Deleting Duplicate Names

I was able to get my Daily Check in and out google sheet to delete everyday at midnight. Now I need the sheet to delete entries after scanning in twice. So like the date and name pop on on the first scan and then on the second name the date and name rows for both the first scan and the second scan get deleted. For example the person scans in in the morning, their name with a timestamp shows up on a form response on google sheets. They leave for the day or go out for lunch they scan out and their name with another timestamp shows up. Which activates a trigger that will delete both entries out of the form responses.

  function deleteDups() {
    const ss=SpreadsheetApp.getActive();
    const sh=ss.getActiveSheet();
    const startrow=1;
    const datarange=sh.getDataRange()
    const lastrow=datarange.getLastRow()
    const vs=datarange.getDisplayValues();//this get the data as string
    let uA=[];//unique array
    let d=0;//delete counter
    vs.forEach(function(r,i){
      let p=r[1]+r[2];//adding two strings together is a string
      //uA is an array of strings
      if(uA.indexOf(p)==-1) {
        uA.push(p);//uA is made up of all the unique strings
      }else{
        sh.deleteRow(i+startrow-d++);//the delete counter makes up for the rows that get deleted because the index into the  data remains the same even though the row is now gone.
      }
    });
  }

This is the current code I have and it'll delete the second entry that gets entered but not the first one as well. So if someone could help me be able to delete both that would be super helpful

This should work

function deleteDups() {
  const ss = SpreadsheetApp.getActive();
  const sh = ss.getActiveSheet();
  const startRow = 2; // Assuming row 1 is the header row, change it if necessary
  const dataRange = sh.getRange(startRow, 1, sh.getLastRow() - startRow + 1, sh.getLastColumn());
  const data = dataRange.getValues();
  const duplicates = [];
  
  data.forEach(function(row, rowIndex) {
    const name = row[1]; // Assuming the name is in column B, change it if necessary
    const timestamp = row[2]; // Assuming the timestamp is in column C, change it if necessary
    const duplicatesIndex = duplicates.findIndex(dup => dup.name === name);
    
    if (duplicatesIndex > -1) {
      const duplicate = duplicates[duplicatesIndex];
      sh.deleteRow(duplicate.rowIndex + startRow - rowIndex);
      sh.deleteRow(rowIndex + startRow - duplicate.rowIndex);
      duplicates.splice(duplicatesIndex, 1);
    } else {
      duplicates.push({ name: name, rowIndex: rowIndex });
    }
  });
}

I believe your goal is as follows.

• You want to delete rows when the columns "B" and "C" are duplicated.
• In your situation, you want to delete all duplicated rows. For example, when the columns "A" to "C" of row 3 and row 5 are a3, b3, c3 and a5, b3, c3, you want to delete those 2 rows.

In this case, how about the following flow?

1. Retrieve all rows by checking the duplicate rows.
2. Delete all duplicated rows.

By this flow, all duplicated rows can be deleted. When this is reflected in your script, how about the following modification?

Pattern 1:

In this pattern, new values are created and the existing sheet is overwritten by the new values.

function deleteDups2() {
  const ss = SpreadsheetApp.getActive();
  const sh = ss.getActiveSheet();
  const datarange = sh.getDataRange();
  const values = [...datarange.getDisplayValues().reduce((m, r) => {
    const k = r[1] + r[2];
    return m.set(k, m.has(k) ? [...m.get(k), r] : [r]);
  }, new Map()).values()].filter(r => r.length == 1).flat(1);
  datarange.clearContent().offset(0, 0, values.length, values[0].length).setValues(values);
}

Pattern 2:

In this pattern, the duplicated rows are deleted.

function deleteDups3() {
  const ss = SpreadsheetApp.getActive();
  const sh = ss.getActiveSheet();
  const datarange = sh.getDataRange();
  const values = [...datarange.getDisplayValues().reduce((m, r, i) => {
    const k = r[1] + r[2];
    return m.set(k, m.has(k) ? [...m.get(k), i + 1] : [i + 1]);
  }, new Map()).values()].filter(r => r.length > 1).flat(1).sort((a, b) => a < b ? 1 : -1);
  values.forEach(e => sh.deleteRow(e));
}

References:

• reduce()
• Map
• forEach()