如何在另一个不相似的列表中添加元素？

Question

1. 第一个数组List<String> first=['101','102','103','104','105'];

2. 第二个数组List<String> first=['101','102','103'];

如何比较两个列表并在另一个数组中添加不匹配的元素

1. 输出List<String> first=['104,'105];

Answer 1

您可以将运行完整代码粘贴粘贴到DartPad

function doPost (e) {
  var lock = LockService.getScriptLock()
  lock.tryLock(10000)

  try {

    var e = SpreadsheetApp.openById("SPREADSHEET ID").getSheetByName("Sort Emails").getRange("B2").getValue();
    var m = SpreadsheetApp.openById("SPREADSHEET ID").getSheetByName("Template").getRange("A1").getValue();
    var subject = 'Welcome ..';
    MailApp.sendEmail(e, subject, m);

    var doc = SpreadsheetApp.getActive();
    var sheet = doc.getSheetByName(sheetName)

    var headers = sheet.getRange(1, 1, 1, sheet.getLastColumn()).getValues()[0]
    var nextRow = sheet.getLastRow() + 1

    var newRow = headers.map(function(header) {
      return header === 'Timestamp' ? new Date() : e.parameter[header]
    })

    sheet.getRange(nextRow, 1, 1, newRow.length).setValues([newRow])
    var y = sheet.getLastRow();
    var x = sheet.getRange(y,2).getValues();

    return ContentService
    .createTextOutput(JSON.stringify({ 'Central Says': 'Successfully subscribed', 'Email' : x}))
    .setMimeType(ContentService.MimeType.JSON)  
  }
  catch (e) {
    return ContentService
    .createTextOutput(JSON.stringify({ 'result': 'error', 'error': e }))
    .setMimeType(ContentService.MimeType.JSON)
  }

  finally {
    lock.releaseLock()
  }
}

输出

void main() {
  List<String> first=['101','102','103','104','105'];
  List<String> second=['101','102','103'];

  List<String> result = [];

  first.forEach((firstElement) { 
                   if (!second.contains(firstElement)){
                     result.add(firstElement);
                   }
                 }
               ); 

  print(result.toString());

}

Answer 2

您可以使用为difference()提供的方法Set

List<String> first = ['101', '102', '103', '104', '105'];
List<String> second = ['101', '102', '103'];
List<String> result = Set.of(first).difference(Set.of(second)).toList()
print(result); // [104, 105]