谷歌表格脚本将完成的行移动到新工作表

Question

I have a script which moves data from one sheet to another when data is added to column "I".

I'd like to adapt the script to perform the action of moving all rows (values) from the 'unpaid' sheet, where column I is not empty, to the paid sheet via a button press rather than 'on edit'. This will allow moving of data from multiple rows at once, rather than with the current script which only moves one row at a time.

Please can someone assist?

function onEdit(event) {

  var ss = SpreadsheetApp.getActiveSpreadsheet();
  var s = event.source.getActiveSheet();
  var r = event.source.getActiveRange();


  if(s.getName() == "Unpaid" && r.getColumn() == 9 && r.getValue() != "") {
    var row = r.getRow();
    var numColumns = 11;
    var targetSheet = ss.getSheetByName("Paid");
    var target = targetSheet.getRange(targetSheet.getLastRow() + 1, 1);
    s.getRange(row, 1, 1, numColumns).copyTo((target),{contentsOnly:true});
    s.deleteRow(row);
  } 
  else if(s.getName() == "Paid" && r.getColumn() == 14 && r.getValue() == false) {
    var row = r.getRow();
    var numColumns = s.getLastColumn();
    var targetSheet = ss.getSheetByName("Unpaid");
    var target = targetSheet.getRange(targetSheet.getLastRow() + 1, 1);
    s.getRange(row, 1, 1, numColumns).moveTo(target);
    s.deleteRow(row);
  }
}

Side note, the "else if" part of the script is designed to move the code back to the original sheet. This isn't needed but I wasn't able to delete that part of the code and maintain functionality.

Answer 1

Explanation:

• The first step is to find the rows in the Unpaid sheet for which column I is not empty. To achieve that, you can apply a ternary operator to every element of column I . In more detail, you can use the map() function to check whether the element is empty or not and then apply a filter() to select only the rows with a non empty value in column I :

   const movRows = iVals.map((c,i)=>c!=''?i:null).filter(f=>f!=null);

• movRows contains the indexes for which column I is not empty. To get the actual row, you need to add 2 to the elements of movRows since we started from the second row I2:I .

• The next step is to iterate over the movRows array and for every element, delete the row in the Unpaid sheet and move it to Paid sheet. We are not moving it directly, but instead pushing the data to the movData array and then use this array to transfer this data for efficiency purposes.

• The iteration needs to happen backwards otherwise every time you delete a row there will be a mismatch between the actual data and the structure of the sheet.

• To create a button in the UI , you can use an onOpen() trigger. The only thing you need to do, is to save the following script to the script editor and then a custom menu button will appear in the UI (spreadsheet). You can click that button to execute the transfer() function.

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Solution:

function transfer() {

const ss = SpreadsheetApp.getActiveSpreadsheet();
const pSh = ss.getSheetByName('Paid');
const upSh = ss.getSheetByName('Unpaid');
const numColumns = 11;
const iVals = upSh.getRange('I2:I'+upSh.getLastRow()).getValues().flat();  
const movData = [];
const movRows =  iVals.map((c,i)=>c!=''?i:null).filter(f=>f!=null);
movRows.reverse().forEach(row=>{ 
    movData.push(upSh.getRange(row+2,1,1,numColumns).getValues()[0]);
    upSh.deleteRow(row+2);                                    
  });   
pSh.getRange(pSh.getLastRow()+1,1,movData.length,movData[0].length).setValues(movData);   
}

and the custom menu button is created by using an onOpen trigger:

function onOpen() {
  SpreadsheetApp.getUi()
  .createMenu('Macros')
  .addItem('Transfer Rows', 'transfer')
  .addToUi();
}