＆＃34;删除行x和行，其中B和D与行x＆＃39; s相同。

function objTst(){
  var ss = SpreadsheetApp.getActiveSpreadsheet();
  var sheet = ss.getSheetByName("Form responses 2");

  var megarng = sheet.getRange(2, 2, sheet.getLastRow(),sheet.getLastColumn()).getValues(); //[][]

  var rodel = [0]; //rows to delete

  for (i = 0; i<megarng.length; i++){
    if (String(megarng[i][1])==("Delete"||"delete")){ //megarng[][x] 0=b, 1=c, 2=d
      rodel.push(i+2) 

      for (g=0; g<megarng.length ; g++){

        if(String(megarng[g][0])===String(megarng[i][0]) && String(megarng[g][2])===String(megarng[i][2])){

          rodel.push(g+2);

          Logger.log(megarng[g][0]);
          Logger.log(megarng[i][0]);
          Logger.log(megarng[g][2]);
          Logger.log(megarng[i][2]);

          Logger.log(i+"i");
          Logger.log(g);
        }
      }
    }
    rodel.sort(function comp(a,b){return a-b});
    Logger.log(rodel);
  }
  for (i=rodel.length-1 ; 0<i; i--){if (rodel[i-1]!==rodel[i]){sheet.deleteRow(rodel[i]);}}
}

通过仅比较某些列上的特定值，删除Google表格上的重复行

1 个答案: