Import 3 google spreadsheet at same time into another - google-apps-script

Question

With this script, I can import one spreadsheet into another spreadsheet in a tab.
function importA() {
var values1 = SpreadsheetApp.openById('xxxurl').
getSheetByName('xxxname').getRange('A1:DE300').getValues();
SpreadsheetApp.getActive().getSheetByName('TAB').
getRange(1,1,values1.length,values1[0].length).setValues(values1)
}
Now, I would like to import 3 spreadsheets one below the other (in the same tab), preserving any increase in the rows of each sheet at the same time.
How could I proceed?

Answer 1

You can try this code and see if it works for you:
function collectAll() {
createNew(); //this can be removed for instances that you would need to run the two functions separately.
SpreadsheetApp.flush();
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sheets = ["TAB1","TAB2","TAB3"];
for( var i=0; i<sheets.length; i++ ) {
if (sheets[i] == "TAB1"){
var sh = ss.getSheetByName('TAB1');
var range1 = sh.getRange(1,1, sh.getLastRow(), sh.getLastColumn()).getValues();
Logger.log(range1);
}
else if (sheets[i] == "TAB2"){
var sh = ss.getSheetByName('TAB2');
var range2 = sh.getRange(1,1, sh.getLastRow(), sh.getLastColumn()).getValues();
Logger.log(range2);
}
else if (sheets[i] = "TAB3"){
var sh = ss.getSheetByName('TAB3');
var range3 = sh.getRange(1,1, sh.getLastRow(), sh.getLastColumn()).getValues();
Logger.log(range3);
}
}
var range = range1.concat(range2,range3);
Logger.log(range);
ss.getSheetByName('TAB').getRange(1,1, range.length, sh.getLastColumn()).setValues(range);
}
function createNew() { //This function creates the Tabs from 3 different sheets using Sheet ID
var sheets = ["ID1","ID2","ID3"]; //Spreadsheet ID goes here
for (i=0; i<sheets.length; i++ ) {
var ssh = SpreadsheetApp.openById(sheets[i]);
var values = ssh.getSheetValues(1,1, ssh.getLastRow(), ssh.getLastColumn());
var create = SpreadsheetApp.getActive().insertSheet('TAB'+(1+i));
create.getRange(1,1, ssh.getLastRow(), ssh.getLastColumn()).setValues(values);
Logger.log(create);
}
}
What this does is that it assigns all of the values of each tab to its own variable range1, range2 and range3 and then it concatenates all values in a single range, then sets the value to the sheet named "TAB".

Answer 2

I've used getSheetByName instead of openByID but you can get the idea.
function test() {
try {
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sheets = ["Sheet1","Sheet2","Sheet3"];
for( var i=0; i<sheets.length; i++ ) {
importA(sheets[i]);
}
}
catch(err) {
console.log(err);
}
}
function importA(name) {
try {
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sh1 = ss.getSheetByName("TAB");
sh1.clear() // if you want to erase old data in TAB
var sh2 = ss.getSheetByName(name);
var r1 = sh1.getDataRange();
var r2 = sh2.getDataRange();
// to prevent a blank line at the top
var i = r1.getNumRows() === 1 ? 0 : 1;
sh1.getRange(r1.getNumRows()+i,1,r2.getNumRows(),r2.getNumColumns()).setValues(r2.getValues());
}
catch(err) {
console.log(err);
}
}

Answer 3

Merge all spreadsheets sheets into one sheet
function mergeSpreadsheetsAllIntoOneSheet() {
const ss = SpreadsheetApp.getActive();
['id1', 'id2', 'id3'].forEach((id, i) => {
let sss = SpreadsheetApp.openById(id);
sss.getSheets().forEach(sh => {
let vs = sh.getDataRange().getValues();
let a = [...Array.from(new Array(vs[0].length).keys(), x => (x == 0)?sss.getName():sh.getName())];
vs.unshift(a);
let ash = ss.getSheets()[0];
if (vs && vs.length > 0) {
ash.getRange(ash.getLastRow() + 1, 1, vs.length, vs[0].length).setValues(vs);
}
});
});
}