有很多代碼片段與我需要的很接近，但沒有什么很適合。我有一些代碼可以復制和重命名選項卡。我只是缺少將選項卡名稱放在每張表的單元格 C2 中的位。例如，如果作業表名稱為 ADM_001.a，則 C2 也包含 ADM_001.a。這是一個簡單的編輯嗎？
function duplicateTemplateN(){
const sh = SpreadsheetApp.getActive();
const template = sh.getSheetByName("TEMPLATE");
const sNames = ["ADM_001.a","ADM_001.b","ADM_002.a","ADM_002.b","ADM_003.a","ADM_003.b","ADM_004.a","ADM_004.b","ADM_005.a","ADM_005.b","ADM_006.a","ADM_006.b","ADM_007.a","ADM_007.b","ADM_008.a","ADM_008.b","ADM_009.a","ADM_009.b","ADM_010.a","ADM_010.b"];
for (let i in sNames){
template.copyTo(sh).setName(sNames[i]);
}
}
uj5u.com熱心網友回復：
嘗試這個
function duplicateTemplateN(){
const ss = SpreadsheetApp.getActive();
const template = ss.getSheetByName("TEMPLATE");
const sNames = ["ADM_001.a","ADM_001.b","ADM_002.a","ADM_002.b","ADM_003.a","ADM_003.b","ADM_004.a","ADM_004.b","ADM_005.a","ADM_005.b","ADM_006.a","ADM_006.b","ADM_007.a","ADM_007.b","ADM_008.a","ADM_008.b","ADM_009.a","ADM_009.b","ADM_010.a","ADM_010.b"];
sNames.forEach(n => {
template.copyTo(ss).setName(n);
ss.getSheetByName(n).getRange("C2").setValue(n);
});
}
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